Question 15 Marks
Draw an angle of $40^\circ$. Copy its supplementary angle.
AnswerIn order to make the copy of the required angle of $40$ degree, we proceed as follows:
Step $1:$ Draw any arbitrary line $AB$ and with the help of a compass draw its bisector $CD.$
Step $2:$ Measure $AE$ and draw a quarter arc with the compass placing on $A.$ Repeat for $EB.$
Step $3:$ With the same measurement draw an arc from $E.$ Join $H$ and $I.$ mark the intersection of $CD$ and $HI$ as $J.$ Join $AJ.$ Measure angle $JAB.$
Supplementary angle:
Step $1:$ Extend $BA$ to $K.$ From $E,$ draw an arc of any radius on the supplement of $40^\circ$.
Step $2:$ Draw another line $OP$ and using the compass with the above length draw an arc with $O$ as centre.
Step $3:$ Measure $ML$. Draw an arc from $Q$ with the measurement of $ML.$ Join to form $OT.$ Measure angle $TOP = 135^\circ$.
View full question & answer→Question 25 Marks
Draw an angle of $70^\circ$. Make a copy of it using only a straight edge and compasses.
AnswerDraw a line $AB.$ Place the centre of the protractor on point $A.$ Coincide $AB$ with the protractor line. Mark C=$70^\circ$. Join $AC.$
Constructing the copy of the above angle goes as follows:
Step $1:$ Draw another line $DE.$
Step $2:$ From $O,$ draw an arc cutting at $F$ and $G.$
Step $3:$ Draw an arc using the compass with the above arc length, placed at $D$ as center.
Step $4:$ Measure $GF.$ Using the same compass length place the point at $I$ and cut the arc.
Step $5:$ Using a ruler draw a line through the cut to get $DK.$ Measure angle $KDE$ using a protractor.
View full question & answer→Question 35 Marks
Draw an angle of measure $135^\circ $ and bisect it.
Answer
$i.\ $Draw any line $PQ$ and take a point $O$ on it.
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line at $A.$
$iii.\ $Without disturbing the radius on the compasses, draw an arc with $A$ as centre which cuts the first arc at $B.$
$iv.\ $Again without disturbing the radius on the compasses and with $B$ as centre, draw an arc which cuts the first arc at $C.$
$v.\ $Join $OB$ and $OC.$
$vi.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle COB.$ Label the points of intersection as $D$ and $E.$
$vii.\ $With $E$ as centre, draw $($in the interior of $\angle COB)$ an arc whose radius is more than half the length $ED$.
$viii.\ $With the same radius and with $D$ as centre, draw another arc in the interior of $\angle COB.$ Let the two arcs intersect at $F.$ Join $\overline{\mathrm{OF}}$ . Then $\overline{\mathrm{OF}}$ is the bisector of $\angle COB,$ i.e. $\angle COF = \angle FOB.$ Now, $\angle FOQ = 90^\circ .$
$ix.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle POF. $ Label the points of intersection as $G$ and $H.$
$x.\ $With $H$ as centre, draw $($in the interior of $\angle POF$ an arc whose radius is more than half the length $HG).$
$xi.\ $With the same radius and with $H$ as centre, draw another arc in the interior of $\angle POF.$ Let the two arcs intersect at I. Join $\overline{\mathrm{OI}}$ . Then $\overline{\mathrm{OI}}$ is the bisector of $\angle POF,$ i.e. $\angle POI = \angle IOF.$ Now $\angle IOQ = 135^\circ .$
$xii.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle IOQ.$ Label the points of intersection as $J$ and $K.$
$xiii.\ $With $K$ as centre, draw $($in the interior of $\angle IOQ)$ an arc whose radius is more than half the length $KJ.$
$xiv.\ $With the same radius and with $J$ as centre, draw another arc in the interior of $\angle IOQ.$ Let the two arcs intersect at $L.$ Join $\overline{\mathrm{OL}}$ . Then $\overline{\mathrm{OL}}$ is the bisector of $\angle IOQ,$ i.e., $\angle IOL = \angle LOQ.$ View full question & answer→Question 45 Marks
Draw an angle of measure $45^\circ $ and bisect it.
AnswerSteps of construction are given as follows:
Step $1:$ Draw a line $AB.$ Place the centre of the protractor on point $A.$ Coincide $AB$ with the protractor line. Mark $C =45^\circ$. Join $AC.$
Step $2:$ Draw an arc cutting $AB$ at $D$ and $AC$ at $E.$
Taking $D$ as centre and length more than the arc $DE$ cut an arc. Repeat for $E.$
Step $3:$ let them intersect at point $X.$ Draw a line joining $AX$ and extend it to $D_1$. Measure angle $D_1AB.$
View full question & answer→Question 55 Marks
Construct with ruler and compass, angle of measure $135^\circ .$
Answer$i.\ $Draw any line $PQ$ and take a point $O$ on it.
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line at $A.$
$iii.\ $Without disturbing the radius on the compasses, draw an arc with $A$ as centre which cuts the first arc at $B.$
$iv.\ $Again without disturbing the radius on the compasses and with $B$ as centre, draw an arc which cuts first arc at $C.$
$v.\ $Join $OB$ and $OC.$
$vi.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle COB.$ Label the points of intersection as $D$ and $E.$
$vii.\ $With $E$ as centre, draw $($in the interior of $\angle COB)$ an arc whose radius is more than half the length $ED.$
$viii.\ $With the same radius and with D as centre, draw another arc in the interior of $\angle COB.$ Let the two arcs intersect at $F.$ Join $\overline {OF}$. Then, $\overline {OF}$ is the bisector of $\angle COB,$ i.e. $\angle COF = \angle FOB.$ Now, $\angle FOQ = 90^\circ .$
$ix.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle POF.$ Label the points of intersection as $G$ and $H.$
$x.\ $With $H$ as centre draw $($in the interior of $\angle POF)$ an arc whose radius is more than half the length $HG.$
$xi.\ $With the same radius and with $G$ as centre, draw another arc in the interior of $\angle POF.$ Let the two arcs intersect at $I.$ Join $OI.$ Then, $\overline {OI}$ is the bisector of $\angle POF,$ i.e. $\angle GOI = \angle IOF.$ Now, $\angle IOQ = 135^\circ .$
View full question & answer→Question 65 Marks
Construct with ruler and compass, angle of measure $45^\circ$
Answer
Construction of $45^\circ$.
In order to construct a $45-$degree angle, first, we will draw a $90$ degree angle using the below-mentioned steps:
$i.\ $Use a ruler to draw a line segment $OB ($of any length$)$
$ii.\ $Now use a compass and open it to any convenient radius. With $O$ as the center, draw an arc which cuts $OB$ at $X$
$iii.\ $With $X$ as the center and the radius as used in step $2,$ draw an arc which cuts the first arc a point $D.$
$iv.\ $With center as $D$ and the radius as used in step $2,$ draw another arc which cuts the first arc at a point $C$
$v.\ $With center as $C$ and $D$ and the radius as used in step $2,$ draw two arcs such that they cut each other at a point $E.$
$vi.\ $Join points $O$ and $E$ and extent $OE$ to a point $A$
$vii.\ $Angle $A O B$ formed above is of $90$ Degree. Inorder, to construct an angle of $45$ degree, we need to construct an angle bisector of angle $AOB,$ using the steps written below.
$viii.\ $Use compass, opened to any radius and with center as $O,$ draw an arc which cuts $OB$ at $P$ and $OA$ at $Q.$
$ix.\ $With center as $P$ and $Q$ and the radius as used in step $8 ,$ draw two arcs such that they cut each other at a point $F$
$x.\ $Join points $O$ and $F$ and extent $OF$ to a point $E.$
$xi.\ EO$ is the bisector of Angle $AOB$ i.e. Angle $AOE=$ Angle$ EOB = 1 \over 2$ of Angle $AOB = 45$ Degree
View full question & answer→Question 75 Marks
Construct with ruler and compass, angle of measure $120^\circ .$
Answer$i.\ $Draw any line $PQ$ and take a point $O$ on it.
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line at $A.$
$iii.\ $Without disturbing the radius on the compasses, draw an arc with $A$ as centre which cuts the first arc at $B.$
$iv.\ $Again without disturbing the radius on the compasses and with $B$ as centre, draw an arc which cuts the first arc at $C.$
$v.\ $Join $OC$ and produce it to any point $R.$ Angle $ROQ = 120^\circ $
View full question & answer→Question 85 Marks
Construct with ruler and compass, angle of measure $90^\circ .$
Answer$i.\ $Draw any line $PQ$ and take a point $O$ on it.
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line at $A.$
$iii.\ $Without disturbing the radius on the compasses, draw an arc with A as centre which cuts the first arc at $B.$
$iv.\ $Again without disturbing the radius on the compasses and with B as centre, draw an arc which cuts the first arc at $C.$
$v.\ $Join $OB$ and $OC.$
$vi.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle COB.$ Label the points of intersection as $D$ and $E.$
$vii.\ $With $E$ as centre, draw $($in the interior of $\angle COB)$ an arc where radius is more than half the length $ED.$
$viii.\ $With the same radius and with $D$ as centre, draw another arc in the interior of $\angle COB.$ Let the two arcs intersect at $F.$ Join $\overline{OF}$. Then, $\overline{OF}$ is the bisector of $\angle COB,$ i.e., $\angle COF = \angle FOB.$ Now, $\angle FOQ = 90^\circ .$
View full question & answer→Question 95 Marks
Construct with ruler and compass, angle of measure $30^\circ .$
Answer$i.\ $Draw a line $\overleftrightarrow {PQ}$ and mark a point $O$ on it.
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line $PQ$ at a point say $A.$
$iii.\ $With the pointer at $A ($as centre$)$, now draw an arc that passes through $O.$
$iv.\ $Let the two arcs intersect at $B.$ Join $OB.$ We get $\angle BOA$ whose measure is $60^\circ .$
$v.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle BOA.$ Label the points of intersection as $D$ and $C.$
$vi.\ $With $C$ as centre, draw $($in the interior of $\angle BOA)$ an arc whose radius is more than half the length $CD.$
$vii.\ $With the same radius and with $D$ as centre, draw another arc in the interior of $\angle BOA.$ Let the two intersect at $E.$ Then, $\overline {OE}$ is the bisector of $\angle BOA,$ i.e. $\angle BOE = \angle EOA = 30^\circ .$
View full question & answer→Question 105 Marks
Construct with ruler and compass, angle of measure $60^\circ .$
Answer$i.\ $Construction of an angle of measure $60^\circ .$
$ii.\ $Draw a line $\overleftrightarrow {PQ}$ and mark a point $O$ on it.
$iii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line $PQ$ at a point say $A.$
$iv.\ $With the pointer at $A ($as centre$),$ now draw an arc that passes through $O.$
$v.\ $Let the two arcs intersect at $B.$ Join $OB.$ We get $\angle BOA$ whose measure is $60^\circ .$
View full question & answer→Question 115 Marks
Draw an angle of measure $153^\circ $ and divide it into four equal parts.
Answer$i.\ $Draw $\overline{O Q}$ of any length.
$ii.\ $Place the centre of the protractor at $O$ and the zero edge along $\overline {O Q}$ .
$iii.\ $Start with $0$ near $Q.$ Mark a point $P$ at $153^\circ .$
$iv.\ $Join $OP.$ Then, $\angle POQ = 153^\circ .$
$v.\ $With $O$ as centre and using compasses, draw an arc that cuts both ray of $\angle POQ.$ Label the points of intersection as $P'$ and $Q'.$
$vi.\ $With $Q'$ as centre, draw $($in interior of $\angle POQ)$ an arc whose radius is more than half the length $Q'P'.$
$vii.\ $With the same radius and with $P'$ as centre, draw another arc in the interior of $\angle POQ.$ Let the two arcs intersect at $R.$ Then, $\overline {O R}$is the bisector of $POQ.$
$viii.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle ROQ.$ Label the points of intersection as $B$ and $A.$
$ix.\ $With $A$ as centre, draw $($in the interior of $\angle ROQ)$ an arc whose radius is more than half the length $AB.$
$x.\ $With the same radius and with $B$ as centre, draw another arc in the interior of $\angle ROQ.$ Let the two arcs intersect at $S.$ Then, $\overline {O S}$ is the bisector of $\angle ROQ.$
$xi.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle POR$. Label the points of intersection as $D$ and $C.$
$xii.\ $With $C$ as centre, draw $($in the interior of $\angle POR)$ an arc whose radius is more than half the length $CD.$
$xiii.\ $With the same radius and with $D$ as centre, draw another arc in the interior of $\angle POR.$ Let the two arcs intersect at $T.$ Then, $\overline {O T}$ is the bisector of $\angle POR.$
Thus, $\overline {O S}$ , $\overline {O R}$ and $\overline {O T}$ divide $\angle POQ = 153^\circ $ into four equal parts.
View full question & answer→Question 125 Marks
Draw a right angle and construct its bisector.
Answer$i.\ $Draw $OQ$ of any length.
$ii.\ $Place the centre of the protractor at $O$ and the zero edge along $OQ.$
$iii.\ $Start with $0$ near $Q.$ Mark point $P$ at $90^\circ .$
$iv.\ $Join $\overline{OP}$. Then, $\angle POQ = 90^\circ .$
$v.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle POQ$. Label the points of intersection as $P'$ and $Q'.$
$vi.\ $With $Q'$ as centre, draw $($in the interior of $\angle POQ)$ an arc whose radius is more than half the length $Q'P'.$
$vii.\ $With the same radius and with $P'$ as centre, draw another arc in the interior of $\angle POQ.$ Let the two arcs intersect at $R$. Then, $\overline{OR}$ is the bisector of $\angle POQ.$
View full question & answer→Question 135 Marks
Draw an angle of measure $147^\circ $ and construct its bisector.
Answer
$i.\ $Draw $\overline {O Q}$ of any length.
$ii.\ $Place the centre of the protractor at $O$ and the zero edge along $\overline {O Q}$ .
$iii.\ $Start with $0$ near $Q.$ Mark a point $P$ at $147^\circ .$
$iv.\ $Join $OP.$ Then, $\angle POQ = 147^\circ .$
$v.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle POQ.$ Label the points of intersection as $P'$ and $Q'.$
$vi.\ $With $Q'$ as centre, draw $($in the interior of $\angle POQ)$ an arc whose radius is more than half the length $Q'P'.$
$vii.\ $With the same radius and with $P'$ as centre, draw another arc in the interior of $\angle POQ.$ Let the two arcs intersect at $R.$ Then, $\overline {O R}$ is the bisector of $\angle POQ.$ View full question & answer→Question 145 Marks
Draw $\angle POQ$ of measure $75^\circ $ and find its line of symmetry.
Answer
$i.\ $Draw $\overline {O Q}$ of any length.
$ii.\ $Place the centre of the protractor at $O$ and the zero edge along .
$iii.\ $Start with $0$ near $Q.$ Mark point $P$ at $75^\circ .$
$iv.\ $Join $\overline {O P}$ . Then $\angle POQ = 75^\circ .$
$v.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle POQ.$ Label the points of intersection as $P'$ and $Q'.$
$vi.\ $With $Q'$ as centre, draw $($in the interior of $\angle POQ)$ an arc whose radius is more than half the length $Q'P'.$
$vii.\ $With the same radius and with $P'$ as centre, draw another arc in the interior of $\angle POQ.$ Let the two arcs intersects at $R.$ Then $\overline {O R}$ is the bisector of $\angle POQ$ which is also the line of symmetry of $\angle POQ$ as $\angle POR = \angle ROQ.$ View full question & answer→Question 155 Marks
Draw any angle with vertex $O.$ Take a point $A$ on one of its arms and $B$ on another such that $OA = OB.$ Draw the perpendicular bisectors of $\overline {OA}$ and $\overline {OB}$. Let them meet at $P.$ Is $PA = PB?$
Answer
$i.\ $Draw any angle $POQ$ with vertex $O.$
$ii.\ $Take a point $A$ on the arm $OQ$ and another point $B$ on the arm $OP$ such that $\overline {OA}$ = $\overline {OB}$.
$iii.\ $With $O$ as centre, using compasses, and radius more than half of line segment $OA$ draw arcs on either side of $\overline {OA}$.
$iv.\ $With the same radius and with $A$ as centre, draw another arcs using compasses. Let it cut the previous arcs at $C$ and $D.$
$v.\ $Join $\overline {CD}$. Then $\overline {CD}$ is the perpendicular bisector of the line segment $\overline {OA}$.
$vi.\ $With $O$ as centre, using compasses, draw arcs. The radius of the arcs should be more than half of the length of $\overline {OB}$.
$vii.\ $With the same radius and with $B$ as centre, draw another arcs using compasses. Let it cut the previous arcs at $E$ and $F.$
$viii.\ $Join $\overline {EF}$. Then $\overline {EF}$ is the perpendicular bisector of the line segment $OB.$ The two perpendicular bisector meet at $P.$
$ix.\ $Join $\overline {PA}$ and $\overline {PB}$. We find that $\overline {PA}$ = $\overline {PB}$. View full question & answer→Question 165 Marks
Draw a circle of radius $4\ cm.$ Draw any two of its chord. Construct the perpendicular bisector of these chords. Where do they meet$?$
Answer
$i.\ $Draw a point with a sharp pencil and mark it as $O.$
$ii.\ $Open the compasses for the required radius $4cm$, by putting the pointer on $0$ and opening the pencil upto $4cm.$
$iii.\ $Place the pointer of the compasses at $O.$
$iv.\ $Turn the compasses slowly to draw the circle.
$v.\ $Draw any two chords $\overline {AB}$ and $\overline {CD}$ of this circle.
$vi.\ $With $A$ as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of $\overline {AB}$.
$vii.\ $With the same radius and with $B$ as centre, draw another circle using compasses. Let it cut the previous circle at $E$ and $F.$
$viii.\ $Join $\overline {EF}$. Then $\overline {EF}$ is the perpendicular bisector of the chord $\overline {AB}$.
$ix.\ $With C as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of $\overline {CD}$.
$x.\ $With the same radius and with $D$ as centre, draw another circle using compasses. Let it cut the previous circle at $G$ and $H.$
$xi.\ $Join $\overline {GH}$. Then $\overline {GH}$ is the perpendicular bisector of the chord $\overline {CD}$.
$xii.\ $We find that the perpendicular bisectors $\overline {EF}$ and $\overline {GH}$ meet at $O,$ the centre of the circle. View full question & answer→Question 175 Marks
Draw a circle with centre $C$ and radius $3.4\ cm.$ Draw any chord $\overline {AB}$ . Construct the perpendicular bisector of $\overline {AB}$ and examine if it passes through $C,$ if $\overline {AB}$ happens to be a diameter.
AnswerLet us draw a circle of radius of $3.4\ cm$ and take any of its diameter.
With $A$ as center and radius $AB$ draw a circle.
With $B$ as a center and same radius draw another circle.
Join their point of intersection and extend it on either side
Thus $EF$ passes through center $C$ View full question & answer→Question 185 Marks
Draw a circle with centre $C$ and radius $3.4\ cm.$ Draw any chord $\overline{AB}$. Construct the perpendicular bisector of $\overline{AB}$ and examine, if it passes through $C.$
Answer
$i.\ $Draw a point with a sharp pencil and mark it as .$C$
$ii.\ $Open the compasses for the required radius $3.4\ cm,$ by putting the pointer on 0 and opening the pencil upto $3.4\ cm.$
$iii.\ $Place the pointer of the compasses at $C.$
$iv.\ $Turn the compasses slowly to draw the circle.
$v.\ $Draw any chord $\overline{AB}$ of this circle.
$vi.\ $With $A$ as centre, using compasses, draw an arc. The radius of this arc should be more than half of the length of $\overline{AB}$.
$vii.\ $With the same radius and with $B$ as centre, draw another arc using compasses. Let it cut the previous arcs at $D$ and $E.$
$viii.\ $Join $\overline {DE}$. Then $\overline {DE}$ is the perpendicular bisector of the line segment $\overline{AB}$. On examinating, we find that it passes through $C.$ View full question & answer→Question 195 Marks
With $\overline{\mathrm{PQ}} $ of length $6.1\ cm$ as diameter draw a circle.
Answer
$1.$Draw a line segment $\overline{\mathrm{PQ}} $ of length $6.1\ cm.$
$2.$With $P$ as centre, using compasses, draw an arc. The radius of this arc should be more than half of the length of $\overline{\mathrm{PQ}} $.
$3.$With the same radius and with $Q$ as centre, draw another arc using compasses. Let it cut the previous arcs at $A$ and $B.$
$4.$Join $\overline{A B}$. It cuts $\overline{\mathrm{PQ}} $ at $C.$ Then $\overline{A B}$ is the perpendicular bisector of the line segment $\overline{\mathrm{PQ}} $.
$5.$Place the pointer of the compasses at $C$ and open the pencil upto $P.$
$6.$Turn the compasses slowly to draw the circle. View full question & answer→Question 205 Marks
Draw a line segment of length $12.8\ cm,$ Using compasses, divide it into four equal parts. Verify by actual measurement.
Answer
$i.\ $Draw a line segment $\overline {A B}$ of length $12.8\ cm.$
$ii.\ $With A as centre, using compasses, draw an arc. The radius of this arc should be more than half of the length of $\overline {A B}$ .
$iii.\ $With the same radius and with $B$ as centre, draw another arc using compasses. Let it cut the previous arc at $C$ and $D.$
$iv.\ $Join $\overline {C D}$ . It cuts $\overline {A B}$ at $E.$ Then $\overline {C D}$ is the perpendicular bisector of the line segment $\overline {A B}$.
$v.\ $With $A$ as centre, using compasses, draw an arc. The radius of this arc should be more than half of the length of $AE.$
$vi.\ $With the same radius and with $E$ as centre, draw another arc using compasses. Let it cut the previous arc at $F$ and $G.$
$vii.\ $Join $\overline {F G}$ . It cuts $\overline {A E}$ at $H.$ Then $\overline {F G}$ is the perpendicular bisector of the line segment $\overline {A E}$ .
$viii.\ $With $E$ as centre, using compasses, draw an arc . The radius of this arc should be more than half of the length of $EB.$
$ix.\ $With the same radius and with $B$ as centre, draw another arc using compasses. Let it cut the previous arc at $I$ and $J.$
$x.\ $Join $\overline {I J}$. It cuts $\overline {E B}$ at $K.$ Then $\overline {I J}$ is the perpendicular bisector of the line segment $\overline {E B}$.
$xi.\ $Now, the points $H, E$ and $K$ divide $AB$ into four equal parts, i.e.,
$\overline{A H}=\overline{H E}=\overline{E K}=\overline{K B}$
By measurement,
$\overline{\mathrm{AH}}=\overline{\mathrm{HE}}=\overline{\mathrm{EK}}=\overline{\mathrm{KB}} = 3.2\ cm.$ View full question & answer→Question 215 Marks
Draw a line $l$ and a point $X$ on it.Through $X,$ draw a line segment $\overline{X Y}$ perpendicular to $l.$ Now draw a perpendicular to $\overline{X Y}$ at $Y. ($use ruler and compasses$)$
Answer
$1.$Given a point $X$ on a line $l .$
$2.$With $X$ as centre and a convenient radius, construct a part circle $($arc intersecting the line l at two point $A$ and $B.)$
$3.$With $A$ and $B$ as centre and a radius greater than $AX$, construct two arcs which cut each other at $Y.$
$4.$Join $\overline{X Y}$. Then $\overline{X Y}$ is perpendicular to $l$ at $X,$ i.e. $\overline{X Y}\ \perp l.$ View full question & answer→Question 225 Marks
Draw any line segment $\overline{P Q}$. Take any point $R$ not on it. Through $R$ draw a perpendicular to $\overline{P Q}$.
Answer
$1.$Let $\overline{P Q}$ be the given line segment and $R$ be a point not on it.
$2.$Place a set-square on $\overline{P Q}$ such that one arm of the right angle aligns along $\overline{P Q}$.
$3.$Place a ruler along the edge opposite of the right angle.
$4.$Hold the ruler fixed. Slide the set-square along the ruler all the point $R$ touches the arm of the set-square.
$5.$Join $RS$ along the edge through $R,$ meeting $\overline{P Q}$ at $S.$ Now $\overline{\mathrm{RS}} \perp \overline{P Q}$. View full question & answer→Question 235 Marks
Draw any line segment $\overline {AB} $. Mark any point $M$ on it. Through $M$ draw a perpendicular to $\overline {AB} . ($Use ruler and compasses$)$
Answer
Steps of construction:
$a.\ $Draw a line segment $\overline {AB} $ and mark a point $M$ on it.
$b.\ $With $M$ as centre and a convenient radius, draw an arc intersecting $\overline {AB} $ at $x$ and $y.$
$c.\ $With $x$ and $y$ as centres and radius greater than half of $\overline {xy} $, draw two arcs which cut each other at $'Q'.$
$d.\ $Join $P, Q.$
Now, $PQ \bot AB$ View full question & answer→Question 245 Marks
Given $\overline {AB}$ of length $7.3 \ cm$ and $\overline {CD}$ of length $3.4\ cm,$ construct a line segment $\overline {XY}$ such that the length of $\overline {XY}$ is equal to the difference between the lengths of $\overline {AB}$ and $\overline {CD}$. Verify by measurement.
Answer
Steps of Construction:
$i.\ $Draw a line $l.$ Mark a point $X$ on line $l.$
$ii.\ $Place the compasses pointer on the $A$ mark of the given line segment $\overline {AB}$. Open it to place the pencil point upto $B$ mark of the given line segment $\overline {AB}$.
$iii.\ $Without changing the opening of the compasses, place the pointer of compasses on $X$ and swing an arc to cut $l$ at $Z.$
$iv.\ $Place the compasses pointer on the $C$ mark of the given line segment $\overline {CD}$. Open it to place the pencil point upto $D$ mark of the given line segment $\overline {CD}$.
$v.\ $Without changing the opening of the compasses, place the pointer of compasses on $Z$ and swing an arc towards $X$ to cut $l$ at $Y.$
$vi.\ \overline {XY}$ is a required line segment of length $=$ the difference between the lengths of $\overline {AB}$ and $\overline {CD}$ i.e $3.9 \ cm.$ View full question & answer→Question 255 Marks
Given $\overline {AB}$ of length $3.9 \ cm,$ construct $\overline {PQ}$ such that the length of $\overline {PQ}$ is twice that of $\overline {AB}$. Verify by measurement. $($Hint : Construct $PX$ such that length of $PX =$ length of $AB ;$ then cut off $\overline {XQ}$ such that $\overline {XQ}$ also has the length of $\overline {AB}.)$
AnswerThe steps of construction are given as follows:
$i.\ $Draw a line segment $AB$ of length $3.9 \ cm$ using a ruler.
$ii.\ $Take a measure of it on the compass and mark a point $X.$
$iii.\ $Draw arcs on both sides of $X$ with the same measure in compass and join them named as $P$ and $Q.$
$iv.\ $Measure $PQ.$
$PQ = 7.8 \ cm$. View full question & answer→Question 265 Marks
Construct $\overline {AB}$ of length $7.8\ cm.$ From this cut off $\overline {AC}$ of length $4.7\ cm.$ Measure $\overline {BC}$.
Answer
Steps of Construction:
$i.\ $Draw a line $l.$ Mark a point $A$ on line $l.$
$ii.\ $Place the compasses pointer on the zero mark on the ruler. Open it to place the pencil point upto the $7.8\ cm$ mark.
$iii.\ $Without changing the opening of the compasses, place the pointer on $A$ and swing an arc to cut $l$ at $B.$
$iv.\ \overline {AB}$ is a line segment of length $7.8\ cm$.
$v.\ $Place the compasses pointer on the zero mark on the ruler. Open it to place the pencil point upto $4.7 \ cm$ mark.
$vi.\ $ Without changing the opening of the compasses, place the pointer on $A$ and swing an arc to cut $l$ at $C.$
$vii.\ \overline {AC}$ is a line segment of length $4.7\ cm.$ On measurement, $\overline {BC} = 3.1\ cm.$ View full question & answer→Question 275 Marks
Construct with ruler and compasses, angles of measure: $45^\circ $
Answer
The below given steps will be followed to construct an angle of $45^\circ .$
$i.\ $Draw a line $l$ and mark a point $P$ on it. Now taking $P$ as centre and with a convenient radius, draw an arc of a circle which intersects line $l$ at $Q.$
$ii.\ $Taking $Q$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $R.$
$iii.\ $Taking $R$ as centre and with the same radius as before, draw an arc intersecting the arc at $S ($see figure$).$
$iv.\ $Taking $R$ and $S$ as centres, draw arcs of same radius to intersect each other at $T.$
$v.\ $Join $PT.$ Let it intersect the major arc at point $U.$
$vi.\ $Taking $Q$ and $U$ as centres, draw arcs with radius more than $\frac{1}{2}QU$ to intersect each other at $V.$ Join $PV.$
$PV$ is the required ray making $45^\circ $ with the given line $l.$ View full question & answer→Question 285 Marks
Construct a rectangle whose adjacent sides are $8\ cm$ and $3\ cm.$
AnswerDraw a line segment $AB$ of length $8\ cm.$ Construct $\angle\text{BAX}=90^{\circ}$ at point A and $\angle\text{ABY}=90^{\circ}$ at point $B.$ Using a compass and ruler, mark a point $D$ on the ray $AX$ such that $AD = 3\ cm.$ Similarly mark the point $C$ on the ray $Y$ such that $BC = 3\ cm.$ Draw the line segment $CD.$
$ABCD$ is the required rectangle.
View full question & answer→Question 295 Marks
Draw any line segment $\overline{\text{AB}}$. Mark any point $M$ on it. Through $M,$ draw a perpendicular to $\overline{\text{AB}}$. $($use ruler and compasses$)$Answer$1.$Draw the given line segment $\overline{\text{AB}}$ and mark any point $M$ on it.
$2.$With $M$ as centre and a convenient radius, construct an arc intersecting the line segment $\overline{\text{AB}}$ at two points $C$ and $D.$
$3.$With $C$ and $D $ as centres and a radius greater than $CM$ construct two arcs. Let these are intersecting each other at $E.$
$4.$Join $EM.$ $\overline{\text{EM}}$ is perpendicular to $\overline{\text{AB}}$.
View full question & answer→Question 305 Marks
Construct the angle with the help of ruler and compasses only: $150^\circ $
AnswerDraw a line $AB$ and take point $O$ at the middle of $AB.$
With a convenient radius and centre at $O,$ draw an arc, which cuts the line $AB$ at $P$ and $Q.$
With the same radius and centre at $Q,$ draw an arc, which cuts the first arc at $R.$
With the same radius and centre at $R,$ draw an arc, which cuts the first arc at $S.$
With the centres $P$ and $S$ and radius more than half of $PS,$
draw two arcs, which cut each other at $T.$
Draw $OT$ and extend it to $C$ to form the ray $OC.$
$\angle\text{BOC}$ is required angle of $150^\circ .$
View full question & answer→Question 315 Marks
Draw a line segment of length $9.5\ cm$ and construct its perpendicular bisector.
View full question & answer→Question 325 Marks
Let $A, B$ be the centres of two circles of equal radii; draw them so that each one of them passes through the centre of the other. Let them intersect at $C$ and $D.$ Examine whether $\overline{\text{AB}}\text{ and}\ \overline{\text{CD}}$ are at right angles.
AnswerLet us draw two circles of same radius which are passing through the centres of the other circle.
Here point $A$ and $B$ are the centres of these circles and these circles are intresecting each other at point $C$ and $D.$
Now in quadrilateral $ADBC,$ we may observe that-
$AD = AC ($radius of circle centred at $A)$
$BC = BD ($radius of circle centred at $B)$
As radius of both circles are equal.
Hence $AD = AC = BC = BD.$
Hence square $ADBC$ is a rhombus and in a rhombus diagonal bisect each other and $90^\circ .$ Hence $\overline{\text{AB}}\text{ and}\ \overline{\text{CD}}$ are at right angles. View full question & answer→Question 335 Marks
If $AB = 7.5\ cm$ and $CD = 2.5\ cm,$ construct a segment whose length is equal to:
$2AB + 3CD$
AnswerGiven: $AB= 7.5cm$ and $CD = 2.5cm$ Draw $AB$ and $CD$
Draw a line/ and take point Eon it. Now, take a divider and open it such that the ends of both its arms are at $A$ and $B.$ Then, we lift the divider and place its one end at $E$ and other end (say $F$) on the line $1,$ as shown in the figure. Again, lift the divider and place its one end at $F$ and another end ($G$) on the line $1$, opposite to $E$. Now, reset the divider in such a way that the ends of its one hand are at $C$ and the end of other hand is at $D.$ Then, we lift the divider and place its one end at $G$ and another end (say $H$) on the line $1,$ opposite to $E$ as shown in the figure. Again, lift the divider and place its one end at $H$ and other end (say $I$) on the line $1,$ opposite to $E$ as shown in the figure. Again, lift the divider and place its one end at $I$ and another end (say $J$) on the line $1,$ opposite to $E$ as shown in the figure. $EG$ is required line segment, whose length is equal to $(2AB + 3CD).$
View full question & answer→Question 345 Marks
Draw a circle with centre at point $O$ and radius $5\ cm.$ Draw its chord $AB,$ draw the perpendicular bisector of line segment $AB.$ Does it pass through the centre of the circle$?$
AnswerDraw a point $O.$ With $O$ as centre and radius equal to $5\ cm,$ draw a circle.
Take any two points $A$ and $B$ on the circumference of the circle and draw a line segment with $A$ and $B$ as its end points.
$AB$ is the chord of the circle.
With $A$ as centre and radius more than half of $AB,$ draw arcs on both sides of $AB.$
With the same radius and $B$ as a centre, draw arcs on both sides of $AB,$ cutting the previous two arcs at $E$ and $F.$
Draw a line passing through $E$ and $F.$
Line $EF$ passes through the centre of the circle $O.$
View full question & answer→Question 355 Marks
Construct the angle with the help of ruler and compasses only:
$105^\circ $
AnswerDraw a ray $OA$ and make an angle $\angle\text{AOB}=90^{\circ}$ and $\angle\text{AOC}=120^{\circ}$
Now bisect $\angle\text{BOC}$ and get the ray $OD.$
$\angle\text{AOD}$ is the required angle of $105^\circ $
View full question & answer→Question 365 Marks
Draw $\overline{\text{AB}}$ of length $7.3\ cm$ and find its axis of symmetry.
View full question & answer→Question 375 Marks
Draw a line segment $AB$ of length $5.8\ cm.$ Draw the perpendicular bisector of this line segment.
AnswerDraw a line segment $AB$ of length $5.8\ cm$ using a ruler. With $A$ as centre and radius more than half of $AB,$ draw arcs on both sides of $AB.$ With the same radius and $B$ as centre, draw arcs on both sides of $AB,$ intersecting the previous arcs at $L$ and $M.$ Draw the line segment $LM$ with $L$ and $M$ as end-points. $LM$ is the required perpendicular bisector of $AB.$ 
View full question & answer→Question 385 Marks
Draw a circle of radius $4\ cm. $ Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Answer
$1.$Mark any point $C$ on the sheet. Now, by adjusting the compasses up to $4\ cm$ and by putting the pointer of compasses at point $C,$ turn the compasses slowly to draw the circle. It is the required circle of $4\ cm$ radius.
$2.$Take any two chords $\overline{\text{AB}}$ and $\overline{\text{AB}}$ in the circle.
$3.$Taking $A$ and $B$ as centres and with radius more than half of $\overline{\text{AB}}$, draw arcs on both sides of $AB,$ intersecting each other at $E, F.$ Join $EF$ which is the perpendicular bisector of $AB.$
$4.$Taking $C$ and $D$ as centres and with radius more than half of $\overline{\text{CD}}$, draw arcs on both sides of $CD,$ intersecting each other at $G, H.$ Join $GH$ which is the perpendicular bisector of $CD.$
Now, we will find that when $EF$ and $GH$ are extended, they meet at the centre of the circle i.e., point $O.$
View full question & answer→Question 395 Marks
Draw a line segment $CD.$ Produce it to $CE$ such that $CE = 3CD.$
AnswerWe draw a line $l$ and take two points $C$ and $D$ on it.
Take a divider and open it such that its end of both arms is at $C$ and $D.$
Then, we lift the divider and place its one end at $D$ and other end on the line $l$ opposite to $C$ as shown in the figure.
Let this point be $A.$
Lift the divider again and place its one end at $A$ and other end on the line $1$ opposite to $C.$
Name this point as $E.$
Here $CD = DE = AE$
Therefore, $CE = CD + DE + AE$
$= CD + CD + CD ($As, $CD ± DE = AE)$
or, $CE = 3CD$
View full question & answer→Question 405 Marks
Construct an angle of $60^\circ $ with the help of compasses and bisect it by paper folding.
AnswerDraw a ray $OA.$ With convenient radius and centre $O,$ draw an arc cutting the ray $OA$ at $P.$ With the same radius and centre at $P, $draw another arc cutting the previous arc at $Q.$ Draw $OQ$ and extend it to $B.$
$\angle\text{AOB}$ is the required angle of $60^\circ .$
We cut the part of paper as sector $OPQ.$ Now, fold the part of paper such that line segments $OP$ and $OQ$ get coincided. Angle made at point $O$ is the required angle, which is half of angle $\angle\text{AOB}.$
View full question & answer→Question 415 Marks
Draw a line segment of length $8.6\ cm.$ Bisect it and measure the length of each part.
AnswerDraw a line segment $AB$ of length $8.6\ cm.$ With $A$ as centre and radius more than half of $AB,$ draw arcs on both sides of $AB.$ With the same radius and $B$ as centre, draw arcs on the both sides of $AB,$ cutting the previous two arcs at $E$ and $F.$ Draw a line segment from $E$ to $F$ intersecting $AB$ at $C.$ On measuring $AC$ and $BC,$ we get: $AC = BC = 4.3\ cm.$
View full question & answer→Question 425 Marks
Construct with ruler and compasses, angles of measure: $60^\circ $
Answer
The below given steps will be followed to construct an angle of $60^\circ .$
$1.$Draw a line $l$ and mark a point $P$ on it. Now, taking $P$ as centre and with a convenient radius, draw an arc of a circle which intersects line $l$ at $Q.$
$2.$Taking $Q$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point $R.$
$3.$Join $PR$ which is the required ray making $60^\circ $ with line $l.$ View full question & answer→Question 435 Marks
Construct with ruler and compasses, angles of measure: $135^\circ $
Answer
The below given steps will be followed to construct an angle of $135^\circ .$
$1.$Draw a line l and mark a point $P$ on it. Now taking $P$ as centre and with a convenient radius, draw a semi-circle which intersects line $l$ at $Q$ and $R.$
$2.$Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S.$
$3.$Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).
$4.$Taking $S$ and $T$ as centre, draw arcs of same radius to intersect each other at $U.$
$5.$Join $PU.$ Let it intersect the arc at $V.$ Now taking $Q$ and $V$ as centres and with radius more than $\frac{1}{2}$$QV,$ draw arcs to intersect each other at $W.$
$6.$Join PW which is the required ray making $135^\circ $ with line $l.$ View full question & answer→Question 445 Marks
Match the following statements:
| |
Column $A$ |
|
Column $B$ |
| $i$ |
Line segment has |
$a$ |
at a point |
| $ii$ |
Two segments may intersect |
$b$ |
if they have equal lengths |
| $iii$ |
Two segments are congruent |
$c$ |
two end-point |
| $iv$ |
Line segment is |
$d$ |
portion of a line |
Answer
| |
Column $A$ |
|
Column $B$ |
| $i$ |
Line segment has |
$c$ |
two end$-$point |
| $ii$ |
Two segments may intersect |
$a$ |
at a point |
| $iii$ |
Two segments are congruent |
$b$ |
if they have equal lengths |
| $iv$ |
Line segment is |
$d$ |
portion of a line |
Solution:
$i.\ $A line segment is a part of a line that is bounded by two distinct end points.
$ii.\ $Two line segments will either not intersect at all or intersect at one point. It can never intersect at more than one point.
$iii.\ $Line segments are congruent if they have the same lengths. If $AB = 6\ cm$ and $CD = 6\ cm$ Then, $AB$ and $CD$ are congruent.
$iv.\ $A line segment is a part of a line that is bounded by two distinct end points.
View full question & answer→Question 455 Marks
Construct with ruler and compasses, angles of measure: $30^\circ $
Answer
The below given steps will be followed to construct an angle of $30^\circ .$
$1.$Draw a line $l$ and mark a point $P$ on it. Now taking $P$ as centre and with convenient radius, draw an arc of a circle which intersects line $l$ at $Q.$
$2.$Taking $Q$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point $R.$
$3.$Now, taking $Q$ and $R$ as centre and with radius more than $\frac{1}{2}RQ,$ draw arcs to intersect each other at $S.$ Join $PS$ which is the required ray making $30^\circ $ with line $l.$ View full question & answer→Question 465 Marks
Construct the angle with the help of ruler and compasses only: $135^\circ $
AnswerDraw the line $AB$ and take the point $O$ at the middle of $AB.$ With a convenient radius and centre at $O,$ draw an arc, which cuts $AB$ at $P$ and $Q,$ respectively. Draw an angle of $90^\circ $ on the ray $OB$ as $\angle\text{BOC}=90^{\circ},$ where ray $OC$ cuts the arc at $R.$ With $Q$ and $R$ as centres and radius more than half of $QR,$ draw two arcs, which cuts each other at $S.$ Draw $OS$ and extend it to form the ray $OD.$
$\angle\text{BOD}$ is required angle of $135^\circ .$
View full question & answer→Question 475 Marks
Draw an angle of measure $135^\circ $ and bisect it.
Answer
The below given steps will be followed to construct an angle of $135^\circ $and its bisector.
$1.\ \text{DPOQ}$ of $135^\circ $ measure can be formed on a line l by using the protractor.
$2.$Draw an arc of a convenient radius, while taking point $O$ as centre. Let it intersect both rays of angle $135^\circ $ at point $A $ and $B.$
$3.$Taking $A$ and $B$ as centres, draw arcs of radius more than $\frac{1}{2}$AB in the interior of angle of $135^\circ .$ Let those intersect each other at $C.$ Join $OC.$
$OC$ is the required bisector of $135^\circ $ angle. View full question & answer→Question 485 Marks
In Fig. $O$ is the centre of the circle.
$a.\ $Name all chords of the circle.
$b.\ $Name all radii of the circle.
$c.\ $Name a chord, which is not the diameter of the circle.
$d.\ $Shade sectors $OAC$ and $\text{OPB}.$
$e.\ $Shade the smaller segment of the circle formed by $CP.$ Answer
A chord of a circle is a straight line segment whose both end points lie on the circle. The longest chord which passes through the centre of the circle is known as diameter. Line segment joining the centre to the point which lie on the circle is known as radius. The portion of a circle enclosed by two radii is known as sector. The segment of a circle is the region bounded by a chord and the arc subtended by the chord.
$a.\ CP$ and $AB$ are the two chords.
$b.\ \text{OA, OB, OC}$ and $OP$ are the radii of the circle.
$c.\ CP$ is a chord which is not the diameter of the circle because it does not pass through the centre.
$d.\ $Shaded sectors $\text{OAC}$ and $\text{OPB}$ are as:
$e.\ $Shaded smaller segment of the circle fromed by $CP $ is as:
View full question & answer→Question 495 Marks
Given $\overline{\text{AB}}$ of length $7.3\ cm$ and $\overline{\text{CD}}$ of length $3.4\ cm,$ construct a line segment $\overline{\text{XY}}$ such that the length of $\overline{\text{XY}}$ is equal to the difference between the lengths of $\overline{\text{AB}}$ and $\overline{\text{CD}}$. Verify by measurement.
Answer$1.$Given that $\overline{\text{AB}}=7.3\text{cm}$ and $\overline{\text{CD}}=3.4\text{cm}$
$2.$Adjust the compasses up to the length of $CD$ and put the pointer of compasses at $A,$ draw an arc to cut $AB$ at $P.$
$3.$Adjust the copasses up to the length of $PB.$ Now draw a line l and mark a point $X$ on it.
$4.$Now putting the pointer of compasses at point $X,$ draw an arc to cut the line at $Y.$
$\overline{\text{XY}}$ is the required line segment.
Now, difference between length of $\overline {\text{AB}}$ and
$\overline {\text{CD}} = 7.3\ cm - 3.4\ cm = 3.9\ cm.$
By ruler we may measure the length of $\overline{\text{XY}}$ which comes to $3.9\ cm.$ View full question & answer→Question 505 Marks
Construct two segments of lengths $4.3\ cm$ and $3.2\ cm.$ Construct a segment whose length is equal to the sum of the lengths of these segments.
AnswerUsing compass and ruler, we construct two segments $AB$ and $CD$ of lengths $4.3\ cm$ and $3.2\ cm,$ respectively. Draw a line $L$ and mark a point $P$ on it. Take a compass and place its metal point at $A$ and adjust it, such that the pencil point reaches point $B.$ Take the compass to line $L,$ such that its metal point is on $P.$ Mark a small mark at $Q$ on the line $L$ corresponding to the pencil point of the compass. Now, reset the compass, such that its metal and pencil points are on $C$ and $D,$ respectively. Take the compass again to line $L,$ such that its metal point is on $Q$ and the pencil point makes a small mark at point $R,$ which opposite to point $P$ on line $L$
$PR$ is the required segment, whose length is equal to the sum of the lengths of these segments.
View full question & answer→Question 515 Marks
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
AnswerObtuse angles are those angles which are greater than $90^\circ $ but less than $180^\circ .$
Draw an obtuse angle $\angle\text{BAC}.$
With an appropriate radius and centre at $A,$ draw an arc such that it intersects $AB$ and $AC$ at $P$ and Q, respectively.
With centre $P$ and radius more than half of $PQ,$ draw an $ARC.$
With the same radius and centre at $Q,$ draw another arc intersecting the previous arc at $R.$
Join $A$ and $R$ and extend it to $X.$
The ray $AX$ is the required bisector of $\angle\text{BAC}.$
If we measure $\angle\text{BAR}$ and $\angle\text{CAR},$
we have $\angle\text{BAR}=\angle\text{CAR}=65^{\circ}$ View full question & answer→Question 525 Marks
Draw an angle of measure $45^\circ $ and bisect it.
Answer
The below given steps will be followed to construct an angle of $45^\circ $ and its bisector.
$1.\angle\text{POQ}$ of $45^\circ $ measure can be formed on a line $l$ by using the protractor.
$2.$Draw an arc of a convenient radius, while taking point $O$ as centre. Let it intersect both rays of angle $45^\circ $ at point $A$ and $B$.
$3.$Taking $A$ and $B$ as centres, draw arcs of radius more than $\frac{1}{2}AB $ in the interior of angle of $45^\circ .$ Let those intersect each other at $C$. Join $OC.$
$OC$ is the required bisector of $45^\circ $ angle.
View full question & answer→Question 535 Marks
Draw any angle with vertex $O.$ Take a point $A$ on one of its arms and $B$ on another such that $OA = OB.$ Draw the perpendicular bisectors of $\overline{\text{OA}}$ and $\overline{\text{OB}}$. Let them meet at $P$. Is $PA = PB ?$
Answer
$1.$ Draw any angle whose vertex is $O.$
$2.$With a convenient radius, draw arcs on both rays of this angle while taking $O$ as centre. Let these points be $A$ and $B.$
$3.$Taking $O$ and $A$ as centres and with radius more than half of $OA,$ draw arcs on both sides of $OA.$ Let these be intersecting at $C$ and $D.$ Join $CD.$
$4.$Similarly, we can find the perpendicular bisector $\overline{\text{EF}}$ of $\overline{\text{OB}}$. These perpendicular bisectors $\overline{\text{CD}}$ and $\overline{\text{EF}}$ will intersect each other at P.
Now, $PA$ and $PB$ can be measured. These are equal in length.
View full question & answer→Question 545 Marks
Construct the angle with the help of ruler and compasses only: $30^\circ $
AnswerDraw a ray $OA.$ With a convenient radius and centre at $O,$ draw an arc, which cuts $OA$ at $P.$ With the same radius and centre at $P,$ draw an arc cutting the previous arc at $P.$ Taking $P$ and $Q$ as centres and radius more than half of $PQ,$ draw two arcs, which cuts each other at $R.$ Draw $OR$ and extend it to $B.$
$\angle\text{AOB}$ is the required angle of $30^\circ .$
View full question & answer→Question 555 Marks
Construct the angle with the help of ruler and compasses only: $90^\circ $
AnswerDraw a ray $OA.$
With a convenient radius and centre at $O,$
draw an arc cutting the ray $OA$ at $P.$
With the same radius and centre at $P,$
draw another arc, which cuts the first arc at $Q.$
With the same radius and centre at $Q,$ draw another arc,
which cuts the first arc at $R.$
With $Q$ and $R$ as centres and radius more than half of $QR,$
which cuts each other at $S.$
Draw $OS$ and extend it to $B$ from the ray $OB.$
$\angle\text{AOB}$ is required angle of $90°.$
View full question & answer→Question 565 Marks
In which of the following figures:
$a.\ $Perpendicular bisector is shown?
$b.\ $Bisector is shown?
$c.\ $Only bisector is shown?
$d.\ $Only perpendicular is shown?
AnswerA bisector is a line which bisects a given line segment into two equal parts. If this bisector is perpendicular to the given line segment, then it is known as perpendicular bisector.
$a.\ $Figure $(ii)$ represents a perpendicular bisector.
$b.\ $Figures $(ii)$ and $(iii)$ represent bisectors.
$c.\ $Figure $(iii)$ represents only bisector.
$d.\ $Figure $(i)$ represents only perpendicular.
View full question & answer→Question 575 Marks
Construct $\overline{\text{AB}}$ of length $7.8\ cm.$ From this, cut off $\overline{\text{AC}}$ of length $4.7\ cm.$ Measure $\overline{\text{BC}}$.
Answer
$1.$Draw a line $l$ and make a point $A$ on it.
$2.$By adjusting the compasses up to $7.8\ cm,$ draw an arc to cut $l$ on $B,$ while putting the pointer of coompasses on point $A.$
$\overline{\text{AB}}$ is the line segment of $7.8\ cm.$
$3.$By adjusting the compasses up to $4.7\ cm,$ draw an arc to cut $l$ on $C, $ while putting the pointer of coompasses on point $A.$
$\overline{\text{AC}}$ is the line segment of $4.7\ cm.$
$4.$Now put the ruler along with this line such that $0$ mark of ruler will match with the point $C.$
Read the position of point $B.$ It comes to $3.1\ cm$
$\overline{\text{BC}}$ is $3.1\ cm.$ View full question & answer→Question 585 Marks
Draw any line segment $\overline{\text{PQ}}$. Without measuring $\overline{\text{PQ}}$, construct a copy of $\overline{\text{PQ}}$.
View full question & answer→Question 595 Marks
Construct with ruler and compasses, angles of measure: $90^\circ $
Answer
The below given steps will be followed to construct an angle of $90^\circ .$
$1.$Draw a line $l$ and mark a point $P$ on it. Now taking $P$ as centre and with a convenient radius, draw an arc of a circle which intersects line $l$ at $Q.$
$2.$Taking $Q$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $R.$
$3.$Taking $R$ as centre and with the same radius as before, draw an arc intersecting the arc at $S ($see figure$).$
$4.$Taking $R$ and $S$ as centre, draw an arc of same radius to intersect each other at $T.$
$5.$ Join $PT,$ which is the required ray making $90^\circ $ with line $l.$ View full question & answer→Question 605 Marks
Construct with ruler and compasses, angles of measure: $120^\circ $
Answer
The below given steps will be followed to construct an angle of $120^\circ .$
$1.$Draw a line $l$ and mark a point $P$ on it. Now taking $P$ as centre and with a convenient radius, draw an arc of a circle which intersects line $l$ at $Q.$
$2.$Taking $Q$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $R.$
$3.$Taking $R$ as centre and with the same radius as before, draw an arc intersecting the arc at $S ($see figure$).$
$4.$Join $PS,$ which is the required ray making $120^\circ $ with line $l.$ View full question & answer→Question 615 Marks
Name the points and then the line segments in each of the following figures:
Answer$1.$Points: $\text{A, B}$ and $C$
Line segments: $\text{AB, BC}$ and $CA$
$2.$Points: $\text{A, B, C}$ and $D$
Line segments: $\text{AB, BC, CD}$ and $DA$
$3.$Points: $\text{A, B, C, D}$ and $E$
Line segments: $\text{AB, BC, CD, DE}$ and $EA$
$4.$Points: $\text{A, B, C, D, E}$ and $F$ Line segments: $\text{AB, CD}$ and $EF$
View full question & answer→Question 625 Marks
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
AnswerTwo angles, which are adjacent and supplementary, are called linear pair of angles.
Draw a line $AB$ and mark a point $O$ on it.
When we draw any angle $\angle\text{AOC},$
we also get another angle $\angle\text{BOC}.$
Bisect $\angle\text{AOC}$ by a compass and a ruler and get the ray $OX.$
Similarly, bisect $\angle\text{BOC}$ and get the ray $OY.$
Now, $\angle\text{XOY}=\angle\text{XOC}+\angle\text{COY}$
$=\frac{1}2{}\angle\text{AOC}+12\angle\text{BOC}$
$=\frac{1}{2}(\angle\text{AOC}+\angle\text{BOC})$
$=\frac{1}{2}\times180^{\circ}=90^{\circ}$ (As $\angle\text{AOC}$ and $\angle\text{BOC}$ are supplementary angles)
View full question & answer→Question 635 Marks
Can we have two acute angles whose sum is:
$a.\ $An acute angle? Why or why not?
$b.\ $A right angle? Why or why not?
$c.\ $An obtuse angle? Why or why not?
$d.\ $A straight angle? Why or why not?
$e.\ $A reflex angle? Why or why not?
Answer$a.\ $Yes, the sum of the two acute angles may be less than a right angle, e.g. $30^\circ $ and $45^\circ $ are acute angles and their sum $(i.e. 30^\circ + 45^\circ = 75^\circ )$ is also an acute angle.
$b.\ $Yes, the sum of two acute angles may be equal to a right angle, $e.g. 30^\circ + 60^\circ = 90^\circ .$
$c.\ $Yes, the sum of two acute angles may be more than a right angle, i.e. obtuse angle, $e.g. 60^\circ + 70^\circ = 130^\circ .$
$d.\ $No, the sum of two acute angles is always less than a straight angle, $i.e. 180^\circ .$
$e.\ $No, the sum of two acute angles is always less than $180^\circ .$ So, their sum cannot be a reflex angle.
View full question & answer→Question 645 Marks
Construct the angle with the help of ruler and compasses only:
$45^\circ $
AnswerTo construct an angle of $45^\circ ,$ construct an angle of $90^\circ $ and bisect it.
Construct the angle $\angle\text{AOB}=90^{\circ},$ where rays $OA$ and $OB$ intersect the arc at points $P$ and $T$ as shown in figure.
With $P$ and $T$ as centres and radius more than half of $PT, $ draw two arcs, which cut each other at $X$
Draw $OX$ and extend it to $C$ to form the ray $OC.$
$\angle\text{AOC}$ is the required angle of $45^\circ .$
View full question & answer→Question 655 Marks
Draw an angle and label it as $\angle\text{BAC}.$ Construct another angle, equal to $\angle\text{BAC}.$
AnswerDraw an angle $\angle\text{BAC}$ also draw a ray $OP.$ With a suitable radius and $A$ as center, draw an arc intersecting $AB$ and $AC$ at $X$ and $Y,$ respectively. With the same radius and $O$ as center, draw an arc to intersect the arc $OP$ at $M.$ Measure $XY$ using the compass. With $M$ as centre and radius equal to $XY,$ draw an arc to intersect the arc drawn from $O$ at $N.$ Join $0$ and $N$ and extend it to $Q.$ $\angle\text{POQ}$ is the required angle.
View full question & answer→Question 665 Marks
Draw a line $l.$ Take a point $A,$ not lying on $l$. Draw a line m such that $\text{m}\perp\text{l}$ and passing through A. Using ruler and a set-square.
AnswerWe draw a line $L$ and take a point $A$ outside it. Place a set square $PQR$ such that its one arm $PQ$ of the right angle is along the line $L.$ Without disturbing the position of set-square, place a ruler along its edge $PR.$ Now, without disturbing the position of the ruler, slide the set-square along the ruler until its arm $QR$ reaches point $A.$ Without disturbing the position of the set-square, draw a line $m.$ Line m is the required line perpendicular to line $L.$
View full question & answer→Question 675 Marks
Draw two acute angles and one obtuse angle without using a protractor. Estimate the measures of the angles. Measure them with the help of a protractor and see how much accurate is your estimate.
AnswerAngles are measured in degrees. The symbol for degrees is a little circle.The $\text{FULL CIRCLE}$ is $360^\circ (360$ degrees$).$
A half circle or a straight angle is $180^\circ .$
A quarter circle or a right angle is $90^\circ .$
Place the midpoint of the protractor on the VERTEX of the angle.
Line up one side of the angle with the zero line of the protractor $($where you see the number $0).$
Read the degrees where the other side crosses the number scale.
$i.\ $Measure the angles.
$ii.\ $Measure the angles. Label each angle as acute or obtuse.
$iii.\ $Tasha measured an acute angle, and got $146^\circ .$ The teacher pointed out that she had read the wrong set of numbers on the protractor.
What is the correct angle measure for the angle she measured?
Measure the following angles using your own protractor. If you need to, make the sides of the angles longer with a ruler.
$iv.\ $Draw four dots, and connect them so that you get a quadrilateral.
Measure all the angles of your quadrilateral. Then add the angle measures.
Did you get $360$ degrees, or close$?$ View full question & answer→Question 685 Marks
Draw a right angle and construct its bisector.
Answer
The below given steps will be followed to construct a right angle and its bisector.
$1.$Draw a line $l $ and mark a point $P$ on it. Draw an arc of convenient radius, while taking point $P$ as centre. Let it intersect line $l$ at $R.$
$2.$Taking $ R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S.$
$3.$Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T ($see figure$).$
$4.$Taking $S$ and $T$ as centres, draw arcs of same radius to intersect each other at $U.$
$5.$Join $PU$. $PU$ is the required ray making $90^\circ $ with line $l.$ Let it intersect the major arc at point $V.$
$6.$Now, taking $R$ and $V$ as centres, draw arcs with radius more than $\frac{1}{2}RV$ to intersect each other at $W.$ Join $PW.$
$PW$ is the required bisector of this right angle. View full question & answer→Question 695 Marks
Draw a circle with centre $C$ and radius $3.4\ cm.$ Draw any chord $\overline{\text{AB}}$. Construct the perpendicular bisector of $\overline{\text{AB}}$ and examine if it passes through $C. $ if $\overline{\text{AB}}$ happens to be a diameter.
View full question & answer→Question 705 Marks
Using protractor, draw a right angle. Bisect it to get an angle of measure $45^\circ .$
AnswerWe know that a right angle is of $90^\circ .$ Draw a ray $OA.$ With the help of a protractor, draw an $\angle\text{AOB}$ of $90^\circ .$ With centre at $O$ and a convenient radius, draw an arc cutting sides $OA$ and $OB$ at $P$ and $Q,$ respectively. With centre at $P$ and radius more than half of $PQ,$ draw an arc. With the same radius and centre at $Q,$ draw another arc intersecting the previous arc at $R.$ Join $O$ and $R$ and extend it to $X.$
$\angle\text{AOX}$ is the required angle of $45^\circ .$
$\angle\text{AOB}=90^{\circ}$
$\angle\text{AOX}=45^{\circ}$
View full question & answer→Question 715 Marks
Draw an angle of $50^\circ $ with the help of protractor. Draw a ray bisecting this angle.
Answer
Steps for construction:
$1.$Draw $\angle\text{BAC}=50^\circ$ with the help of protractor.
$2.$With $A$ as the centre and any convenient radius, draw an arc cutting $AB$ and $AC$ at $Q$ and $P,$ respectively.
$3.$With $P$ as the centre and radius more than half of $PQ$, draw an arc.
$4.$With $Q$ as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point $S.$
$5.$Draw $SA$ and produce it to point $R.$
Then, ray $AR$ bisects $\angle\text{BAC}.$ View full question & answer→Question 725 Marks
Draw a line segment $PQ = 6.2\ cm$. Draw the perpendicular bisector of $PQ.$
Answer
Steps for construction:
$1.$Draw a line segment $PQ,$ which is equal $6.2\ cm.$
$2.$With $P$ as the centre and radius more than half of $PQ,$ draw arcs, one on each side of $PQ.$
$3.$With $Q$ as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at $A$ and $B,$ respectively.
$4.$Draw $AB,$ meeting $PQ$ at $R.$ View full question & answer→Question 735 Marks
Draw a line segment $AB = 5.6\ cm.$ Draw the right bisector of $AB.$
Answer
Steps for construction:
$1.$Draw a line segment $AB,$ which is equal to $5.6\ cm.$
$2.$With $A$ as the centre and radius more than half of $AB,$ draw arcs, one on each side of $AB.$
$3.$With $B$ as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at $M$ and $N,$ respectively.
$4.$Draw $MN,$ meeting $AB$ at $R.$ View full question & answer→Question 745 Marks
Using your protractor, draw an angle of measure $108^\circ .$ With this angle as given, draw an angle of $54^\circ .$
AnswerDraw a ray $OA.$ With the help of a protractor, construct an angle $\angle\text{AOB}$ of $108^\circ .$ Since, $\frac{108}{2}=54^{\circ}$ Therefore, $54^\circ $ is half of $108^\circ .$ To get the angle of $54^\circ ,$ we need to bisect the angle of $108^\circ .$ With centre at $O$ and a convenient radius, draw an arc cutting sides $OA$ and $OB$ at $ P$ and $Q,$ respectively. With centre at $P$ and radius more than half of $PQ,$ draw an arc. With the same radius and centre at $Q,$ draw another arc intersecting the previous arc at $R.$ Join $O$ and $R$ and extend it to $X. $
$\angle\text{AOX}$ is the required angle of $54^\circ .$
View full question & answer→Question 755 Marks
Draw a line segment $AB = 5.6\ cm.$ Draw the perpendicular bisector of $AB.$
Answer
Steps for construction:
$1.$Draw a line segment $AB = 5.6\ cm.$
$2.$With $A$ as the centre and radius more than half of $AB,$ draw arcs, one on each side of $AB.$
$3.$With $B$ as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at $P$ and $Q,$ respectively.
$4.$Draw $PQ,$ meeting $AB$ at $R.$ View full question & answer→Question 765 Marks
Given some line segment $\overline{\text{AB}}$, whose length you do not know, construct $\overline{\text{PQ}}$ such that the length of $\overline{\text{PQ}}$ is twice that of $\overline{\text{AB}}$.
View full question & answer→Question 775 Marks
Draw any line segment $\overline{\text{PQ}}$. Take any point $R$ not on it. Through $R,$ draw a perpendicular to $\overline{\text{PQ}}. ($use ruler and set$-$square$)$
View full question & answer→Question 785 Marks
Draw an angle of $60^\circ ,$ using a pair of compasses. Bisect it to make an angle of $30^\circ .$
Answer
$1.$Draw a ray $QP.$
$2.$Wth $Q$ as the centre and any convenient radius,draw an arc cutting $QP$ at $N$.
$3.$With $N$ as the centre and radius same as before, draw another arc to cut the previous arc at $M.$
$4.$Draw $QM$ and produce it to $R.$
$\angle\text{PQR}$ is an angle of $60^\circ $. $\angle\text{PQR}$ is an angle of $60°$.
$5.$With $M$ as the centre and radius more than half of $MN$, draw an arc.
$6.$With $N$ as the centre and radius same as in step $(5),$ draw another arc, cutting the previously drawn arc at point $X.$
$7.$Draw $QX$ and produce it to point $S.$
Ray $QS$ is the bisector of $\angle\text{PQR}.$ View full question & answer→Question 795 Marks
Using ruler and compasses only, draw a right angle.
AnswerDraw a ray $OA.$ With a convenient radius and centre at $O,$ draw an arc $PQ$ with the help of a compass intersecting the ray $OA$ at $P.$ With the same radius and centre at $P,$ draw another arc intersecting the arc $PQ$ at $R.$ With the same radius and centre at $R,$ draw an arc cutting the arc $PQ$ at $C,$ opposite $P.$ Taking $C$ and $R$ as the centre, draw two arcs of radius more than half of $CR$ that intersect each other at $S.$ Join $O$ and $S$ and extend the line to $B.$
$\angle\text{AOB}$ is the required angle of $90^\circ .$
View full question & answer→Question 805 Marks
Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?
AnswerWe may draw a circle of any convenient radius, also having its centre as $O.$ Let $AB$ and $CD$ are two diameters of this circle. When we join the ends of these diameters, a quadrilateral $ABCD$ is formed.
As we know that the diameters of a circle are equal in length, hence quadrilateral so formed will be having its diagonals of equal length. Also, $OA = OB = OC = OD =$ radius $r$ and if a quadrilateral has its diagonals of same length and bisecting each other. It will be a rectangle. Let $DE$ and $FG$ be two diameters of this circle such that these are perpendicular to each other. Now we may find that a quadrilateral is formed by joining the ends of these diameters. 
We may find that $OD = OE = OF = OG =$ radius r. In this quadrilateral $DFEG,$ diagonals are equal and perpendicular to each other. Also these are bisecting each other, so it ill be a square. We may measure the length of sides of quadrilaterals so formed to check our answer. View full question & answer→Question 815 Marks
Draw an angle of measure $153^\circ $ and divide it into four equal parts.
Answer
The below given steps will be followed to construct an angle of $153^\circ $ measure and its bisector.
$1.$Draw a line $l$ and mark a point $O$ on it. Place the centre of the protractor at point $O$ and the zero edge along line $l.$
$2.$Mark a point $A$ at $153^\circ .$ Join $OA.$
$OA$ is the required ray making $153^\circ $ with line $l.$
$3.$Draw an arc of convenient radius, while taking point $O$ as centre. Let it intersect both rays of angle $153^\circ $ at point $A$ and $B.$
$4.$Taking $A$ and $B$ as centres, draw arcs of radius more than $\frac{1}{2}$$AB$ in the interior of angle of $153^\circ .$ Let those intersect each other at $C.$ Join $OC.$
$5.$Let $OC$ intersect the major arc at point $D.$ Now, with radius more than $\frac{1}{2}$$AD,$ draw arcs while taking $A$ and $D$ as centres, and $D$ and $B$ as centres. Let these be intersecting each other at point $E$ and $F$ respectively. Join $OE, OF.$
$OF, OC, OE$ are the rays dividing $153^\circ $ angle in $4$ equal parts. View full question & answer→Question 825 Marks
Draw a circle with centre at point $O.$ Draw its two chords $AB$ and $CD$ such that $AB$ is not parallel to $CD$. Draw the perpendicular bisectors of $AB$ and $CD.$ At what point do they intersect$?$
AnswerDraw a circle with centre at $0.$ We draw two chords $AB$ and $CD$ as shown in the figure.
$i.\ $With $A$ as centre and radius more than half of $AB,$ draw arcs on both sides of $AB$.
$ii.\ $With the same radius and $B$ as centre, draw arcs cutting the arcs of step $(i)$ at $P$ and $Q.$
$iii.\ $Join $P$ and $Q.$
$iv.\ $With $C$ as centre and radius more than half of $CD,$ draw arcs on both sides of $CD.$
$v.\ $With the same radius and $D$ as centre, draw arcs cutting the arcs of step $(iv)$ at $R$ and $S.$
$vi.$Join $R$ and $S.$
We draw the line segments of perpendicular bisector of $AB$ and $CD.$ We see that the perpendicular bisector of $AB$ and $CD$ meet at $0,$ the centre of the circle.
View full question & answer→Question 835 Marks
Using a pair of compasses construct the following angles: $60^\circ $
Answer
Steps of construction:
$1.$Draw a ray $QP.$
$2.$With $Q$ as the centre and any convenient radius, draw an arc cutting $QP$ at $N.$
$3.$With $N$ as the centre and the same radius as before, draw another arc to cut the previous arc at $M.$
$4.$Draw $QM$ and produce it to $R.$
$\angle\text{PQR}$ is the required angle of $60^\circ .$ View full question & answer→Question 845 Marks
Draw a line segment of length $10\ cm$ and bisect it. Further bisect one of the equal parts and measure its length.
AnswerDraw a line segment $AB$ of length $10\ cm$ and bisect it.
$i.\ $With $A$ as centre and radius more than half of $AB,$ draw arcs on both sides of $AB.$
$ii.\ $With the same radius and $B$ as centre, draw arcs cutting the arcs of step $(i)$ at $P$ and $Q,$ respectively.
$iii.\ $Join $ P$ and $Q$. Line $PQ$ intersects line $AB$ at $C.$
$iv.\ $With $A$ as centre and radius more than half of $AC,$ draw arcs on both sides of $AB.$
$v.\ $With the same radius and $C$ as centre, draw arcs cutting the arcs of step $(iv)$ at R and $S, $ respectively.
$vi.\ $ Join $R$ and $S.$
Line $RS$ intersects $AC $ at $D$. If we measure $AD$ with the ruler, we have $AD = 2.5\ cm$
View full question & answer→Question 855 Marks
Draw a line segment $AB$ and bisect it. Bisect one of the equal parts to obtain a line segment of length $\frac{1}{2}(\text{AB}).$
AnswerDraw a line segment $AB.$
$i.\ $With $A$ as centre and radius more than half of $AB,$ draw arcs on both sides of $AB.$
$ii.\ $With the same radius and $B$ as centre, draw arcs cutting the arcs drawn in step $(i)$ at $P$ and $Q.$
$iii.\ $Join $P $ and $Q.$
$PQ$ intersects $AB$ at $C.$
$iv.\ $With $A$ as centre and radius more than half of $AC,$ draw arcs on both sides of $AC.$
$v.\ $ With the same radius and $C$ as centre, draw arcs cutting the arcs drawn in step $(iv)$ at $R$ and $S.$
$vi.\ $Join $R$ and $S. RS$ intersects $AB$ at $D.$
Now, $AC$ and $CB$ are equal. Both are $\frac{1}{2}(\text{AB}).$ Again, divide $AC$ at $D. $ So, $AD$ and $AC$ are of same length, i.e., $\frac{1}{4}\text{(AB)}.$
View full question & answer→Question 865 Marks
Construct $\angle\text{AOB}=85^\circ$ with the help of a protractor. Draw a ray $OX$ bisecting $\angle\text{AOB}.$
Answer
Steps for construction:
$1.$Draw $\angle\text{AOB}=85^\circ$ with the help of a protractor.
$2.$With $O$ as the centre and any convenient radius, draw an arc cutting $OA$ and $OB$ at $P $ and $Q, $ respectively.
$3.$With $P$ as the centre and radius more than half of $PQ,$ draw an arc.
$4.$With $Q$ as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point $R.$
$5.$Draw $RO$ and produce it to point $X.$
Then, ray $OX$ bisects $\angle\text{AOB}.$ View full question & answer→Question 875 Marks
Draw an angle of $70^\circ .$ Make a copy of it using only a straight edge and compasses.
Answer
The below given steps will be followed to construct an angle of $70^\circ $ measure and its copy.
$1.$Draw a line $l$ and mark a point $O$ on it. Place the centre of the protractor at point $O$ and the zero edge along line $l.$
$2.$Mark a point $A$ at $70^\circ $. Join $OA. OA$ is the ray making $70^\circ $ with line l. Draw an arc of convenient radius in the interior of $70^\circ $ angle, while taking point $O$ as centre. Let it intersect both rays of angle $70^\circ $ at point $B$ and $C.$
$3.$Draw a line m and mark a point $P$ on it. With the same radius as used before, again draw an arc while taking point $P$ as centre. Let it cut the line m at point $D.$
$4.$Now, adjust the compasses up to the length of $BC.$ With this radius, draw an arc while taking $D$ as centre, which will intersect the previously drawn arc at point $E.$
$5.$Join $PE. PE$ is the required ray which makes the same angle $($i.e. $70^\circ )$ with line m. View full question & answer→Question 885 Marks
Draw a line $l$ and a point $X$ on it. Through $X,$ draw a line segment $\overline{\text{XY}}$ perpendicular to $l.$ Now draw a perpendicular to $\overline{\text{XY}}$ at $Y. ($use ruler and compasses$)$
Answer$1.$Draw a line $l$ and mark a point $X$ on it.
$2.$Taking $X$ as centre and with a convenient radius, draw an arc intersecting line l at two points $A$ and $B.$
$3.$With $A$ and $B$ as centres and a radius more than $AX,$ construct two arcs intersecting each other at $Y.$
$4.$Join $XY.$ $\overline{\text{XY}}$ is perpendicular to $l.$
Similarly, a perpendicular to $\overline{\text{XY}}$ at the point $Y$ can be drawn. The line $\overline{\text{ZY}}$ is perpendicular to $\overline{\text{XY}}$ at $Y.$
View full question & answer→Question 895 Marks
Draw an angle of $40^\circ .$ Copy its supplementary angle.
Answer
The below given steps will be followed to construct an angle of $40^\circ $ measure and the copy of its supplementary angle.
$1.$Draw a line segment $\overline{\text{PQ}}$ and mark a point $O$ on it. Place the centre of the protractor at point $O$ and the zero edge along line segment$\overline{\text{PQ}}$.
$2.$Mark a point $A $ at $40^\circ $. Join $OA. OA$ is the required ray making $40^\circ $ with $\overline{\text{PQ}}$. $Ð$
$POA$ is the supplementary angle of $40^\circ .$
$3.$Draw an arc of convenient radius in the interior of $Ð\ POA,$ while taking point $O$ as centre. Let it intersect both rays of $Ð\ POA$ at point $B$ and $C.$
$4.$Draw a line m and mark a point $S$ on it. With the same radius as used before, again draw an arc while taking point $S$ as centre. Let it cut the line m at point $T.$
$5.$Now, adjust the compasses up to the length of $BC.$ With this radius, draw an arc while taking $T$ as centre, which will intersect the previously drawn arc at point $R.$
$6.$Join $RS. RS$ is the required ray which makes the same angle with line $m,$ as the supplementary of $40^\circ $ is $140^\circ .$ View full question & answer→Question 905 Marks
Draw an angle of measure $147^\circ $ and construct its bisector.
Answer
The below given steps will be followed to construct an angle of $147^\circ $ measure and its bisector.
$1.$Draw a line $l $ and mark a point $O$ on it. Place the centre of the protractor at point $O$ and the zero edge along line $l.$
$2.$Mark a point $A$ at $147^\circ .$ Join $OA. OA$ is the required ray making $147^\circ $ with line $l.$
$3.$Draw an arc of convenient radius, while taking point $O$ as centre. Let it intersect both rays of angle $147^\circ $ at point $A$ and $B.$
$4.$Taking $A$ and $B$ as centres, draw arcs of radius more than $\frac{1}{2}AB$ in the interior of angle of $147^\circ .$ Let those intersect each other at $C.$ Join $OC.$
$OC$ is the required bisector of $147^\circ $ angle. View full question & answer→Question 915 Marks
Draw a line $AB.$ Take a point $P$ outside it. Draw a line passing through $P$ and parallel to $AB.$
Answer
Steps for construction:
$1.$Draw a line $AB$.
$2.$Take a point $P$ outside $AB$ and another point $O$ on $AB.$
$3.$Draw $PO.$
$4.$Draw $\angle\text{FPO}$ such that $\angle\text{FPO}$ is equal to $AOP.$
$5.$Extend $FP$ to $E.$
Then, the line $EF$ passes through the point $P$ and $EF \| AB.$ View full question & answer→Question 925 Marks
With $\overline{\text{PQ}}$ of length $6.1\ cm$ as diameter, draw a circle.
View full question & answer→Question 935 Marks
Draw a line $l.$ Take a point $A,$ not lying on $l.$ Draw a line $m$ such that $\text{m}\perp\text{l}$ and passing through $A.$ Using ruler and compasses.
AnswerWith $A$ as centre, draw an arc $PQ,$ which intersects line $L$ at points $P$ and $Q.$ Without disturbing the compass and taking $P$ and $Q$ as centres, we construct two arcs such that they intersect each other. The point where both arcs intersect is $B.$ Join points $A$ and $B$ and extend it in both directions. This is the required line.
View full question & answer→Question 945 Marks
Draw a line $AB$ and take two points $C$ and $E$ on opposite sides of $AB.$ Through $C,$ draw $\text{CD}\perp\text{AB}$ and through $E$ draw $\text{EF}\perp\text{AB}.$ Using ruler and compassed.
AnswerDraw a line $AB$ and take two points $C$ and $E$ on its opposite sides.
With $C$ as centre, draw an arc $PQ,$ which intersects line $AB$ at $P$ and $Q.$
Taking $P$ and $Q$ as centres, construct two arcs, such that they intersect each other at $H.$
Join points $H $ and $C. HC$ crosses AB at $D.$
We have $\text{CD}\perp\text{AB}.$
Similarly, take $E$ as centre and draw an arc $RS.$
Taking $R$ and $S$ as centres, draw two arcs which intersect each other at $G.$
Join points $G$ and $E.$ $GE$ crosses $AB$ at $F.$
We have $\text{EF}\bot\text{AB}.$
View full question & answer→Question 955 Marks
Draw the perpendicular bisector of a given line segment $AB$ of length $6\ cm.$
Answer
Steps for construction:
$1.$Draw a line segment $AB,$ which is equal to $6 \ cm.$
$2.$With $A$ as the centre and radius more than half of $AB,$ draw arcs, one on each side of $AB.$
$3.$With $B$ as the centre and radius same as before, draw arcs, cutting the perviously drawn arcs at $M$ and $N,$ respectively.
$4.$Draw $MN$ meeting $AB$ at $D.$
$MN$ is the required perpendicular bisector of $AB.$ View full question & answer→Question 965 Marks
Draw a circle with centre $C$ and radius $3.4\ cm.$ Draw any chord $\overline{\text{AB}}$. Construct the perpendicular bisector of $\overline{\text{AB}}$ and examine if it passes through $C.$
Answer
$1.$Mark any point $C$ on the sheet.
$2.$By adjusting the compasses up to $3.4\ cm$ and by putting the pointer of the compasses at point $C,$ turn the compasses slowly to draw the circle. It is the required circle of $3.4\ cm$ radius.
$3.$Now, mark any chord $\overline{\text{AB}}$ in the circle.
$4.$Taking $A$ and $B$ as centres, draw arcs on both sides of$\overline{\text{AB}}$. Let these intersect each other at $D$ and $E.$
$5.$Join $DE,$ which is the perpendicular bisector of $AB.$
When $\overline{\text{DE}}$ is extended, it will pass through point $C.$
View full question & answer→Question 975 Marks
Draw a line segment $AB$ of length $10\ cm.$ Mark a point $P$ on $AB$ such that $ AP = 4\ cm.$ Draw a line through $P$ perpendicular to $AB.$
AnswerWe draw line $L$ and take a point $A$ on it. Using a ruler and a compass, we mark a point $B, 10\ cm$ from $A,$ on the line $L.$
$AB$ is the required line segment of $10\ cm.$ Again, we mark a point $P,$ which is $4\ cm$ from $A,$ in the direction of $B.$ With $P$ as centre, take a radius of $4\ cm$ and construct an arc intersecting the line $L$ at two points $A$ and $E.$ With $A$ and $E$ as centres, take a radius of $6\ cm$ and construct two arcs intersecting each other at $R.$ We join $PR$ and extend it. $PR$ is the required line, which is perpendicular to $AB.$
View full question & answer→Question 985 Marks
Draw a line segment $AB$ and by ruler and compasses, obtain a line segment of length $\frac{3}{4}(\text{AB}).$
AnswerDraw a line segment $AB$ using the ruler.
$i.\ $With $A$ as centre and radius more than half of $AB,$ draw arcs on both sides of $AB.$
$ii.\ $With the same radius and $B$ as centre, draw arcs cutting the arcs drawn in step $(i)$ at $P$ and $Q.$
$iii.\ $Join $P$ and $Q. PQ$ intersects $AB$ at $C.$
$iv.\ $With $A$ as centre and radius more than half of $AB,$ draw arcs on both sides of $AC.$
$v.\ $With the same radius and $C$ as centre, draw arcs cutting the arcs drawn in step $(iv)$ at $R$ and $S.$
$vi.\ $Join $R$ and $S. RS$ intersects $AB$ at $D.$
Bisect $AC$ again and mark the point of bisection as $D.$
So, we have: $AD =\frac{1}{4}\text{(AB)},$
$DC =\frac{1}{4}\text{(AB)}$ and
$CB =\frac{1}{2}\text{(AB)}$
Therefore, $DB =\frac{1}{4}(\text{AB})+\frac{1}{2}(\text{AB})=\frac{3}{4}(\text{AB})$
Thus, $DB$ is the required line segment of length $\frac{3}{4}(\text{AB}).$
View full question & answer→Question 995 Marks
Draw a line $AB$ and take two points $C$ and $E$ on opposite sides of $AB.$ Through $C,$ draw $\text{CD}\perp\text{AB}$ and through $E$ draw $\text{EF}\perp\text{AB}.$ ruler and set-squares.
AnswerDraw a line $AB$ and take two points $C$ and $E$ on the opposite sides of the line $AB.$ On the side of $E,$ place a set-square $PQR,$ such that its one arm $PQ$ of the right angle is along the line $AB.$ Without disturbing the position of the set-square, place a ruler along its edge $PR.$ Now, without disturbing the position of the ruler, slide the set square along the ruler until the arm $QR$ reaches point $C.$ Without disturbing the position of the set-square, draw a line $CD$, where $D$ is a point on $AB. CD$ is the required line and $\text{CD}\perp\text{AB}.$ We repeat the same process starting with taking set-square on the side of $E,$ we draw a line $\text{EF}\perp\text{AB}.$
View full question & answer→Question 1005 Marks
Use a pair of compasses and construct the following angles: $22\frac{1}{2}^\circ$
Answer
Steps of Construction:
$1.$Draw a ray $OA.$
$2.$With $O$ as centre and any suitable radius draw an arc above $OA,$ cutting it at $B.$
$3.$With $B$ as centre and same radius cut the previous arc at $C $ and then with $C$ as centre and same radius cut the arc at $D.$
$4.$With $C$ as centre and radius more than half $CD$ draw an arc.
$5.$With $D$ as centre and same radius draw another arc to cut the previous arc at $E.$
$6.$Join $OE.$ Then $\angle AOE = 90^\circ .$
$7.$Draw the bisector $OF$ of $\angle\text{AOE}.$
$8.$Draw the bisector $OG$ of $\angle\text{AOF}.$
Then, $\angle\text{AOG}=22\frac{1}{2}^\circ$ is the required angle. View full question & answer→Question 1015 Marks
Use a pair of compasses and construct the following angles: $45^\circ $
Answer
Steps of Construction:
$1.$Draw a ray $OA.$
$2.$With $O$ as centre and any su itable radius draw an arc above $OA$ to cut it at $B.$
$3.$With $B$ as centre and same radius cut the previous arc at $C$ and then with $C $ as centre and same radius cut the arc at $D.$
$4.$With $C$ as centre and radius more than half $CD,$ draw an arc.
$5.$With $D$ as centre and same radius draw another arc to cut the previous arc at $E.$
$6.$Join $OE.$ Then $\angle\text{AOE}=90^\circ.$
$7.$Draw the bisector $OF$ of angle $\angle\text{AOE}.$
Then, $\angle\text{AOF}=45^\circ$ is the required angle. View full question & answer→Question 1025 Marks
Use a pair of compasses and construct the following angles:$135^\circ $
Answer
Steps of Construction:
$1.$Draw a ray $OA.$
$2.$With $O$ as centre and any suitable radius draw an arc above $OA,$ cutting it at $B.$
$3.$With $B$ as centre and same radius as before draw another arc to cut the previous arc at $C.$ With $C$ as centre and same radius draw the arc to cut it at $D.$ Again with $D$ as centre and same radius cut the arc at $E.$
$4.$Join $OD$ and produce it to $G.$ Then $\angle\text{AOG}=120^\circ.$
$5.$With $D$ as centre and radius more than half $DE$ draw an arc.
$6.$With $E$ as centre and same radius draw another arc to cut the previous arc at $F$ Join $OF.$
$7.$Draw the bisector $OH$ of $\angle\text{GOF}.$
Then, $\angle\text{AOH}=135^\circ$ is the required angle. View full question & answer→Question 1035 Marks
Draw a line $AB.$ Take a point $P$ outside it. Deaw a line passing through $P$ and perpendicular to $AB.$
Answer
Steps for construction:
$1.$Draw a line $AB.$
$2.$Take a point $P$ outside $AB.$
$3.$With $P$ as the centre and a convenient radius, draw an arc intersecting $AB$ at $M$ and $N,$ respectively.
$4.$With $M$ as the centre and radius more than half of $MN,$ draw an arc.
$5.$With $N$ as the centre and the same radius, draw an arc cutting the previously drawn arc at $Q.$
$6.$Draw $PQ$ meeting $AB$ at $S.$
$PQ$ is the required line that passes through $P$ and is perpendicular to $AB.$ View full question & answer→Question 1045 Marks
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line.
AnswerDraw two lines $AB$ and $CD$ intersecting each other at $O.$
We know that the vertically opposite angles are equal.
Therefore, $\angle\text{BOC}=\angle\text{AOD}$ and $\angle\text{AOC}=\angle\text{BOD}.$
We bisect angle $AOC$ and draw the bisecting ray as $OX.$
Similarly, we bisect angle $BOD$ and draw the bisecting ray as $OY.$
Now, $\angle\text{XOA}+\angle\text{AOD}+\angle\text{DOY}$
$=\frac{1}{2}\angle\text{AOC}+\angle\text{AOD}+\frac{1}{2}\angle\text{BOD}$
$=\frac{1}{2}\angle\text{BOD}+\angle\text{AOD}+\frac{1}{2}\angle\text{BOD}$
$[\text{As,}\angle\text{AOC}=\angle\text{BOD}]$
$=\angle\text{AOD}+\angle\text{BOD}$
Since, $AB$ is a line.
Therefore, $\angle\text{AOD}$ and $\angle\text{BOD}$ are supplementary angles and the sum of these two angles will be $180^\circ $.
Therefore, $\angle\text{XOA}+\angle\text{AOD}+\angle\text{DOY}=180^{\circ}$
We know that the angles on one side of a straight line will always add to $180^\circ .$
Also, the sum of the angles is $180^\circ .$
Therefore, $XY$ is a straight line.
Thus, $OX$ and $OY$ are in the same line.
View full question & answer→Question 1055 Marks
Draw a line $AB.$ Take a point $P$ on it. Draw a line passing through $P$ and perpendicular to $AB.$
Answer
Steps for construction:
$1.$Draw a line $AB$.
$2.$Take a point $P$ on line $AB.$
$3.$With $P$ as the centre, draw an arc of any radius, which intersects line $AB$ at $M$ and $N,$ respectively.
$4.$With $M$ as the centre and radius more than half of $MN,$ draw an arc.
$5.$With $N$ as the centre and the same radius as in step $(4),$ draw an arc that cuts the previously drawn arc at $R.$
$6.$Draw $PR. PR$ is the required line, which is perpendicular to $AB.$ View full question & answer→Question 1065 Marks
Construct the following angles with the help of a protractor: $45^\circ , 67^\circ , 38^\circ , 110^\circ , 179^\circ , 98^\circ , 84^\circ $
Answer$45^\circ $ We draw a ray $OA.$ We place the protractor on $OA$ such that its centre coincides with the point $O$ and the diameter of the protractor coincides with $OA.$ We mark a point $B$ against the mark of $45^\circ $ on the protractor. We remove the protractor and draw $OB.$
$\angle\text{AOB}$ is the required angle of $45^\circ .$
Similarly, we draw the angles $67^\circ, 38^\circ, 110^\circ, 179^\circ, 98^\circ$ and $84^\circ$.
View full question & answer→Question 1075 Marks
Draw a line segment of length $12.8\ cm.$ Using compasses, divide it into four equal parts. Verify by actual measurement.
View full question & answer→Question 1085 Marks
Using a protractor, draw $\angle\text{BAC}$ of measure $70^\circ .$ On side $AC,$ take a point $P,$ such that$ AP = 2\ cm.$ From $P$ draw a line perpendicular to $AB.$
AnswerDraw a line segment $AC$ on a line $L$
$i.\ $Take a protractor and place it on the segment $AC$ such that segment $AC$ coincides with the line of diameter of protractor and middle of this line coincides with point $A.$
$ii.\ $Counting from the right side, mark the point as $B$ at the point of $70^\circ $ of the protractor and draw $AB.$
$iii.\ $Now, measuring $2\ cm$ from $A$ on $AC,$ mark a point $P.$
$iv.\ $With $P$ as centre, draw an arc intersecting line $1 $ at points $E$ and $F.$
$v.\ $Using the same radius and $E$ and $F$ as centres, construct two arcs that intersect at point $G$ on the other side.
$vi.\ $Join $PG$.
View full question & answer→Question 1095 Marks
Draw $\angle\text{POQ}$ of measure $75^\circ $ and find its line of symmetry.
Answer
The below given steps will be followed to construct an angle of $75^\circ $ and its line of symmetry.
$1.$Draw a line $l$ and mark two points $O$ and $Q$ on it, as shown in the figure. Draw an arc of convenient radius, while taking point $O$ as centre. Let it intersect line $l$ at $R.$
$2.$Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S.$
$3.$Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T ($see figure$).$
$4.$Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U.$
$5.$Join $OU.$ Let it intersect the arc at $V.$ Now, taking $S$ and $V $ as centres, draw arcs with radius more than $\frac{1}{2}$$SV.$ Let those intersect each other at $P.$ Join $OP,$ which is the ray making $75^\circ $ with the line $l.$
$6.$Let this ray be intersecting our major arc at point $W.$ Now, taking $R$ and $W$ as centres, draw arcs with radius more than $\frac{1}{2}$RW in the interior of angle of $75^\circ .$ Let these be intersecting each other at $X.$ Join
$7. OX.$
$OX$ is the line of symmetry for $ÐPOQ = 75^\circ .$
View full question & answer→Question 1105 Marks
Use a pair of compasses and construct the following angles:$15^\circ $
Answer
Steps of Construction:
$1.$Draw a ray $OA.$
$2.$With $O$ as centre and any suitable radius draw an arc above $OA,$ cutting it at $B.$
$3.$With $B$ as centre and same radius as before draw another arc to cut the previous arc at $C.$ Join $OC$ and prouce it to $D.$
$4.$Draw the bisector $OE $ of $\angle\text{AQD}.$ Then $\angle\text{AOE}=30^\circ.$
$5.$Draw the bisector $OF$ of $\angle\text{AOE}.$
Then, $\angle\text{AOF}=15^\circ$ is the required angle. View full question & answer→Question 1115 Marks
Use a pair of compasses and construct the following angles: $105^\circ $
Answer
Steps of Construction:
$1.$Draw a ray $OA.$
$2.$With $O$ as centre and any suitable radius draw an arc cutting $OA$ at $B.$
$3.$With $B$ as centre and same radius cut the previous arc at $C$ and then with $C$ as centre and same radius cut the arc at $D.$
$4.$With $C $ as centre and radius more than half $CD$ draw an arc.
$5.$With $D$ as centre and same radius draw another arc to cut the previous arc at $E.$
$6.$Join $OE.$ Also join $OD$ and produce it to $F.$
$7.$Draw the bisector $OG$ of $\angle\text{EOF}.$
Thus, $\angle\text{AOG}=105^\circ$ is the required angle. View full question & answer→Question 1125 Marks
Draw a line segment $AB = 6\ cm.$ Take a point $C$ on $AB$ such that $AC = 2.5\ cm.$ Draw $CD$ perpendicular to $AB.$
Answer
Steps of constructions:
$1.$Draw a line segment $AB,$ which is equal to $6\ cm.$
$2.$Take a point $C$ on $AB$ such that $AC$ is equal to $2.5\ cm$.
$3.$With $C$ as the centre, draw an arc cutting $AB$ at $M$ and $N.$
$4.$With $M$ as the centre and radius more than half of $MN,$ draw an arc.
$5.$With $N$ as the centre and the same radius as before, draw another arc cutting the perviously drawn arc at $S.$
$6.$Draw $SC$ and produce it to $D.$ View full question & answer→Question 1135 Marks
Draw a line segment $PQ$ of length $12\ cm.$ Mark a point $O$ outside this segment. Draw a line through $O$ perpendicular to $PQ.$
AnswerDraw a line $L$ and take a point $P$ on it. Using a ruler and a compass, mark a point $Q$ on the line $L,$ where $PQ = 12\ cm.$ Mark a point $Q$ outside $PQ.$ Now, with $O$ as centre, draw an arc of appropriate radius such that the arc cuts the line at points $A$ and $B.$ Taking $A$ and $B$ as centres, construct two arcs such that they intersect each other at $C.$ Join $OC. OC$ is the required line, which is perpendicular to $PQ.$
View full question & answer→Question 1145 Marks
Use a pair of compasses and construct the following angles: $150^\circ $
Answer
Steps of Construction:
$1.$Draw a ray $OA.$
$2.$With $O$ as centre and any suitable radius draw an arc cutting $OA$ at $G.$
$3.$With $G$ as centre and same radius cut the arc at $B$ and then $B$ as centre and same radius cut the arc at $C.$ Again, with $C$ as centre and same radius cut the arc at $D.$
$4.$With $C$ as centre and radius more than half $CD$ draw an arc.
$5.$With $D$ as centre and same radius draw another arc to cut the previous arc at $E.$
$6.$Join $OE$ and produce it to $F.$
Then, $\angle\text{AOF}=150^\circ.$ View full question & answer→Question 1155 Marks
Draw a line segment $AB$ of length $8\ cm.$ At each end of this line segment, draw a line perpendicular to $AB.$ Are these two lines parallel?
Answer$i.\ $Take a convenient radius with $A$ as centre and draw an arc intersecting the line at points $W$ and $X.$
$ii.\ $With $W$ and $X$ as centres and radius greater than $AW,$ construct two arcs intersecting each other at $M.$
$iii.\ $Join $AM$ and extend it in both directions to $P$ and $Q.$
$iv.\ $Take a convenient radius with $B$ as centre and draw an arc intersecting the line at points $Y$ and $Z.$
$v.\ $With $Y$ and $Z$ as centres and a radius greater than $YB,$ construct two arcs intersecting each other at $N.$
$vi.\ $Join $BN$ and extend it in both directions to $S$ and $R.$
Let the lines perpendicular at $A$ and $B$ be $PQ$ and $RS$, respectively.
Since, $\angle\text{QAB}=90^{\circ}$ and $\angle\text{ABR}=90^{\circ}$
Therefore, $\angle\text{QAB}=\angle\text{ABR}.$
When two parallel lines are intersected by a third line, the two alternate interior angles are equal.
Since, $\angle\text{QAB}=\angle\text{ABR}$
Therefore, $PQ$ and $RS$ are parallel.
View full question & answer→Question 1165 Marks
Draw an angle equal to $\triangle\text{AOB},$ given in the adjoining figure.
Answer
Here $\angle\text{AOB}$ is given.
Steps for construction:
$1.$Draw a ray $QP.$
$2.$With $O$ as the centre and any suitable radius, draw an arc cutting $OA$ and $OB$ at $C$ and $E,$ respectively.
$3.$With $Q$ as the centre and the same radius as in step $(2),$ draw an arc cutting $QP$ at $D.$
$4.$With $D$ as the centre and radius equal to $CE,$ cut the arc through $D$ at $F.$
$5.$Draw $QF$ and produce it to point $R.$
$\therefore\ \angle\text{PQR}=\angle\text{AOB}$ View full question & answer→Question 1175 Marks
Construct an angle of $120^\circ $ and bisect it.
Answer
Steps of construction:
$1.$Draw a ray $QP.$
$2.$With $Q$ as the centre and any convenient radius, draw an arc cutting $QP$ at $N.$
$3.$With $N$ as the centre and the same radius, cut the arc at $A.$ Again, with $A$ as the centre and the same radius, cut the arc at $M.$
$4.$Draw $QM$ and produce it to $R.$
$\angle\text{PQR is 120}^\circ.$
$5.$With $M$ as the centre and radius more than half of $MN,$ draw an arc.
$6.$With $N$ as the centre and the same radius mentioned in step $(5),$ draw another arc, cutting the previously drawn arc at point $X.$
$7.$Draw $QX$ and produce it to point $S.$
Ray $QS$ is a bisector of $\angle\text{PQR}.$ Ray $QS$ is a bisector of $\angle\text{PQR}.$ View full question & answer→Question 1185 Marks
Using a protractor, draw an angle of measure $72^\circ .$ With this angle as given, draw angles of measure $36^\circ $ and $54^\circ .$
AnswerDraw a ray $OA.$ With the help of a protractor, draw an angle $\angle\text{AOB}$ of $72^\circ . $
With a convenient radius and centre at $O,$ draw an arc cutting sides $OA$ and $OB$ at $P$ and $Q,$ respectively.
With $P$ and $Q$ as centres and radius more than half of $PQ,$ draw two arcs cutting each other at $R.$
Join $O$ and $R$ and extend it to $X.$
$OR$ intersects arc $PQ$ at $C.$ With $C$ and $Q$ as centres and radius more than half of $CQ,$
draw two arcs cutting each other at $T. $
Join $O$ and $T$ and extend it to $Y.$
Now, $OX $ bisects $\angle\text{AOB}$
Therefore, $\angle\text{AOX}=\angle\text{BOX}=\frac{72}{2}=36^{\circ}$
Again, $OY$ bisects $\angle\text{BOX}$
Therefore, $\angle\text{XOY}=\angle\text{BOY}=\frac{36}{2}=18^{\circ}$
Therefore, $\angle\text{AOX}$ is the required angle of $36^\circ $ and $\angle\text{AOY}=\angle\text{AOX}+\angle\text{XOY}=36^{\circ}+18^{\circ}=54^{\circ}$
Therefore, $\angle\text{AOY}$ is the required angle of $54^\circ .$
View full question & answer→Question 1195 Marks
Use a pair of compasses and construct the following angles: $75^\circ $
Answer
Steps of Construction:
$1.$Draw a ray $OA.$
$2.$With $O$ as centre and any suitable radius draw an arc cutting $OA$ at $B.$
$3.$With $B$ as centre and same radius* cut the previous arc at $C$ and then with $C$ as centre cut the arc at $D.$
$4.$With $C$ as centre and radius more than half $CD$ draw an arc.
$5.$With $D$ as centre and same radius draw another arc to cut the previous arc at $E.$
$6.$Join $OE.$ Also join $OC$ and produce it to $G.$
$7.$Draw the bisector $OF$ of $\angle\text{EOG}.$
Then, $\angle\text{AOF}=75^\circ$ is the required angle. View full question & answer→Question 1205 Marks
Use a pair of compasses and construct the following angles: $67\frac{1}{2}^\circ$
Answer
Steps of Construction:
$1.$Draw a ray $OA.$
$2.$With $O$ as centre and any suitable radius draw an arc above $OA$ to cut it $B.$
$3.$With $B$ as centre and same radius cut the previous arc at $C$ and then with $C$ as centre and same radius cut the arc at $D.$
$4.$With $C$ as centre and radius more than half $CD$ draw an arc.
$5.$With $D$ as centre and same radius draw another arc to cut the previous arc at $E.$
$6.$Join $OE.$ Then $\angle\text{AOE}=90^\circ.$
$7.$Draw the bisector $OF$ of $\angle\text{AOE}.$
$8.$Draw the bisector $OG$ of $\angle\text{EOF}.$
Then, $\angle\text{AOG}=67\frac{1}{2}^\circ$ is the required angle. View full question & answer→Question 1215 Marks
Using ruler and compasses only, draw an angle of measure $135^\circ .$
AnswerWe draw a line $AB$ and mark a point $O$ on it. With a convenient radius and centre at $O,$ draw an arc $PQ$ with the help of a compass intersecting the line $AB$ at $P$ and $Q.$ With the same radius and centre at $P,$ draw another arc intersecting the arc $PQ$ at $R.$ With the same radius and centre at $Q,$ draw one more arc intersecting the arc $PQ$ at $S,$ opposite to $P.$ Taking $S$ and $R$ as centres and radius more than half of $SR,$ draw two arcs intersecting each other at $T.$ Join $O$ and $T$ intersecting the arc $PQ$ at $C$. Taking $C$ and $Q$ as centres and radius more than half of $CQ,$ draw two arcs intersecting each other at $D.$ Join $O$ and $D$ and extend it to $X$ to form the ray $OX.$
$\angle\text{AOX}$ is the required angle of measure $135^\circ .$
View full question & answer→Question 1225 Marks
Draw $\angle\text{ABC}$ os measure $60^\circ $ such that $AB = 4.5\ cm$ and $BC = 5\ cm$ and $BC = 5\ cm.$ Through C draw a line parallel to $AB$ and through $B$ draw a line parallel to $AC.$ Intersecting each other at $D.$ Measure $BD$ and $CD.$
Answer
Steps for construction:
$1.$Draw a line $BX$ and take a point $A,$ such that $AB$ is equal to $4.5\ cm.$
$2.$Draw $\angle\text{ABP}=60^\circ$ with the help of protractor.
$3.$With $A$ as the centre and a radius of $5\ cm,$ draw an arc cutting $PB$ at $C.$
$4.$Draw $AC.$
$5.$Now, draw $\angle\text{BCY}=60^\circ$
$6.$Then, draw $\angle\text{ABW},$ such that $\angle\text{ABW}$ is equal to $\angle\text{CAX},$ which cut the ray $CY$ at $D.$
$7.$Draw $BD.$
When we measure $BD$ and $CD.$ We have, $BD = 5\ cm$ and $CD = 4.5\ cm.$ View full question & answer→Question 1235 Marks
Using a protractor, draw $\angle\text{BAC}$ of measure $45^\circ .$ Take a point $P$ in the interior of $\angle\text{BAC}.$ From $P$ draw line segments $PM$ and $PN$ such that $\text{PM}\perp\text{AB}$ and $\text{PN}\perp\text{AC},$ Measure $\angle\text{MPN}.$
Answer$i.\ $Draw a line segment $A$ on the line $L .$
$ii.\ $Take a protractor and place it on the segment $AC$ such that $AC$ coincides with the line of the diameter of the protractor and the middle point of the line coincides with point $A$.
$iii.\ $Counting from the right side, mark a point as $B$ at the point of $45^\circ $ of protractor and draw a line segment $AB.$
$iv.\ $Take a convenient radius with $P$ as centre, construct an arc intersecting the line segments $AB$ at $T$ and $Q$ and $AC$ at $R$ and $S.$
$v.\ $Using the same radius and with $T$ and $Q$ as centres, construct two arcs intersecting at $G$ on the other side.
$vi.\ $Using the same radius and with $R$ and $S$ as centres, construct two arcs intersecting at $H$ on the other side.
$vii.\ $Join $PG$ and $PH$ which intersects $AB$ and $AC$ at $M$ and $N,$ respectively.
On measuring $\angle\text{MPN}$ using a protractor, we get it equal to $135^\circ .$
View full question & answer→Question 1245 Marks
Draw a reactangle whose two adjacent sides are $5.4\ cm$ and $3.5\ cm.$
Answer
Steps of construction:
$1.$Draw a ray $AX.$
$2.$With $A$ as the centre, cut the ray $XA$ at $B,$ such that $AB$ is equal to $3.5\ cm.$
$3.$With $B$ as the centre and with any convenient radius, draw an arc cutting $AX$ at $M$ and $N.$
$4.$With $N$ as the centre and with radius more than half of $MN,$ draw an arc.
$5.$With $M$ as the centre and with the radius same as before, draw another arc to cut the previous arc at $Y.$
$6.$Draw $BY$ and produced it to $W.$
$7.$With $B$ as the centre and a radius of $5.4\ cm,$ cut ray $BW$ at point $C.$
$8.$With $C$ as the centre and a radius $3.5\ cm,$ draw an arc on the right side of $BC.$
$9.$With $A$ as the centre and a radius $5.4\ cm,$ draw an arc cutting the previous arc at $D.$
$10.$Join $CD$ and $AD.$
$\text{ABCD}$ is the required rectangle.
View full question & answer→Question 1255 Marks
Draw an angle and label it as $\angle\text{BAC}.$ Draw its bisector ray $AX$ and take a point $P$ on it. From $P$ draw line segments $PM$ and $PN,$ such that $\text{PM}\perp\text{AB}$ and $\text{PN}\perp\text{AC},$ where $M$ and $N$ are respectively points on rays $AB$ and $AC.$ Measure $PM$ and $PN.$ Are the two lengths equal$?$
Answer$i.\ $Draw $\angle\text{BAC}$ on the line segment $AC.$
With a convenient radius and $A$ as centre, draw an arc from $AB$ and $AC$.
$ii.\ $The points where arc cuts $AB$ and $AC,$ take both points as centres and draw two small arcs intersecting at $X.$ Now, draw $AX.$
$iii.\ $Take a point $P$ on the ray $AX.$
$iv.\ $Take a convenient radius with $P$ as centre and construct an arc intersecting the line segments $AB$ at $T$ and $Q$ and $AC$ at $R$ and $S,$ respectively.
$v.\ $Using the same radius and with $T$ and $Q$ as centres, construct two arcs intersecting at $G$ on the other side.
$vi.\ $Using the same radius and with $R$ and $S$ as centres, construct two arcs intersecting at $H$ on the other side.
$vii.\ $Join $PG$ and $PH,$ which intersects $AB$ and $AC$ at $M$ and $N,$ respectively.
On measuring $PM$ and $PN$ using a ruler, we find that both are equal.
View full question & answer→Question 1265 Marks
Draw an angle of $45^\circ ,$ using a pair of compasses.
Answer
Steps of constructions:
$1.$Draw a ray $OA.$
$2.$With centre $O$ and a suitable radius draw an arc meeting $OA$ at $E.$
$3.$With centre $E$ and with same radius, cut the first arc firstly at $F$ and then from $F$ with same radius cut act at $G.$
$4.$With centres $F$ and $G,$ with suitable radius, draw arcs intersecting each other at $H.$
$5.$Join $OH$ intersecting the first arc at $L$ and produce it to $C.$
$6.$With centre $E$ and $L$ and with suitable radius draw arcs intersecting each other at $M.$
$7.$Join $OM$ and produce it to $B.$
Then, $\angle\text{AOB}=45^\circ$ View full question & answer→Question 1275 Marks
Draw a square, each of whose sides is $5\ cm.$ Use a pair of compasses and a ruler in your constribuction.
Answer
Steps of Construction:
$1.$With the help of a ruler draw a line segment $AB = 5\ cm.$
$2.$With $A$ as centre and any suitable radius draw an arc cutting $AB$ at $C.$
$3.$With $C $ as centre and same radius cut the previous arc at $D$ and then with $D$ as centre and same radius cut the arc at $E.$
$4.$With $D$ as centre and radius more than half $DE$ draw an arc.
$5.$With $E$ as centre and same radius draw another arc to cut the previous arc at $F.$
$6.$Join $AF$ and produce it to $G$ such that $AG = 5\ cm.$
$7.$With $G$ as centre and radius equal to $AB$ draw an arc. With $B$ as centre and same radius draw another arc to cut the previous arc at $H.$
$8.$Join $GH$ and $BH.$
Then, $\text{ABHG}$ is the required square. View full question & answer→Question 1285 Marks
Construct an angle of $90^\circ $ and bisect it.
Answer
Construction steps:
$1.$Draw a line $OA.$
$2.$Take a point $B$ on $OA$. With $B$ as the centre and any convenient radius, draw an arc cutting $OA $ at $M$ and $N.$
$3.$With $N$ as the centre and radius more than half of $MN,$ draw an arc.
$4.$With $M$ as the centre and the same radius as before, draw another arc to cut the previous arc at $W.$
$5.$Draw $WB,$ meeting the arc at $S.$ Produce it to $C.$
$\angle\text{ABC}$ is the required angle of $90^\circ \angle\text{ABC}$ is the required angle of $90^\circ .$
$6.$With $S$ as the centre and radius more than half of $SN,$ draw an arc.
$7.$With $N$ as centre and the same radius as in step $(6),$ draw another arc, cutting the previously drawn arc at point $X.$
$8.$Draw $BX$ and produce it to point $D.$ Ray $BD$ is the angle bisctor of $\angle\text{ABC}.$ Ray $BD$ is the angle bisctor of $\angle\text{ABC}.$ Ray $BD$ is the angl bisctor of $\angle\text{ABC}.$ Ray $BD$ is the angle bisctor of $\angle\text{ABC}.$ View full question & answer→
