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→
