Question 15 Marks
Draw an angle of $50^\circ $ with the help of protractor. Draw a ray bisecting this angle.
Answer
Steps for construction:
$i.\ $Draw $\angle\text{BAC}=50^\circ$ with the help of protractor.
$ii.\ $With $A$ as the centre and any convenient radius, draw an arc cutting $AB$ and $AC$ at $Q$ and $P,$ respectively.
$iii.\ $With $P$ as the centre and radius more than half of $PQ$, draw an arc.
$iv.\ $With $Q$ as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point $S$.
$v.\ $Draw $SA$ and produce it to point $R.$
Then, ray $AR$ bisects $\angle\text{BAC}.$ View full question & answer→Question 25 Marks
Draw a line segment $PQ = 6.2\ cm$. Draw the perpendicular bisector of $PQ.$
Answer
Steps for construction:
$i.\ $Draw a line segment $PQ$, which is equal $6.2\ cm.$
$ii.\ $With $P$ as the centre and radius more than half of $PQ$, draw arcs, one on each side of $PQ.$
$iii.\ $With $Q$ as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at $A$ and $B,$ respectively.
$iv.\ $Draw $AB$, meeting $PQ$ at $R.$ View full question & answer→Question 35 Marks
Using a pair of compasses construct the following angles: $120^\circ $
Answer
Steps for construction:
$i.\ $Draw a ray $QP.$
$ii.\ $With $Q$ as the centre and any convenient radius, draw an arc cutting $QP$ at $N.$
$iii.\ $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 $M$.
$iv.\ $Draw $QM$ and produce it to $R.$
$\angle\text{PQR}$ is the required angle of $120^\circ .$ View full question & answer→Question 45 Marks
Draw a line segment $AB = 5.6\ cm$. Draw the right bisector of $AB.$
Answer
Steps for construction:
$i.\ $Draw a line segment $AB$, which is equal to $5.6\ cm.$
$ii.\ $With $A$ as the centre and radius more than half of $AB$, draw arcs, one on each side of $AB.$
$iii.\ $With $B$ as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at $M$ and $N,$ respectively.
$iv.\ $Draw $MN$, meeting $AB$ at $R.$ View full question & answer→Question 55 Marks
Draw a line segment $AB = 5.6\ cm$. Draw the perpendicular bisector of $AB.$
Answer
Steps for construction:
$i.\ $Draw a line segment $AB = 5.6\ cm.$
$ii.\ $With $A$ as the centre and radius more than half of $AB$, draw arcs, one on each side of $AB.$
$iii.\ $With $B$ as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at $P$ and $Q,$ respectively.
$iv.\ $Draw $PQ$, meeting $AB$ at $R.$ View full question & answer→Question 65 Marks
Using a pair of compasses construct the following angles: $90^\circ $
Answer
Steps for construction:
$i.\ $Draw a line $PX.$
$ii.\ $Take a point $Q$ on $AC.$ With $Q$ as the centre and any convenient radius, draw an arc cutting $AX$ at $M$ and $N.$
$iii.\ $With $N$ as the centre and radius more than half of $MN$, draw an arc.
$iv.\ $With $M$ as the centre and the same radius as before, draw another arc to cut the previous arc at $W.$
$v.\ $Draw $QW$ produce it to $R.$
$\angle\text{PQR}$ is required angle of $90^\circ .$ View full question & answer→Question 75 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 $R.$
$5.\angle\text{PQR}$ is an angle of $60^\circ $. $\angle\text{PQR}$ is an angle of $60^\circ .$
$6.$With $M$ as the centre and radius more than half of $MN$, draw an arc.
$7.$With $N$ as the centre and radius same as in step $(5)$, draw another arc, cutting the previously drawn arc at point $X.$
$8.$Draw $QX$ and produce it to point $S$.
$9.$Ray $QS$ is the bisector of $\angle\text{PQR}.$ View full question & answer→Question 85 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 95 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 105 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 115 Marks
Draw the perpendicular bisector of a given line segment $AB$ of length 6cm.
Answer
Steps for construction:
1. Draw a line segment $A B$, which is equal to $6 \ cm .$
2. With $A$ as the centre and radius more than half of $A B$, draw arcs, one on each side of $A B$.
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 $M N$ meeting $A B$ at $D$.
$M N$ is the required perpendicular bisector of $A B$. View full question & answer→Question 125 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 $O A$, 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 $O E$. Then $\angle A O E=90^{\circ}$.
7. Draw the bisector $OF$ of $\angle AOE$.
8. Draw the bisector $OG$ of $\angle AOF$.
Then, $\angle AOG =22 \frac{1}{2}^{\circ}$ is the required angle. View full question & answer→Question 135 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 $O A$ 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 $C D$, draw an arc.
5. With $D$ as centre and same radius draw another arc to cut the previous arc at $E$.
6. Join $O E$. Then $\angle A O E=90^{\circ}$.
7. Draw the bisector OF of angle $\angle AOE$.
Then, $\angle\text{AOF}=45^\circ$ is the required angle. View full question & answer→Question 145 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 $O A$, 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 $\angle A O G=120^{\circ}$.
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 $\angle GOF$.
Then, $\angle\text{AOH}=135^\circ$ is the required angle. View full question & answer→Question 155 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 $A B$.
2. Take a point $P$ outside $A B$.
3. With $P$ as the centre and a convenient radius, draw an arc intersecting $A B$ at $M$ and $N$, respectively.
4. With $M$ as the centre and radius more than half of $M N$, 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 $A B$ at $S$.
$P Q$ is the required line that passes through $P$ and is perpendicular to $A B$. View full question & answer→Question 165 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 $A B$.
2. Take a point $P$ on line $A B$.
3. With $P$ as the centre, draw an arc of any radius, which intersects line $A B$ at $M$ and $N$, respectively.
4. With $M$ as the centre and radius more than half of $M N$, 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 $P R . P R$ is the required line, which is perpendicular to $A B$. View full question & answer→Question 175 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 185 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 195 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 $2.5cm.$
$3.$ 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 205 Marks
Use a pair of compasses and construct the following angles: $150^\circ $
Answer
Steps of Construction:
1. Draw a ray $O A$.
2. With $O$ as centre and any suitable radius draw an arc cutting $O A$ 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 $C D$ draw an arc.
5. With $D$ as centre and same radius draw another arc to cut the previous arc at $E$.
6. Join $O E$ and produce it to $F$.
Then, $\angle\text{AOF}=150^\circ.$ View full question & answer→Question 215 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 $O A$ and $O B$ 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 $C E$, cut the arc through $D$ at $F$.
5. Draw $Q F$ and produce it to point $R$.
$\therefore\ \angle\text{PQR}=\angle\text{AOB}$ View full question & answer→Question 225 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 $Q M$ and produce it to $R$.
$\angle PQR$ is $120^{\circ}$.
5. With $M$ as the centre and radius more than half of $M N$, 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 $Q X$ and produce it to point $S$.
Ray $QS$ is a bisector of $\angle PQR$. Ray $QS$ is a bisector of $\angle PQR$. View full question & answer→Question 235 Marks
Use a pair of compasses and construct the following angles: $75^\circ $
Answer
Steps of Construction:
1. Draw a ray $O A$.
2. With $O$ as centre and any suitable radius draw an arc cutting $O A$ 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 $C D$ 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 $\angle EOG$.
Then, $\angle\text{AOF}=75^\circ$ is the required angle. View full question & answer→Question 245 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 $O A$ 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 $C D$ 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 $\angle AOE$.
8. Draw the bisector $OG$ of $\angle EOF$.
Then, $\angle\text{AOG}=67\frac{1}{2}^\circ$ is the required angle. View full question & answer→Question 255 Marks
Draw a rectangle whose two adjacent sides are $5\ cm$ and $3.5\ cm$. Make use of a pair of compasses and a ruler only.
Answer
Steps of Construction:
1. Draw a line-segment $A B=5 cm$ with the help of a rular.
2. With Aas centre and suitable radius draw an arc cutting $A B$ 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 $D E$ 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 $AG =3.5 cm$. Then $\angle BAG =90^{\circ}$.
7. With $G$ as centre and radius equal to $A B$ draw an arc. With $B$ as centre and radius equal to $A G$ draw another arc to cut the previous arc at $H.$
8. Join $G H$ and $B H$. Then, $A B H G$ is the required rectangle. View full question & answer→Question 265 Marks
Draw $\angle A B C$ os measure $60^{\circ}$ such that $A B=4.5 \ cm$ and $B C=5 \ cm$ and $B C=5 \ cm$. Through $C$ draw a line parallel to $A B$ and through $B$ draw a line parallel to $A C$. 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 275 Marks
Draw a reactangle whose two adjacent sides are $5.4\ cm$ and $3.5\ cm.$
Answer
Steps of construction:
$1.$ Draw a ray $A X$.
$2.$ With $A$ as the centre, cut the ray $X A$ at $B$, such that $A B$ is equal to $3.5 \ cm .$
$3.$ With $B$ as the centre and with any convenient radius, draw an arc cutting $A X$ 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 $B W$ at point $C$.
$8.$ With $C$ as the centre and a radius $3.5 \ cm$ , draw an arc on the right side of $B C$.
$9.$ With $A$ as the centre and a radius $5.4 \ cm$ , draw an arc cutting the previous arc at $D$.
$10 .$ Join $CD$ and $AD.$
$ABCD$ is the required rectangle. View full question & answer→Question 285 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 $O A$ 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 $O M$ and produce it to $B$.
Then, $\angle\text{AOB}=45^\circ$ View full question & answer→Question 295 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 $A B=5 \ cm$.
$2.$ With $A$ as centre and any suitable radius draw an arc cutting $A B$ 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 $D E$ draw an arc.
$5.$ With $E$ as centre and same radius draw another arc to cut the previous arc at $F$.
$6.$ Join $A F$ and produce it to $G$ such that $A G=5 \ cm$.
$7.$ With $G$ as centre and radius equal to $A B$ 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, $ABHG$ is the required square. View full question & answer→Question 305 Marks
Construct an angle of $90^\circ $ and bisect it.
Answer
Construction steps:
$1.$ Draw a line $O A$.
$2.$ Take a point $B$ on $O A$. With $B$ as the centre and any convenient radius, draw an arc cutting $O A$ 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 $C.$
$\angle ABC$ is the required angle of $90^{\circ} \angle ABC$ is the required angle of $90^{\circ}$.
$6.$ With $S$ as the centre and radius more than half of $S N$, 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 $B X$ and produce it to point $D$. Ray $B D$ is the angle bisctor of $\angle A B C$. Ray $B D$ is the angle bisctor of $\angle A B C$. Ray $B D$ is the angl bisctor of $\angle ABC$. Ray $BD$ is the angle bisctor of $\angle ABC$. View full question & answer→