RS Aggarwal 2017 Solutions for Class 8 Math Chapter 17 Construction Of Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Construction Of Quadrilaterals are extremely popular among class 8 students for Math Construction Of Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2017 Book of class 8 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2017 Solutions. All RS Aggarwal 2017 Solutions for class 8 Math are prepared by experts and are 100% accurate.

#### Page No 198:

#### Question 1:

Construct a quadrilateral *ABCD* in which *AB* = 4.2 cm, *BC* = 6 cm, *CD* = 5.2 cm, *DA* = 5 cm and *AC* = 8 cm.

#### Answer:

Steps of construction:

Step 1: Draw $\mathrm{AB}=4.2\mathrm{cm}$.

Step 2: With A as the centre and radius equal to $8\mathrm{cm}$, draw an arc.

Step 3: With B as the centre and radius equal to $6\mathrm{cm}$, draw another arc, cutting the previous arc at C.

Step 4: Join BC.

Step 5: With A as the centre and radius equal to $5\mathrm{cm},$ draw an arc.

Step 6: With C as the centre and radius equal to $5.2\mathrm{cm}$, draw another arc, cutting the previous arc at D.

Step 7: Join AD and CD.

Thus, ABCD is the required quadrilateral.

#### Page No 198:

#### Question 2:

Construct a quadrilateral *PQRS* in which *PQ* = 5.4 cm, *QR* = 4.6 cm, *RS* = 4.3 cm, *SP* = 3.5 cm and diagonal *PR* = 4 cm.

#### Answer:

Steps of construction:

Step 1: Draw $\mathrm{PQ}=5.4\mathrm{cm}$.

Step 2: With P as the centre and radius equal to $4\mathrm{cm}$, draw an arc.

Step 3: With Q as the centre and radius equal to $4.6\mathrm{cm}$, draw another arc, cutting the previous arc at R.

Step 4: Join QR.

Step 5: With P as the centre and radius equal to $3.5\mathrm{cm},$ draw an arc.

Step 6: With R as the centre and radius equal to $4.3\mathrm{cm}$, draw another arc, cutting the previous arc at S.

Step 7: Join PS and RS.

Thus, PQRS is the required quadrilateral.

#### Page No 198:

#### Question 3:

Construct a quadrilateral *ABCD* in which *AB* = 3.5 cm, *BC* = 3.8 cm, *CD* = *DA* = 4.5 cm and diagonal *BD* = 5.6 cm.

#### Answer:

Steps of construction:

Step 1: Draw $\mathrm{AB}=3.5\mathrm{cm}$.

Step 2: With B as the centre and radius equal to $5.6\mathrm{cm}$, draw an arc.

Step 3: With A as the centre and radius equal to $4.5\mathrm{cm}$, draw another arc, cutting the previous arc at D.

Step 4: Join BD and AD.

Step 5: With D as the centre and radius equal to $4.5\mathrm{cm},$ draw an arc.

Step 6: With B as the centre and radius equal to $3.8\mathrm{cm}$, draw another arc, cutting the previous arc at C.

Step 7: Join BC and CD.

Thus, ABCD is the required quadrilateral.

#### Page No 198:

#### Question 4:

Construct a quadrilateral *ABCD* in which *AB* = 3.6 cm, *BC* = 3.3 cm, *AD* = 2.7 cm, diagonal *AC* = 4.6 cm and diagonal *BD* = 4 cm.

#### Answer:

Steps of construction:

Step 1: Draw $\mathrm{AB}=3.6\mathrm{cm}$.

Step 2: With B as the centre and radius equal to $4\mathrm{cm}$, draw an arc.

Step 3: With A as the centre and radius equal to $2.7\mathrm{cm}$, draw another arc, cutting the previous arc at D.

Step 4: Join BD and AD.

Step 5: With A as the centre and radius equal to $4.6\mathrm{cm},$ draw an arc.

Step 6: With B as the centre and radius equal to $3.3\mathrm{cm}$, draw another arc, cutting the previous arc at C.

Step 7: Join AC, BC and CD.

Thus, ABCD is the required quadrilateral.

#### Page No 198:

#### Question 5:

Construct a quadrilateral *PQRS* in which *QR* = 7.5 cm, *PR* = *PS* = 6 cm, *RS* = 5 cm and *QS* = 10 cm. Measure the fourth side.

#### Answer:

Steps of construction:

Step 1: Draw $QR=7.5\mathrm{cm}.$

Step 2: With *Q* as the centre and radius equal to $10\mathrm{cm}$, draw an arc.

Step 3: With* R* as the centre and radius equal to $5\mathrm{cm}$, draw another arc, cutting the previous arc at* S*.

Step 4: Join *QS* and *RS*.

Step 5: With *S *as the centre and radius equal to $6\mathrm{cm},$ draw an arc.

Step 6: With *R* as the centre and radius equal to $6\mathrm{cm}$, draw another arc, cutting the previous arc at *P*.

Step 7: Join *PS* and *PR*.

Step 8: *PQ *= 4.9 cm

Thus,* PQRS* is the required quadrilateral.

#### Page No 198:

#### Question 6:

construct a quadrilateral *ABCD* in which *AB* =3.4 cm, *CD* = 3 cm, *DA* = 5.7 cm, *AC* = 8 cm and *BD* = 4 cm.

#### Answer:

Steps of construction:

Step 1: Draw $AB=3.4\mathrm{cm}.$

Step 2: With *B* as the centre and radius equal to $4\mathrm{cm}$, draw an arc.

Step 3: With* A* as the centre and radius equal to $5.7\mathrm{cm}$, draw another arc, cutting the previous arc at *D.*

Step 4: Join *BD* and *AD*.

Step 5: With *A* as the centre and radius equal to $8\mathrm{cm},$ draw an arc.

Step 6: With *D* as the centre and radius equal to $3\mathrm{cm}$, draw another arc, cutting the previous arc at *C*.

Step 7: Join *AC, CD *and *BC.*

Thus, *ABCD* is the required quadrilateral.

#### Page No 198:

#### Question 7:

Construct a quadrilateral *ABCD* in which *AB = BC* = 3.5 cm, *AD = CD* = 5.2 cm and ∠*ABC* = 120°.

#### Answer:

Steps of construction:

Step 1: Draw *AB*= $3.5\mathrm{cm}$.

Step 2: Make $\angle ABC={120}^{\circ}$.

Step 3: With B as the centre, draw an arc $3.5\mathrm{cm}$ and name that point *C*.

Step 4: With *C* as the centre, draw an arc $5.2\mathrm{cm}$.

Step 5: With *A* as the centre, draw another arc $5.2\mathrm{cm}$, cutting the previous arc at *D*.

Step 6: Join *CD* and *AD.*

Thus, $ABCD$ is the required quadrilateral.

#### Page No 198:

#### Question 8:

Construct a quadrilateral *ABCD* in which *AB* = 2.9 cm, *BC* = 3.2 cm, *CD* = 2.7 cm, *DA* = 3.4 cm and ∠*A* = 70°.

#### Answer:

Steps of construction:

Step 1: Draw *AB*= $2.9cm$

Step 2: Make $\angle A={70}^{\circ}$

Step 3: With* A* as the centre, draw an arc of $3.4cm$. Name that point as *D*.

Step 4: With *D *as the centre, draw an arc of $2.7cm$.

Step 5: With* B* as the centre, draw an arc of 3.2 cm, cutting the previous arc at* C*.

Step 6: Join *CD* and *BC*.

Then, $ABCD$ is the required quadrilateral.

#### Page No 198:

#### Question 9:

Construct a quadrilateral *ABCD* in which *AB* = 3.5 cm, *BC* = 5 cm, *CD* = 4.6 cm, ∠*B* = 125° and ∠*C* = 60°.

#### Answer:

Steps of construction:

Step 1: Draw *BC*= $5cm$

Step 2: Make $\angle B={125}^{\circ}and\angle C={60}^{\circ}$

Step 3: With *B* as the centre, draw an arc of $3.5cm$. Name that point as *A.*

Step 4: With *C* as the centre, draw an arc of $4.6cm$. Name that point as *D.*

Step 5: Join *A *and* D.*

Then, $ABCD$ is the required quadrilateral.

#### Page No 198:

#### Question 10:

Construct a quadrilateral *PQRS* in which *PQ* = 6 cm, *QR* = 5.6 cm, *RS* = 2.7 cm, ∠*Q* = 45° and ∠*R* = 90°.

#### Answer:

Steps of construction:

Step 1: Draw *QR*= $5.6cm$

Step 2: Make $\angle Q={45}^{\circ}\mathrm{and}\angle R={90}^{\circ}$

Step 3: With *Q* as the centre, draw an arc of $6cm$. Name that point as *P*.

Step 4: With *R* as the centre, draw an arc of $2.7cm$. Name that point as* S*.

Step 6: Join* P *and* S*.

Then, $PQRS$ is the required quadrilateral.

#### Page No 198:

#### Question 11:

Construct a quadrilateral *ABCD* in which *AB* = 5.6 cm, *BC* = 4 cm, ∠*A* = 50°, ∠*B* = 105° and ∠*D* = 80°.

#### Answer:

Steps of construction:

Step 1: Draw *AB*= $5.6cm$

Step 2: Make $\angle A={50}^{\circ}and\angle B={105}^{\circ}$

Step 3: With *B* as the centre, draw an arc of $4cm$.

Step 3: Sum of all the angles of the quadrilateral is ${360}^{\circ}$.

$\angle A+\angle B+\angle C+\angle D={360}^{\circ}\phantom{\rule{0ex}{0ex}}{50}^{\circ}+{105}^{\circ}+\angle C+{80}^{\circ}={360}^{\circ}\phantom{\rule{0ex}{0ex}}{235}^{\circ}+\angle C={360}^{\circ}\phantom{\rule{0ex}{0ex}}\angle C={360}^{\circ}-{235}^{\circ}\phantom{\rule{0ex}{0ex}}\angle C={125}^{\circ}$

Step 5: With *C* as the centre, make $\angle C\mathrm{equal}\mathrm{to}\angle {125}^{\circ}$.

Step 6: Join *C *and* D*.

Step 7: Measure $\angle D={80}^{\circ}$

Then, $ABCD$ is the required quadrilateral.

#### Page No 199:

#### Question 12:

Construct a quadrilateral *PQRS* in which *PQ* = 5 cm, *QR* = 6.5 cm, ∠*P* = ∠*R* = 100° and ∠*S* = 75°.

#### Answer:

Steps of construction:

Step 1: Draw *PQ*= $5cm$

Step 2:

$\angle P+\angle Q+\angle R+\angle S={360}^{\circ}\phantom{\rule{0ex}{0ex}}{100}^{\circ}+\angle Q+{100}^{\circ}+{75}^{\circ}={360}^{\circ}\phantom{\rule{0ex}{0ex}}{275}^{\circ}+\angle Q={360}^{\circ}\phantom{\rule{0ex}{0ex}}\angle Q={360}^{\circ}-{275}^{\circ}\phantom{\rule{0ex}{0ex}}\angle Q={85}^{\circ}$

Step 3: Make $\angle P={100}^{\circ}and\angle Q={85}^{\circ}$

Step 3: With *Q *as the centre, draw an arc of $6.5cm$.

Step 4: Make $\angle R={100}^{\circ}$

Step 6: Join *R *and* S.*

Step 7: Measure $\angle S={75}^{\circ}$

Then, $PQRS$ is the required quadrilateral.

#### Page No 199:

#### Question 13:

Construct a quadrilateral *ABCD* in which *AB* = 4 cm, *AC* = 5 cm, *AD* = 5.5 cm and ∠*ABC* = ∠*ACD* = 90°.

#### Answer:

Steps of construction:

Step 1: Draw $AB=4cm$

Step 2: $Make\angle B={90}^{\circ}$

Step 3: $A{C}^{2}=A{B}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}{5}^{2}={4}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}25-16=B{C}^{2}\phantom{\rule{0ex}{0ex}}BC=3cm$

With *B* as the centre, draw an arc equal to 3 cm.

Step 4: Make $\angle C={90}^{\circ}$

Step 5: With *A* as the centre and radius equal to $5.5cm$, draw an arc and name that point as *D*.

Then, $ABCD$ is the required quadrilateral.

#### Page No 201:

#### Question 1:

Construct a parallelogram *ABCD* in which *AB* = 5.2 cm, *BC* = 4.7 cm and *AC* = 7.6 cm.

#### Answer:

Steps of construction:

Step 1: Draw *AB *= $5.2cm$

Step 2: With *B* as the centre, draw an arc of $4.7cm$.

Step 3: With *A* as the centre, draw another arc of $7.6cm$, cutting the previous arc at C.

Step 4: Join *A *and* C.*

Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with *C* as the centre, draw an arc of $5.2cm$.

Step 6: With* A* as the centre, draw another arc of $4.7cm$, cutting the previous arc at* D*.

Step 7: Join *CD* and *AD*.

Then, *ABCD *is the required parallelogram.

#### Page No 201:

#### Question 2:

Construct a parallelogram *ABCD* in which *AB* = 4.3 cm, *AD* = 4 cm and *BD* = 6.8 cm.

#### Answer:

Steps of construction:

Step 1: Draw AB= $4.3cm$

Step 2: With *B* as the centre, draw an arc of $6.8cm$.

Step 3: With *A* as the centre, draw another arc of $4cm$, cutting the previous arc at *D*.

Step 4: Join *BD* and *AD*.

Step 5: We know that the opposite sides of a parallelogram are equal.

Thus, with *D* as the centre, draw an arc of $4.3cm$.

Step 6: With *B* as the centre, draw another arc of $4cm$, cutting the previous arc at *C*.

Step 7: Join *CD* and *BC*.

then, *ABCD* is the required parallelogram.

#### Page No 201:

#### Question 3:

Construct a parallelogram *PQRS* in which *QR* = 6 cm, *PQ* = 4 cm and *∠PQR* = 60° cm.

#### Answer:

Steps of construction:

Step 1: Draw* PQ*= 4 cm

Step 2: Make $\angle PQR={60}^{\circ}$

Step 2: With *Q* as the centre, draw an arc of 6 cm and name that point as *R*.

Step 3: With *R* as the centre, draw an arc of 4 cm and name that point as* S.*

Step 4: Join *SR* and *PS.*

Then, *PQRS *is the required parallelogram.

#### Page No 201:

#### Question 4:

Construct a parallelogram *ABCD* in which *BC* = 5 cm, *∠BCD* = 120° and *CD* = 4.8 cm.

#### Answer:

Steps of construction:

Step 1: Draw BC= $5cm$

Step 2: Make an $\angle BCD={120}^{\circ}$

Step 2: With *C* as centre draw an arc of $4.8\mathrm{cm}$, name that point as* D*

Step 3: With D as centre draw an arc $5\mathrm{cm}$, name that point as *A*

Step 4: With *B *as centre draw another arc $4.8\mathrm{cm}$ cutting the previous arc at *A*.

Step 5: Join *AD* and *AB*

then, *ABCD *is a required parallelogram.

#### Page No 201:

#### Question 5:

Construct a parallelogram, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm. Measure the other side.

#### Answer:

We know that the diagonals of a parallelogram bisect each other.

Steps of construction:

Step 1: Draw *AB*= $4.4cm$

Step 2: With *A* as the centre and radius $2.8cm$, draw an arc.

Step 3: With *B* as the centre and radius $3.5cm$, draw another arc, cutting the previous arc at point *O*.

Step 4: Join *OA* and *OB.*

Step 5: Produce *OA* to *C,* such that *OC= AO*. Produce *OB *to *D*, such that* OB=OD*.

Step 5: Join *AD, BC,* and *CD*.

Thus,* ABCD *is the required parallelogram. The other side is 4.5 cm in length.

#### Page No 201:

#### Question 6:

Construct a parallelogram *ABCD* in which *AB* = 6.5 cm, *AC* = 3.4 cm and the altitude *AL* from *A* is 2.5 cm. Draw the altitude from *C* and measure it.

#### Answer:

Steps of construction:

Step 1: Draw *AB*= 6.5cm

Step 2: Draw a perpendicular at point *A. N*ame that ray as *AX*. From point *A,* draw an arc of length 2.5 cm on the ray *AX* and name that point as L.

Step 3: On point *L, *make a perpendicular. Draw a straight line YZ passing through L, which is perpendicular to the ray AX.

Step 4: Cut an arc of length 3.4 cm on the line *YZ *and name it as *C.*

Step 5: From point *C,* cut an arc of length 6.5 cm on the line *YZ. N*ame that point as* D.*

Step 6: Join *BC* and* AD. *

Therefore, quadrilateral *ABCD* is a parallelogram.

The altitude from C measures 2.5 cm in length.

#### Page No 201:

#### Question 7:

Construct a parallelogram *ABCD,* in which diagonal* AC* = 3.8 cm, diagonal* BD *= 4.6 cm and the angle between *AC* and *BD* is 60°.

#### Answer:

We know that the diagonals of a parallelogram bisect each other.

Steps of construction:

Step 1: Draw *AC*= $3.8\mathrm{cm}$

Step 2: Bisect *AC* at O.

Step 3: Make $\angle COX={60}^{\circ}$

Produce *XO* to *Y.*

Step 4:

$OB=\frac{1}{2}(4.6)\mathrm{cm}\phantom{\rule{0ex}{0ex}}OB=2.3\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{and}OD=\frac{1}{2}(4.6)\mathrm{cm}\phantom{\rule{0ex}{0ex}}OD=2.3\mathrm{cm}$

Step 5: Join *AB, BC, CD* and *AD*.

Thus, *ABCD* is the required parallelogram.

#### Page No 201:

#### Question 8:

Construct a rectangle *ABCD* whose adjacent sides are 11 cm and 8.5 cm.

#### Answer:

Steps of construction:

Step 1: Draw* AB* = $11cm$

Step 2: Make $\angle A={90}^{\circ}\phantom{\rule{0ex}{0ex}}\angle B={90}^{\circ}$

Step 3: Draw an arc of 8.5 cm from point* A* and name that point as *D. *

Step 4: Draw an arc of 8.5 cm from point *B* and name that point as *C.*

Step 5: Join *C* and* D.*

Thus, *ABCD* is the required rectangle.

#### Page No 201:

#### Question 9:

Construct a square, each of whose sides measures 6.4 cm.

#### Answer:

All the sides of a square are equal.

Steps of construction:

Step 1: Draw *AB *= $6.4cm$

Step 2: Make $\angle A={90}^{\circ}\phantom{\rule{0ex}{0ex}}\angle B={90}^{\circ}$

Step 3: Draw an arc of length 6.4 cm from point *A* and name that point as *D. *

Step 4: Draw an arc of length 6.4 cm from point *B *and name that point as *C.*

Step 5: Join *C *and* D*.

Thus, *ABCD* is a required square.

#### Page No 201:

#### Question 10:

Construct a square, each of whose diagonals measures 5.8 cm.

#### Answer:

We know that the diagonals of a square bisect each other at right angles.

Steps of construction:

Step 1: Draw *AC*= $5.8\mathrm{cm}$

Step 2: Draw the perpendicular bisector *XY *of *AC,* meeting *it* at *O.*

Step 3:

: $FromO:\phantom{\rule{0ex}{0ex}}OB=\frac{1}{2}(5.8)\mathrm{cm}=2.9\mathrm{cm}\phantom{\rule{0ex}{0ex}}OD=\frac{1}{2}(5.8)\mathrm{cm}=2.9\mathrm{cm}$

Step 4: Join *AB, BC, CD* and *DA*.

*ABCD* is the required square.

#### Page No 201:

#### Question 11:

Construct a rectangle *PQRS* in which *QR* = 3.6 cm and diagonal *PR* = 6 cm. Measure the other side of the rectangle.

#### Answer:

Steps of construction:

Step 1: Draw *QR* = $3.6cm$

Step 2: Make $\angle Q={90}^{\circ}\phantom{\rule{0ex}{0ex}}\angle R={90}^{\circ}$

Step 3:

$P{R}^{2}=P{Q}^{2}+Q{R}^{2}\phantom{\rule{0ex}{0ex}}{6}^{2}=P{Q}^{2}+3.{6}^{2}\phantom{\rule{0ex}{0ex}}P{Q}^{2}=36-12.96\phantom{\rule{0ex}{0ex}}P{Q}^{2}=23.04\phantom{\rule{0ex}{0ex}}PQ=4.8\mathrm{cm}$

Step 3: Draw an arc of length 4.8 cm from point *Q* and name that point as* P*.

Step 4: Draw an arc of length 6 cm from point *R*, cutting the previous arc at* P*.

Step 5: Join* PQ*

Step 6: Draw an arc of length 4.8 cm from point *R.
F*rom point P, draw an arc of length 3.6 cm, cutting the previous arc. Name that point as

*S*.

Step 7: Join

*P*and

*S.*

Thus,

*PQRS*is the required rectangle. The other side is 4.8 cm in length.

#### Page No 201:

#### Question 12:

Construct a rhombus the lengths of whose diagonals are 6 cm and 8 cm.

#### Answer:

We know that the diagonals of a rhombus bisect each other.

.Steps of construction:

Step 1: Draw AC= $6cm$

Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O.

Step 3:

$OB=\frac{1}{2}\left(8\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}OB=4\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{and}OD=\frac{1}{2}\left(8\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}OD=4\mathrm{cm}$

Draw an arc of length 4 cm on *OX *and name that point as *B.*

Draw an arc of length 4 cm on *OY* and name that point as* D*.

Step 4 : Join *AB, BC, CD* and *AD*.

Thus, *ABCD* is the required rhombus, as shown in the figure.

#### Page No 201:

#### Question 13:

Construct a rhombus *ABCD* in which *AB* = 4 cm and diagonal *AC* is 6.5 cm.

#### Answer:

Steps of construction:

Step 1: Draw *AB* = $4cm$

Step 2: With *B* as the centre, draw an arc of $4\mathrm{cm}$.

Step 3: With *A* as the centre, draw another arc of $6.5\mathrm{cm}$, cutting the previous arc at* C.*

Step 4: Join *AC* and* BC.*

Step 5: With *C* as the centre, draw an arc of 4 cm.

Step 6: With* A* as the centre, draw another arc of $4\mathrm{cm}$, cutting the previous arc at *D*.

Step 7: Join *AD* and* CD*.

*ABCD* is the required rhombus.

#### Page No 201:

#### Question 14:

Draw a rhombus whose side is 7.2 cm and one angle is 60°.

#### Answer:

Steps of construction:

Step1: Draw *AB* = $7.2\mathrm{cm}$

Step2: Draw $\angle ABY=60\xb0\phantom{\rule{0ex}{0ex}}\angle BAX=120\xb0$

Sum of the adjacent angles is 180°.

$\angle BAX+\angle ABY=180\xb0\phantom{\rule{0ex}{0ex}}=>\angle BAX=180\xb0-60\xb0=120\xb0$

Step 3:

$\mathrm{Set}\mathrm{off}AD(7.2\mathrm{cm})\mathrm{along}AX\mathrm{and}BC(7.2\mathrm{cm})\mathrm{along}BY.$

Step 4: Join *C* and *D*.

Then, *ABCD *is the required rhombus.

#### Page No 201:

#### Question 15:

Construct a trapezium *ABCD* in which *AB* = 6 cm, *BC* = 4 cm, *CD* = 3.2 cm, ∠*B* = 75° and *DC*||*AB*.

#### Answer:

Steps of construction:

Step 1: Draw *AB*=$6\mathrm{cm}$

Step 2: Make $\angle ABX={75}^{\circ}$

Step 3: With *B* as the centre, draw an arc at $4cm$. Name that point as *C.*

Step 4: $AB\parallel CD$

$\therefore \angle ABX+\angle BCY=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle BCY=180\xb0-75\xb0=105\xb0$

Make $\angle BCY=105\xb0$

At* C*, draw an arc of length $3.2\mathrm{cm}$.

Step 5: Join A and D.

Thus, *ABCD* is the required trapezium.

#### Page No 201:

#### Question 16:

Draw a trapezium *ABCD* in which *AB*||*DC*, *AB* = 7 cm, *BC *= 5 cm, *AD *= 6.5 cm and ∠*B* = 60°.

#### Answer:

Steps of construction :

Step1: Draw *AB* equal to 7 cm.

Step2: Make an angle, $\angle ABX,\mathrm{equal}\mathrm{to}60\xb0.$

Step3: With *B *as the centre, draw an arc of $5\mathrm{cm}$. Name that point as *C*. Join *B *and *C*.

Step4:

$AB\parallel DC\phantom{\rule{0ex}{0ex}}\therefore \angle ABX+\angle BCY=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle BCY=180\xb0-60\xb0=120\xb0$

Draw an angle, $\angle BCY,\mathrm{equal}\mathrm{to}120\xb0.$

Step4: With *A* as the centre, draw an arc of length $6.5\mathrm{cm}$, which cuts *CY*. Mark that point as *D*.

Step5: Join *A *and* D*.

Thus, *ABCD* is the required trapezium.

#### Page No 202:

#### Question 1:

Define the terms:

(i) Open curve

(ii) Closed curve

(iii) Simple closed curve

#### Answer:

( i) Open curve: An open curve is a curve where the beginning and end points are different.

Example: Parabola

(ii) Closed Curve: A curve that joins up so there are no end points.

Example: Ellipse

(iii) Simple closed curve: A closed curve that does not intersect itself.

#### Page No 202:

#### Question 2:

The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. Find the measure of each angle.

#### Answer:

Let the angles be $\left(x\right)\xb0,\left(2x\right)\xb0,\left(3x\right)\xb0\mathrm{and}\left(4x\right)\xb0.$

Sum of the angles of a quadrilateral is ${360}^{\circ}$.

$x+2x+3x+4x=360\phantom{\rule{0ex}{0ex}}10x=360\phantom{\rule{0ex}{0ex}}x=\frac{360}{10}\phantom{\rule{0ex}{0ex}}x=36$

$\left(2x\right)\xb0=(2\times 36{)}^{\circ}={72}^{\circ}\phantom{\rule{0ex}{0ex}}(3x)\xb0=(3\times 36{)}^{\circ}={108}^{\circ}\phantom{\rule{0ex}{0ex}}\left(4x\right)\xb0=(4\times 36{)}^{\circ}={144}^{\circ}$

The angles of the quadrilateral are ${36}^{\circ},{72}^{\circ},{108}^{\circ}\mathrm{and}{144}^{\circ}.$

#### Page No 202:

#### Question 3:

Two adjacent angles of a parallelogram are in the ratio 2 : 3. Find the measure of each of its angles.

#### Answer:

$\mathrm{Let}\mathrm{the}\mathrm{two}\mathrm{adjacent}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{parallelogram}\mathrm{b}\mathrm{e}\left(2x\right)\xb0\mathrm{and}\left(3x\right)\xb0.$

Sum of any two adjacent angles of a parallelogram is ${180}^{\circ}$.

$\therefore 2x+3x=180\phantom{\rule{0ex}{0ex}}\Rightarrow 5x=180\phantom{\rule{0ex}{0ex}}\Rightarrow x=36$

$\left(2x\right)\xb0=(2\times 36)\xb0={72}^{\circ}\phantom{\rule{0ex}{0ex}}\left(3x\right)\xb0=(3\times 36)\xb0={108}^{\circ}$

Measures of the angles are ${72}^{\circ}and{108}^{\circ}$.

#### Page No 202:

#### Question 4:

The sides of a rectangle are in the ratio 4 : 5 and its perimeter is 180 cm. Find its sides.

#### Answer:

Let the length be $4x$ cm and the breadth be $5x$ cm.

Perimeter of the rectangle =180 $cm$

Perimeter of the rectangle=$2(l+b)$

$2(l+b)=180\phantom{\rule{0ex}{0ex}}\Rightarrow 2(4x+5x)=180\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(9x\right)=180\phantom{\rule{0ex}{0ex}}\Rightarrow 18x=180\phantom{\rule{0ex}{0ex}}\Rightarrow x=10$

$\therefore \mathrm{Length}=4x\mathrm{cm}=4\times 10=40\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Breadth}=5x\mathrm{cm}=5\times 10=50\mathrm{cm}$

#### Page No 202:

#### Question 5:

Prove that the diagonals of a rhombus bisect each other at right angles.

#### Answer:

Rhombus is a parallelogram.

$\u2206AOB\mathrm{and}\u2206COD\phantom{\rule{0ex}{0ex}}\angle OAB=\angle OCD(\mathrm{alternate}\mathrm{angle})\phantom{\rule{0ex}{0ex}}\angle ODC=\angle OBA(\mathrm{alternate}\mathrm{angle})\phantom{\rule{0ex}{0ex}}\angle DOC=\angle AOB(\mathrm{vertically}\mathrm{opposite}\mathrm{angles})\phantom{\rule{0ex}{0ex}}\u2206AOB\cong COB\phantom{\rule{0ex}{0ex}}\therefore AO=CO\phantom{\rule{0ex}{0ex}}OB=OD$

Therefore, the diagonals bisects at O.

Now, let us prove that the diagonals intersect each other at right angles.

Consider $\u2206COD\hspace{0.17em}and\u2206COB$:

$CD=CB\hspace{0.17em}(\mathrm{all}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{rhombus}\mathrm{are}\mathrm{equal})\phantom{\rule{0ex}{0ex}}CO=CO(\mathrm{common}\mathrm{side})\phantom{\rule{0ex}{0ex}}OD=OB\hspace{0.17em}(p\mathrm{oint}O\mathrm{bisects}BD)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

∴ $\u2206COD\cong \u2206COB$

∴ $\angle COD=\angle COB$ (corresponding parts of congruent triangles)

Further, $\angle COD+\angle COB=180\xb0(l\mathrm{inear}\mathrm{pair})$

∴ $\angle COD=\angle COB=90\xb0$

It is proved that the diagonals of a rhombus are perpendicular bisectors of each other.

#### Page No 202:

#### Question 6:

The diagonals of a rhombus are 16 cm and 12 cm. Find the length of each side of the rhombus.

#### Answer:

All the sides of a rhombus are equal in length.

The diagonals of a rhombus intersect at ${90}^{\circ}$.

The diagonal and the side of a rhombus form right triangles.

In $\u25b3AOB$:

$A{B}^{2}=AO{}^{2}+O{B}^{2}\phantom{\rule{0ex}{0ex}}={8}^{2}+{6}^{2}\phantom{\rule{0ex}{0ex}}=64+36\phantom{\rule{0ex}{0ex}}=100\phantom{\rule{0ex}{0ex}}AB=10\mathrm{cm}$

Therefore, the length of each side of the rhombus is 10 cm.

#### Page No 202:

#### Question 7:

**Mark (✓) against the correct answer:**

Two opposite angles of a parallelogram are (3*x* − 2)° and (50 − *x*)°. The measures of all its angles are

(a) 97°, 83°, 97°, 83°

(b) 37°, 143°, 37°, 143°

(c) 76°, 104°, 76°, 104°

(d) none of these

#### Answer:

(b) 37^{o}, 143^{o}, 37^{o} 143^{o}

Opposite angles of a parallelogram are equal.

$\phantom{\rule{0ex}{0ex}}\therefore 3x-2=50-x\phantom{\rule{0ex}{0ex}}\Rightarrow 3x+x=50+2\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=52\phantom{\rule{0ex}{0ex}}\Rightarrow x=13\phantom{\rule{0ex}{0ex}}$

Therefore, the first and the second angles are:

${\left(3x-2\right)}^{\circ}={\left(2\times 13-2\right)}^{\circ}={37}^{\circ}\phantom{\rule{0ex}{0ex}}{\left(50-x\right)}^{\circ}={\left(50-13\right)}^{\circ}={37}^{\circ}\phantom{\rule{0ex}{0ex}}$

Sum of adjacent angles in a parallelogram is ${180}^{\circ}$.

Adjacent angles = ${180}^{\circ}-{37}^{\circ}={143}^{\circ}$

#### Page No 202:

#### Question 8:

**Mark (✓) against the correct answer:**

The angles of quadrilateral are in the ratio 1 : 3 : 7 : 9. The measure of the largest angle is

(a) 63°

(b) 72°

(c) 81°

(d) none of these

#### Answer:

(d) none of the these

Let the angles be $\left(x\right)\xb0,\left(3x\right)\xb0,\left(7x\right)\xb0\mathrm{and}\left(9x\right)\xb0$.

Sum of the angles of the quadrilateral is ${360}^{\circ}$.

$x+3x+7x+9x=360\phantom{\rule{0ex}{0ex}}20x=360\phantom{\rule{0ex}{0ex}}x=18$

$\mathrm{Angles}:\phantom{\rule{0ex}{0ex}}\left(3x\right)\xb0=(3\times 18)={54}^{\circ}\phantom{\rule{0ex}{0ex}}\left(7x\right)\xb0=(7\times 18)\xb0={126}^{\circ}\phantom{\rule{0ex}{0ex}}\left(9x\right)\xb0=(9\times 18)\xb0={162}^{\circ}$

#### Page No 202:

#### Question 9:

**Mark (✓) against the correct answer:**

The length of a rectangle is 8 cm and each of its diagonals measures 10 cm. The breadth of the rectangle is

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 9 cm

#### Answer:

(b) 6 cm

Let the breadth of the rectangle be* x* cm.

Diagonal =10 cm

Length= 8 cm

The rectangle is divided into two right triangles.

$Diagona{l}^{2}=Lengt{h}^{2}+Breadt{h}^{2}\phantom{\rule{0ex}{0ex}}{10}^{2}={8}^{2}+{x}^{2}\phantom{\rule{0ex}{0ex}}100-64={x}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}=36\phantom{\rule{0ex}{0ex}}x=6cm$

Breadth of the rectangle = 6 cm

#### Page No 202:

#### Question 10:

**Mark (✓) against the correct answer:**

In a square *PQRS*, if *PQ* = (2*x* + 3) cm and *QR* = (3*x* − 5) cm then

(a) *x* = 4

(b) *x* = 5

(c) *x* = 6

(d) *x* = 8

#### Answer:

(d) *x* = 8

All sides of a square are equal.

$PQ=QR\phantom{\rule{0ex}{0ex}}(2x+3)=(3x-5)\phantom{\rule{0ex}{0ex}}=>2x-3x=-5-3\phantom{\rule{0ex}{0ex}}=>x=8\mathrm{cm}$

#### Page No 202:

#### Question 11:

**Mark (✓) against the correct answer:**

The bisectors of two adjacent angles of a parallelogram intersect at

(a) 30°

(b) 45°

(c) 60°

(d) 90°

#### Answer:

(d) 90°

We know that the opposite sides and the angles in a parallelogram are equal.

Also, its adjacent sides are supplementary, i.e. sum of the sides is equal to 180.

Now, the bisectors of these angles form a triangle, whose two angles are:

$\frac{A}{2}\mathrm{and}\frac{B}{2}\mathrm{or}\frac{A}{2}=(90-\frac{A}{2})\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\xb0.\phantom{\rule{0ex}{0ex}}\frac{\angle A}{2}+90-\frac{\angle A}{2}+\angle O={180}^{\circ}\phantom{\rule{0ex}{0ex}}\angle O=180-90\phantom{\rule{0ex}{0ex}}\angle O={90}^{\circ}\phantom{\rule{0ex}{0ex}}$

Hence, the two bisectors intersect at right angles.

#### Page No 202:

#### Question 12:

**Mark (✓) against the correct answer:**

How many diagonals are there in a hexagon?

(a) 6

(b) 8

(c) 9

(d) 10

#### Answer:

(c) 9

Hexagon has six sides.

$\mathrm{Number}\mathrm{of}\mathrm{diagonals}=\frac{n(n-3)}{2}(\mathrm{where}n\mathrm{is}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{side}s)\phantom{\rule{0ex}{0ex}}=\frac{6(6-3)}{2}\phantom{\rule{0ex}{0ex}}=9$

#### Page No 202:

#### Question 13:

**Mark (✓) against the correct answer:**

Each interior angle of a polygon is 135. How many sides does it have?

(a) 10

(b) 8

(c) 6

(d) 5

#### Answer:

(b) 8

$\mathrm{Interior}\mathrm{angle}=\frac{180(n-2)}{n}\phantom{\rule{0ex}{0ex}}\Rightarrow 135=\frac{180(n-2)}{n}\phantom{\rule{0ex}{0ex}}\Rightarrow 135n=180n-360\phantom{\rule{0ex}{0ex}}\Rightarrow 360=180n-135n\phantom{\rule{0ex}{0ex}}\Rightarrow n=8$

It has 8 sides.

#### Page No 202:

#### Question 14:

**Fill in the blanks.**

For a convex polygon of *n* sides, we have:

(i) Sum of all exterior angles = .........

(ii) Sum of all interior angles = .........

(iii) Number of diagonals = .........

#### Answer:

(i) Sum of all exterior angles = ${360}^{\circ}$

(ii) Sum of all interior angles = $(n-2)\times 180\xb0\phantom{\rule{0ex}{0ex}}$

(iii) Number of diagonals = $\frac{n(n-3)}{2}$

#### Page No 202:

#### Question 15:

**Fill in the blanks.**

For a regular polygon of *n* sides, we have:

(i) Sum of all exterior angles = .........

(ii) Sum of all interior angles = .........

#### Answer:

(i) Sum of all exterior angles of a regular polygon is ${360}^{\circ}$.

(ii) Sum of all interior angles of a polygon is $(n-2)\times 180\xb0,\mathrm{where}n\mathrm{is}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{sides}.$

#### Page No 202:

#### Question 16:

**Fill in the blanks.**

(i) Each interior angle of a regular octagon is (.........)°.

(ii) The sum of all interior angles of a regular hexagon is (.........)°.

(iii) Each exterior angle of a regular polygon is 60°. This polygon is a .........

(iv) Each interior angle of a regular polygon is 108°. This polygon is a .........

(v) A pentagon has ......... diagonals.

#### Answer:

(i) Octagon has 8 sides.

$\therefore \mathrm{Interior}\mathrm{angle}=\frac{180\xb0n-360\xb0}{n}\phantom{\rule{0ex}{0ex}}\mathrm{Int}\mathrm{erior}\mathrm{angle}=\frac{(180\xb0\times 8)-360\xb0}{8}=135\xb0$^{}

(ii) Sum of the interior angles of a regular hexagon = $(6-2)\times {180}^{\circ}={720}^{\circ}$

(iii) Each exterior angle of a regular polygon is ${60}^{\circ}$.

$\therefore \frac{360}{60}=6$

Therefore, the given polygon is a hexagon.

(iv) If the interior angle is ${108}^{\circ}$, then the exterior angle will be ${72}^{\circ}$. (interior and exterior angles are supplementary)

Sum of the exterior angles of a polygon is 360°.

Let there be *n *sides of a polygon.

$72n=360\phantom{\rule{0ex}{0ex}}n=\frac{360}{72}\phantom{\rule{0ex}{0ex}}n=5$

Since it has 5 sides, the polygon is a pentagon.

(v) A pentagon has 5 diagonals.

$\mathrm{If}n\mathrm{is}\mathrm{the}\mathrm{number}\mathrm{of}s\mathrm{ides,}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{diagonals}=\frac{n(n-3)}{2}\phantom{\rule{0ex}{0ex}}\frac{5(5-3)}{2}\phantom{\rule{0ex}{0ex}}=5$

#### Page No 203:

#### Question 17:

**Write 'T' for true and 'F' for false for each of the following:**

(i) The diagonals of a parallelogram are equal.

(ii) The diagonals of a rectangle are perpendicular to each other.

(iii) The diagonals of a rhombus bisect each other at right angles.

(iv) Every rhombus is a kite.

#### Answer:

(i) F

The diagonals of a parallelogram need not be equal in length.

(ii) F

The diagonals of a rectangle are not perpendicular to each other.

(iii) T

(iv) T

Adjacent sides of a kite are equal and this is also true for a rhombus. Additionally, all the sides of a rhombus are equal to each other.

#### Page No 203:

#### Question 18:

Construct a quadrilateral *PQRS* in which *PQ* = 4.2 cm, *∠PQR* = 60°, *∠QPS* = 120°, *QR* = 5 cm and *PS* = 6 cm.

#### Answer:

Steps of construction:

Step 1: Take *PQ* = 4.2 cm

Step 2: $\mathrm{Make}\angle XPQ={120}^{\circ},\angle YQP={60}^{\circ}\phantom{\rule{0ex}{0ex}}$

Step 3: Cut an arc of length 5 cm from point *Q*. Name that point as *R*.

Step 4: From *P,* make an arc of length 6 cm. Name that point as S.

Step 5: Join *P* and* S*.

Thus, *PQRS* is a quadrilateral.

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