Rs Aggarwal 2019 2020 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 2019 2020 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 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class 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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