RS Aggarwal 2020 2021 Solutions for Class 8 Maths 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 Maths Construction Of Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2020 2021 Book of class 8 Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RS Aggarwal 2020 2021 Solutions. All RS Aggarwal 2020 2021 Solutions for class 8 Maths are prepared by experts and are 100% accurate.

#### Page No 198:

#### Question 1:

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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 â€‹$5.2\mathrm{cm}$, cutting the previous arc at *D*.

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

Thus, $ABCD$ is the required quadrilateral.

#### Page No 198:

#### Question 8:

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 â€‹$5.2\mathrm{cm}$, cutting the previous arc at *D*.

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

Thus, $ABCD$ is the required quadrilateral.

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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:

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.

#### 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*.

â€‹then, *ABCD* is the required parallelogram.

#### Page No 201:

#### Question 3:

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*.

â€‹then, *ABCD* is the required parallelogram.

#### 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:

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.

#### 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*

â€‹then, *ABCD *is a required parallelogram.

#### Page No 201:

#### Question 5:

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*

â€‹then, *ABCD *is a required parallelogram.

#### 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:

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.

#### 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:

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.

#### 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*.

â€‹Thus, *ABCD* is the required parallelogram.

#### Page No 201:

#### Question 8:

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*.

â€‹Thus, *ABCD* is the required parallelogram.

#### 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:

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.

#### 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*.

â€‹Thus, *ABCD* is a required square.

#### Page No 201:

#### Question 10:

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*.

â€‹Thus, *ABCD* is a required square.

#### 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:

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.

#### 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*.

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

â€‹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:

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*.

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

â€‹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.

#### 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*.

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

#### Page No 201:

#### Question 13:

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*.

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

#### 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.*

â€‹Step 4: Join *AC* and* BC.*

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

Step 6: â€‹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:

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.*

â€‹Step 4: Join *AC* and* BC.*

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

Step 6: â€‹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.

#### 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:

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.

#### 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:

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.

#### 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*.

â€‹Thus, *ABCD* is the required trapezium.

#### Page No 202:

#### Question 1:

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*.

â€‹Thus, *ABCD* is the required trapezium.

#### 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:

( 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.

#### 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:

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 ${36}^{\circ},{72}^{\circ},{108}^{\circ}\mathrm{and}{144}^{\circ}.$

#### 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:

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

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 ${72}^{\circ}and{108}^{\circ}$.

#### 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:

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

Perimeter of the rectangle =180 $cm$

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

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

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

#### 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:

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.

#### 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:

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.

#### 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:

(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 = ${180}^{\circ}-{37}^{\circ}={143}^{\circ}$

#### 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:

(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 ${360}^{\circ}$.

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

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

#### 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:

(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

#### 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:

(d) *x* = 8

All sides of a square are equal.

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

#### 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:

(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.

#### 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:

(c) 9

Hexagon has six sides.

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

#### 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:

(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.

#### 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:

(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 = $\frac{n(n-3)}{2}$

#### 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:

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

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

#### 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:

(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.

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

#### 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:

(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.

#### 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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