# RD Sharma Solutions Class 8 Practical Geometry Exercise 18.5

## RD Sharma Solutions Class 8 Chapter 18 Exercise 18.5

#### Exercise 18.5

1. Construct a quadrilateral ABCD given that AB = 4 cm, BC = 3 cm, $\angle A=75°\;$, $\angle B=80°\;$ and $\angle C=120°\;$.

Steps of construction:

Step I: Draw AB = 4 cm.

Step II: Construct $\angle XAB=75°\;$ at A and $\angle ABY=80°\;$ at B.

Step III: With B as the center and radius 3 cm, cut off BC = 3 cm.

Step IV: At C, draw $\angle BCZ=120°\;$ such that it meets AX at D.

2. Construct a quadrilateral ABCD, where AB = 5.5 cm, BC = 3.7 cm, $\angle A=60°\;$, $\angle B=105°\;$ and $\angle D=90°\;$.

We know that the sum of all the angles in a quadrilateral is 360.

i.e. $\angle A+\angle B+\angle C+\angle D=360°\;$

$\angle C=105°\;$

Steps of construction:

Step I: Draw AB = 5.5 cm.

Step II: Construct $\angle XAB=60°\;$ at A and $\angle ABY=105°\;$.

Step III: With B as the center and radius 3.7 cm, cut off BC = 3.7 cm

Step IV: At C, draw $\angle BCZ=105°\;$ such that it meets AX at D.

3. Construct a quadrilateral PQRS, where PQ = 3.5 cm, QR = 6.5 cm, $\angle P=\angle R=105°\;$ and $\angle S=75°\;$.

We know that the sum of all the angles in a quadrilateral is 360.

i.e., $\angle P+\angle Q+\angle R+\angle S=360°\;$

$\angle Q=75°\;$

Steps of construction:

Step I: Draw PQ = 3.5 cm.

Step II: Construct $\angle XPQ=75°\;$ at P and $\angle PQY=75°\;$ at Q.

Step III: With Q as the center and radius 6.5 cm, cut off QR = 6.5

Step IV: At R, draw $\angle QRZ=105°\;$ such that it meets PX at S.

4. Construct a quadrilateral ABCD when BC = 5.5 cm, CD = 4.1 cm, $\angle A=70°\;$, $\angle B=110°\;$ and $\angle D=85°\;$.

We know that the sum of all the angles in a quadrilateral is 360. i.e.

i.e. $\angle A+\angle B+\angle C+\angle D=360°\;$

$\angle C=95°\;$

Steps of construction:

Step I: Draw BC = 5.5 cm.

Step II: Construct $\angle XBC=110°\;$ at A and $\angle BCY=95°\;$.

Step III: With C as the center and radius 4.1 cm, cut off CD = 4.1 cm.

Step IV: At D, draw $\angle CDZ=85°\;$ such that it meets BY at A.

5. Construct a quadrilateral ABCD, where $\angle A=65°\;$, $\angle B=105°\;$, $\angle C=75°\;$, BC = 5.7 cm and CD = 6.8 cm.

We know that the sum of all the angles in a quadrilateral is 360. i.e

i.e. $\angle A+\angle B+\angle C+\angle D=360°\;$

$\angle D=115°\;$

Steps of Construction:

Step I: Draw BC = 5.7 cm.

Step II: Construct $\angle XBC=105°\;$ at B and $\angle BCY=105°\;$ at C.

Step III: With C as the center and radius 6.8 cm, cut off CD = 6.8 cm.

Step IV: At D, draw $\angle CDZ=115°\;$ such that it meets BY at A.

6. Construct a quadrilateral PQRS, in which PQ = 4 cm, QR = 5 cm, $\angle P=50°\;$, $\angle Q=110°\;$ and $\angle R=70°\;$.

Steps of construction:

Step I: Draw PQ = 4 cm.

Step II: Construct $\angle XPQ=50°\;$ at P and $\angle PQY=110°\;$ at Q.

Step III: With Q as the center and radius 5 cm, cut off QR = 5 cm.

Step IV: At R, draw $\angle QRZ=70°\;$ such that it meets PX at S.