RD Sharma Solutions Class 8 Practical Geometry Exercise 18.5

RD Sharma Solutions Class 8 Chapter 18 Exercise 18.5

RD Sharma Class 8 Solutions Chapter 18 Ex 18.5 PDF Free Download

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.

The quadrilateral so obtained is the required quadrilateral.

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.

The quadrilateral so obtained is the required quadrilateral.

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.

The quadrilateral so obtained is the required quadrilateral.

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.

The quadrilateral so obtained is the required quadrilateral.

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.

The quadrilateral so obtained is the required quadrilateral.

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.

The quadrilateral so obtained is the required quadrilateral.