These NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise 4.3

Question 1.

Construct the following quadrilaterals:

(i) Quadrilateral MORE

MO = 6 cm

OR = 4.5 cm

∠M = 60°

∠O = 105°

∠R = 105°

(ii) Quadrilateral PLAN

PL = 4 cm

∠A = 6.5 cm

∠P = 90°

∠A = 110°

∠N = 85°

(iii) Parallelogram HEAR

HE = 5 cm

EA = 6 cm

∠R = 85°

(iv) Rectangle OKAY

OK = 7 cm

KA = 5 cm

Solution:

(i)

Steps of construction:

I. Draw a line segment MO = 6 cm.

II. At M, draw \(\overrightarrow{\mathrm{MX}}\), such that ∠OMX = 60°

III. At O, draw \(\overrightarrow{\mathrm{OY}}\), such that ∠MOY = 105°

IV. From \(\overrightarrow{\mathrm{OY}}\), cut off OR = 4.5 cm.

V. At R, draw RZ, such that ∠ORZ = 105°. Let \(\overrightarrow{\mathrm{RZ}}\) intersects \(\overrightarrow{\mathrm{MX}}\) at E. Thus, MORE is the required quadrilateral.

(ii)

Steps of construction:

I. Draw a line segment PL = 4 cm

II. At L, draw \(\overrightarrow{\mathrm{LX}}\) such that ∠PLX = 75°

III. From \(\overrightarrow{\mathrm{LX}}\), cut off LA = 6.5 cm

IV. At A, draw \(\overrightarrow{\mathrm{AY}}\) such that ∠LAY = 110°

V. At P, draw \(\overrightarrow{\mathrm{PZ}}\) such that ∠LPZ = 90°. Let \(\overrightarrow{\mathrm{PZ}}\) and \(\overrightarrow{\mathrm{AY}}\) intersect at N. Thus, PLAN is required quadrilateral.

(iii)

Steps of Construction:

I. Draw a line segment \(\overrightarrow{\mathrm{HE}}\) = 5 cm

II. At E, draw \(\overrightarrow{\mathrm{EX}}\) such that ∠HEA = 85°

III. From \(\overrightarrow{\mathrm{EX}}\), cut-off EA = 6 cm.

IV. With centre at A and radius = 5 cm, draw an arc.

V. With centre at H and radius = 6 cm, draw an arc such that it intersects the previous arc at R.

VI. Join RA and RH. Thus, HEAR is the required parallelogram.

(iv)

Steps of Construction:

I. Draw a line segment OK = 7 cm.

II. At O, draw \(\overrightarrow{\mathrm{OP}}\) such that ∠KOP = 90°.

III. From \(\overrightarrow{\mathrm{OP}}\), cut-off \(\overrightarrow{\mathrm{OY}}\) = 5 cm.

IV. At K, draw \(\overrightarrow{\mathrm{KQ}}\) such that ∠OKQ = 90°

V. From \(\overrightarrow{\mathrm{KQ}}\), cut-off \(\overrightarrow{\mathrm{KA}}\) = 5 cm.

VI. Join AY. Thus, OKAY is the required rectangle.