Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 4 Practical Geometry Ex 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

LA = 6.5 cm

∠P = 90°

∠A = 110°

∠N = 85°

(iii) Parallelogram

OKAY

HE = 5 cm

EA = 6 cm

∠R = 85°

(iv) Rectangle

HEAR

OK = 7 cm

KA = 5 cm

Solution:

(i) Steps of constructions:

I. Draw a Line segment MO = 6 cm

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

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

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

V. At R, draw \(\overrightarrow { RZ } \) , such that ∠ORZ = 105°

Let \(\overrightarrow { RZ } \) intersects \(\overrightarrow { MX } \) at E.

Thus, MORE is the required quadrilateral.

(ii) Steps of construction:

I. Draw a line segment AL = 6.5 cm.

II. At A, draw \(\overrightarrow { AX } \) such that ∠LAX = 110°

III. At L, draw \(\overrightarrow { LY } \) such that ∠ALY = 75°.

Note: ∠L = 75° is not given, but we can determine it using angle sum property

∵ Sum of three given angles

= 110° + 90° + 85° = 285°

∴ The fourth angle ∠L = 360° – 285° = 75°

IV. rom \(\overrightarrow { LY } \) , cut-off LP = 4 cm.

V At P, draw \(\overrightarrow { PZ } \) such that ∠LPZ = 90°

Let \(\overrightarrow { PZ } \) and \(\overrightarrow { AX } \) intersect at N.

Thus, PLAN is the required quadrilateral.

(iii) Steps of construction:

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

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

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

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

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. draw \(\overrightarrow { OP } \) such that ∠KOP = 90°.

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

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

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

VI. Join A and Y.

Thus, OKAY is the required rectangle