# GSEB Solutions Class 8 Maths Chapter 4 Practical Geometry Ex 4.3

Gujarat BoardĀ 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.
MORE
MO = 6 cm
OR = 4.5 cm
ā M = 60Ā°
ā O = 105Ā°
ā R = 105Ā°

PLAN
PL = 4 cm
LA = 6.5 cm
ā P = 90Ā°
ā A = 110Ā°
ā N = 85Ā°

(iii) Parallelogram
OKAY
HE = 5 cm
EA = 6 cm
ā R = 85Ā°

(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 ā OMX = 60Ā°
III. At O, draw $$\overrightarrow { OY }$$ , such that ā MOY = 105Ā°
IV. From $$\overrightarrow { OY }$$ , cut off OR = 4.5 cm
V. At R, draw $$\overrightarrow { RZ }$$ , such that ā ORZ = 105Ā°
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 ā LAX = 110Ā°
III. At L, draw $$\overrightarrow { LY }$$ such that ā ALY = 75Ā°.
Note: ā L = 75Ā° is not given, but we can determine it using angle sum property
āµ Sum of three given angles
= 110Ā° + 90Ā° + 85Ā° = 285Ā°
ā“ The fourth angle ā L = 360Ā° – 285Ā° = 75Ā°

IV. rom $$\overrightarrow { LY }$$ , cut-off LP = 4 cm.
V At P, draw $$\overrightarrow { PZ }$$ such that ā LPZ = 90Ā°
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 ā HEA = 85Ā°.
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 ā KOP = 90Ā°.
III. From $$\overrightarrow { OP }$$ , cut-off $$\overrightarrow { OY }$$ = 5 cm.
IV. At K, draw $$\overrightarrow { KQ }$$ such that ā OKQ = 90Ā°.
V. From $$\overrightarrow { KQ }$$ cut-off KA = 5 cm.
VI. Join A and Y.
Thus, OKAY is the required rectangle