Frank ICSE Solutions for Class 9 Maths – Constructions of Quadrilaterals
Ex No: 20.1
Solution 1:
Solution 2:
Solution 3:
Solution 4:
Solution 5:
Solution 6:
Solution 7:
Solution 8:
Solution 9:
Solution 10:
Solution 11:
Solution 12:
Solution 13:
Solution 14(a):
Steps of construction:
Draw AD = 3.2 cm
Draw ∠XAD = 90°.
With D as centre and radius BD = 5.5 cm, draw an arc to cut AX at point B.
Join BD.
With B as centre and radius 3.2 cm draw an arc and with D as centre and radius = AB, draw another arc to cut the previous arc at C.
Join BC and CD.
Thus, ABCD is the required rectangle.
CD = 4.5 cm
Solution 14(b):
Steps of construction:
Draw BC = 6.2 cm
Through B, draw BP such that ∠B = 90°
From BP, cut BA = 5 cm
With A and C as centres and radii 6.2 cm and 5 cm respectively, draw arcs cutting each other at D.
Join AD and CD.
Thus, ABCD is the required triangle.
Solution 15:
Solution 16:
Solution 17:
Solution 18(a):
Since area of rectangle = 21 cm2
And, length = 4.2 cm
Breadth = Area ÷ Length = 21 ÷ 4.2 = 5 cm
Steps of construction:
Draw BC = 5 cm
Through B, draw BP such that ∠B = 90°
From BP, cut BA = 4.2 cm
With A and C as centres and radii 5 cm and 4.2 cm respectively, draw arcs cutting each other at D.
Join AD and CD.
Thus, ABCD is the required triangle.
Solution 18(b):
Since area of rectangle = 33.8 cm2
And, breadth = 6.5 cm
Length = Area ÷ Breadth = 33.8 ÷ 6.5 = 5.2 cm
Steps of construction:
Draw BC = 6.5 cm
Through B, draw BP such that ∠B = 90°
From BP, cut BA = 5.2 cm
With A and C as centres and radii 6.5 cm and 5.2 cm respectively, draw arcs cutting each other at D.
Join AD and CD.
Thus, ABCD is the required triangle.
Solution 19:
Solution 20:
