**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 cm^{2}

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 cm^{2}

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:**