Question

In the trapezium ABCD, AD || BC, AB = 25 cm, BC = 76 cm, CD = 39 cm, AD = 20 cm, find its area.

Answer 1

We are given a trapezium ABCD with:

• AD || BC → AD and BC are the parallel sides
• AB = 25 cm
• BC = 76 cm
• CD = 39 cm
• AD = 20 cm

We need to find the area of the trapezium.

Step-by-step:

Use the area formula for a trapezium:

Area = `1/2`× sum of parallel × height

We know:

Parallel sides: AD = 20 cm, BC = 76 cm

We must now find the height of the trapezium.

Step 1: Use Pythagoras theorem in triangle ΔABX and ΔCDY

Let perpendiculars from A and D fall on BC at points X and Y, forming height h.

So, total base BC = BX + XY + YC = 76

Let:

• BX = x
• YC = y
• So, x + y + 20 = 76
  ⇒ x + y = 56

Use right triangles:

1. In ΔABX:

AB² = h² + x²

⇒ 25² = h² + x²   ...(1)

2. In ΔCDY:

CD² = h² + y²

⇒ 39² = h² + y²   ...(2)

Now subtract (1) from (2):

39² – 25² = y² – x²

= (y + x)(y – x)

⇒ (y + x)(y – x) = 1521 – 625

⇒ (y + x)(y – x) = 896

⇒ 56(y – x) = 896

⇒ `y - x = 896/56`

⇒ y – x = 16

 Now solve:

• x + y = 56
• y – x = 16

Add both:

2y = 72

⇒ y = 36, x = 20

Now plug into (1):

25² = h² + 20²

⇒ 625 = h² + 400

⇒ h² = 225

⇒ h = 15 cm

Step 2: Area calculation

Area = `1/2` × (AD + BC) × h

= `1/2 xx (20 + 76) xx 15`

= `(96 xx 15)/2`

= 720 cm²