Question

Find the area of isosceles trapezium ABCD where AB = 10 cm, AD = BC = 20 cm and DC = 42 cm.

Answer 1

Given:

• ABCD is an isosceles trapezium.
• AB = 10 cm   ...(Top base)
• DC = 42 cm   ...(Bottom base)
• AD = BC = 20 cm   ...(Equal non-parallel sides)

Step-wise calculation:

1. Let the trapezium be drawn such that AD and BC are the equal sides.

2. Calculate the height (h) of trapezium using Pythagoras theorem by dropping perpendiculars from A and B to the base DC.

The difference in the bases is 42 – 10 = 32 cm.

Half of this difference on each side is `32/2` = 16 cm.

Imagine D to foot of perpendicular from A as x and similarly from B to C as the same length 16 cm isosceles.

3. Using triangle formed by height, side 20 cm and half difference 16 cm:

`h = sqrt(20^2 - 16^2)`

= `sqrt(400 - 256)`

= `sqrt(144)`

= 12 cm

4. The area (A) of trapezium formula:

`A = 1/2 xx (AB + DC) xx h`

Substitute values:

`A = 1/2 xx (10 + 42) xx 12`

= `1/2 xx 52 xx 12`

= 26 × 12

= 312 cm²

The area of the isosceles trapezium ABCD is 312 cm².