Question

Calculate the area of a trapezium, the sides of which, taken in order, are 50, 17, 25 and 12 cm, respectively and the first being parallel to the third.

Answer 1

Given:

Trapezium with sides in order: 50 cm, 17 cm, 25 cm, 12 cm.

The first side 50 cm is parallel to the third 25 cm) so bases a = 50 cm and b = 25 cm; legs are 17 cm and 12 cm.

Step-wise calculation:

1. Let the height be h. Drop perpendiculars from the ends of the shorter base to the longer base; let the horizontal projections of the two legs onto the longer base be x and y.

So, x + 25 + y = 50

⇒ x + y = 25

2. By Pythagoras on the two right triangles:

17² = x² + h² 

⇒ 289 = x² + h²

12² = y² + h²

⇒ 144 = y² + h²

3. Subtract the second from the first:

289 – 144 = x² – y²

= (x – y)(x + y)

145 = (x – y) × 25

⇒ x – y = `145/25`

= 5.8

4. Solve for x and y:

`x = ((x + y) + (x - y))/2` 

= `(25 + 5.8)/2`

= 15.4 cm

y = 25 – x

= 9.6 cm

5. Find h from 17² = x² + h²:

h² = 289 – (15.4)²

= 289 – 237.16

= 51.84

h = `sqrt(51.84)`

= 7.2 cm

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

= `1/2 xx (50 + 25) xx 7.2`

= `1/2 xx 75 xx 7.2`

= 37.5 × 7.2

= 270

Area = 270 cm².