Question

The parallel sides of a Trapezium are 52 cm and 27 cm and the other two sides are 25 cm and 30 cm find the area of the trapezium

Answer 1

See attachment

The trapezium can be split into a parallelogram and a triangle.

Find the perimeter of ΔBCE:

Perimeter = 25 + 25 + 30 = 80 cm

Find the semiperimeter of ΔBCE:

SemiPerimeter = 80 ÷ 2 = 40 cm

Find the Area of ΔBCE:

Area = √p(p - a)(p - b)(p - c)

Area = √40(40 - 25)(40 - 25)(40 - 30) = √90000 = 300 cm²

Find the height of the trapezium:

Area of  ΔBCE = 1/2 x base x height

300 = 1/2 x 25 x height

300 = 12.5 x height

Height = 300 ÷ 12.5 = 24 cm

Find the area of the trapezium:

Area = 1/2 (a + b) h

Area = 1/2 ( 52 + 27 ) x 24

Area = 948 cm²

Answer: The area of the trapezium is 948 cm²

Answer 2

Area of a trapezium = (a + b ) / 2 h, where a and b are the parallel sides of the trapezium and h is the height.

Since height is not known here, the line parallel to one dissimilar side is drawn to create a triangle inside the trapezium.

The area of that triangle = square root of s(s-a)(s-b)(s-c), where a is 52 – 27 = 25, b is 25 and c is 30.

Thus, s = (a + b + c) / 2 = 40. Thus, area of triangle = square root of 40 x 15 x 15 x 10 = 300 cm2.

Normally, area of triangle = ½ x base x height. So here, 300 = ½ x 25 x h. Thus, h = 24 cm.

So the area of trapezium = [(52 + 27) / 2] x 24 = 948 cm2.