Question

Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as the difference of the area of a rectangle and the sum of the areas of two triangles.

Answer 1

Given:
Length of the parallel sides of a trapezium are 10 cm and 15 cm.
The distance between them is 6 cm.
Let us extend the smaller side and then draw perpendiculars from the ends of both sides.

In this case, the figure will look as follows:

Area of trapezium ABCD=(Area of rectangle ABGH)-[(Area of triangle AHD)+(Area of triangle BGC)]
\[ = (15 \times 6) - [(\frac{1}{2} \times DH \times 6) + (\frac{1}{2} \times GC \times 6)]\]
\[ = 90 - [3\times DH + 3 \times GC]\]
 = 90 - 3[DH+GC]
Here, HD+DC+CG=15 cm
DC=10 cm
HD+10+CG=15
HD+GC=15-10=5 cm
Putting this value in the above equation:
\[ {\text{ Area of the trapezium }=90-3(5)=90-15=75 cm}^2\]