#### 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 sum of the areas of two triangles and one rectangle.

#### Solution

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.

Area of trapezium ABCD =(Area of rectangle EFCD)+(Area of triangle AED+Area of triangle BFC)

\[=(10\times 6)+[(\frac{1}{2}\times AE\times ED)+(\frac{1}{2}\times BF\times FC)]\]

\[=60+[(\frac{1}{2}\times AE\times6)+(\frac{1}{2}\times BF\times6)]\]

\[=60+[3AE+3 BF]\]

\[=60+3\times(AE+BF)\]

\[\text{ Here, AE+EF+FB }= 15cm\]

And EF = 10 cm

\[ \therefore AE+10+BF=15\]

Or, AE+BF=15-10=5 cm

Putting this value in the above formula:

\[ {\text{ Area of the trapezium }=60+3\times(5)=60+15=75 cm}^2\]