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.

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.

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