#### 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.

#### 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.

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:

\[ {\text{ Area of the trapezium }=90-3(5)=90-15=75 cm}^2\]

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:

\[ {\text{ Area of the trapezium }=90-3(5)=90-15=75 cm}^2\]

Is there an error in this question or solution?

Solution 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 Concept: Area of Trapezium.