# 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.

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

(i) the sum of the areas of two triangles and one rectangle.

(ii) 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 $10 \mathrm{~cm}$ and $15 \mathrm{~cm}$.

The distance between them is $6 \mathrm{~cm}$.

Let us extend the smaller side and then draw perpendiculars from the ends of both sides.

(i) Area of trapezium $\mathrm{ABCD}=($ Area of rectangle EFCD $)+($ Area of triangle $\mathrm{AED}+$ Area of triangle $\mathrm{BFC})$

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

$=60+\left[\left(\frac{1}{2} \times \mathrm{AE} \times 6\right)+\left(\frac{1}{2} \times \mathrm{BF} \times 6\right)\right]$

$=60+[3 \mathrm{AE}+3 \mathrm{BF}]$

$=60+3 \times(\mathrm{AE}+\mathrm{BF})$

Here, $\mathrm{AE}+\mathrm{EF}+\mathrm{FB}=15 \mathrm{~cm}$

And $\mathrm{EF}=10 \mathrm{~cm}$

$\therefore \mathrm{AE}+10+\mathrm{BF}=15$

Or, $A E+B F=15-10=5 \mathrm{~cm}$

Putting this value in the above formula:

Here, $\mathrm{AE}+\mathrm{EF}+\mathrm{FB}=15 \mathrm{~cm}$

And $\mathrm{EF}=10 \mathrm{~cm}$

$\therefore \mathrm{AE}+10+\mathrm{BF}=15$

Or, $A E+B F=15-10=5 \mathrm{~cm}$

Putting this value in the above formula:

Area of the trapezium $=60+3 \times(5)=60+15=75 \mathrm{~cm}^{2}$

(ii) In this case, the figure will look as follows:

Area of trapezium $\mathrm{ABCD}=($ Area of rectangle $\mathrm{ABGH})-[($ Area of triangle $\mathrm{AHD})+($ Area of triangle BGC $)$

$=(15 \times 6)-\left[\left(\frac{1}{2} \times \mathrm{DH} \times 6\right)+\left(\frac{1}{2} \times \mathrm{GC} \times 6\right)\right]$

$=90-[3 \times \mathrm{DH}+3 \times \mathrm{GC}]$

$=90-3[\mathrm{DH}+\mathrm{GC}]$

Here, $\mathrm{HD}+\mathrm{DC}+\mathrm{CG}=15 \mathrm{~cm}$

$\mathrm{DC}=10 \mathrm{~cm}$

$\mathrm{HD}+10+\mathrm{CG}=15$

$\mathrm{HD}+\mathrm{GC}=15-10=5 \mathrm{~cm}$

Putting this value in the above equation:

Area of the trapezium $=90-3(5)=90-15=75 \mathrm{~cm}^{2}$