Question

Find the approx. numerical value of the area of the above trapezium.

• A. $259.65c{m}^2$
• B. $359.65c{m}^2$
• C. $219.65c{m}^2$
• D. $59.65c{m}^2$

Answer 1

The correct option is A $259.65c{m}^2$

The trapezium can be broken into a parallelogram and a triangle
The sides of the triangle are $10cm.16cm$ and $(32-20)cm$ i.e $10cm,16cm$ and $12cm$
First calculate the area of the triangle with the help of Heron's Formula
$a=10cm,b=16cm,c=12cm$
$S=\frac{(a+b+c)}{2}=\frac{(10+16+12)}{2}=19cm$
Hence area $A=\sqrt{\left(19(19-10)(19-16)(19-12)\right)}=59.93$
Taking the base on the parallel sides i.e $12cm$
Height of the triangle, $H=\frac{2A}{c}=\frac{2\times 59.93}{12}=9.98cm$
Area of the parallelogram $=\text{base}\times \text{height}=209.98=199.6c{m}^2$
Total area of the trapezium $=199.6+59.93=259.53c{m}^2$