Question

Find the area of the figure given below.

• A. $h\frac{(b+a)}{2}sq.units$
• B. $h(a+b)$sq. units
• C. $2h(a+b)$ sq.units
• D. Can't be determined

Answer 1

The correct option is A

$h\frac{(b+a)}{2}sq.units$

Area of trapezium = Area of rectangle + Area of the two triangles
A $=(base\times height)+2\times \frac{1}{2}\times base\times height$
$=b\times h+2\times \frac{1}{2}\times h\times \left(\frac{a-b}{2}\right)$
$=b\times h+\frac{h(a-b)}{2}$
$=h[b+\frac{a-b}{2}]$
$=h\left[\frac{2b+a-b}{2}\right]$
Area of trapezium $=h\frac{(b+a)}{2}$sq.units