Question

In the given figure, the side of the square is $10cm$. $EF=2.5cm$ and C and D are half way between the top and bottom sides of the figure. The area of the shaded portion of the figure is:

• A. $43.75c{m}^2$
• B. $56.25c{m}^2$
• C. $55.25c{m}^2$
• D. $50.25c{m}^2$

Answer 1

The correct option is A $43.75c{m}^2$

We know that area of rectangle $=length\times breadth$

& area of trapezium $=\frac{1}{2}\times sumofparallelsides\times height.$

From the given figure, we have

$AB=10cm$
$CD=EF=2.5cm$
$EC=FD=5cm$
Trapezium ABCD has height $=10-5=5cm$
Area of the shaded part $=$ Area of rectangle $CEFD+$ Area of trapezium $ABCD$

$=2.5\times 5+\frac{1}{2}\times (2.5+10)\times 5=12.5+31.25=43.75{cm}^2$