Question

In given pentagon area of trapezium BCDE is 2 times the area of $△$ABC. If AF = 3 cm, FG = 4 cm, and ED = 5 cm, then find area of pentagon.

• A. 15 $c{m}^2$
• B. 30 $c{m}^2$
• C. 45 $c{m}^2$
• D. 60 $c{m}^2$

Answer 1

The correct option is C 45 $c{m}^2$

According to given condition
$\frac{\text{area of}△\text{ABC}}{\text{area of trapezium BCDE}}$ = $\frac{1}{2}$
$\frac{\frac{1}{2}\times \text{base}\times \text{height}}{\frac{1}{2}\times \text{(sum of parallel sides)}\times \text{height}}$ = $\frac{1}{2}$
$\frac{\frac{1}{2}\times \text{BC}\times \text{AF}}{\frac{1}{2}\times \text{(BC+ED)}\times \text{FG}}$ = $\frac{1}{2}$
$\frac{\text{BC}\times \text{3}}{\text{(BC+5)}\times \text{4}}$ = $\frac{1}{2}$

3$\times$2$\times$BC = 4$\times$(BC+5)
6BC = 4BC+20
2BC=20
BC = $\frac{20}{2}$ = 10 cm

Area of $△$ABC = $\frac{1}{2}$$\times$BC$\times$AF
=$\frac{1}{2}$$\times$10$\times$3 = 5$\times$3 = 15 $c{m}^2$

area of trapezium BCDE
= $\frac{1}{2}$$\times$(BC+ED)$\times$(FG)
= $\frac{1}{2}$$\times$(10+5)$\times$4 = 15$\times$2 = 30 $c{m}^2$

Area of pentagon = Area of $△$ABC + Area of trapezium BCDE
Area of pentagon = 15+30 = 45 $c{m}^2$