Question

If BC = 8 cm, ED = 6 cm, AF = 3 cm and AG = 7 cm, then find the area of pentagon ABEDC.

• A. 12 $c{m}^2$
• B. 28 $c{m}^2$
• C. 40 $c{m}^2$
• D. 56 $c{m}^2$

Answer 1

The correct option is C 40 $c{m}^2$

Area of pentagon = Area of $△$ABC + Area of trapezium BCDE

Area of $△$ABC = $\frac{1}{2}$$\times$base$\times$height
= $\frac{1}{2}$$\times$BC$\times$AF = $\frac{1}{2}$$\times$8$\times$3
= 4$\times$3 = 12 $c{m}^2$

Area of trapezium BCDE
= $\frac{1}{2}$$\times$(Sum of parallel sides)$\times$height
= $\frac{1}{2}$$\times$(BC+ED)$\times$(FG)
​​​​​​​ = $\frac{1}{2}$$\times$(8+6)$\times$(AG-AF)
= $\frac{1}{2}$$\times$(8+6)$\times$(7-3)
= $\frac{1}{2}$$\times$14$\times$4
= 7$\times$4 = 28 $c{m}^2$

Area of pentagon = 12+28 = 40 $c{m}^2$