Question

Find the area of the polygon given below.

Answer 1

Here, we can see that there are 3 triangles and one trapezium. So, we first begin with the area of ∆𝐴𝐹𝐵.

We know that area of a triangle
= (1/2) × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
= (1/2) × 2 × 3 = 3 $c{m}^2$

Now, area of ∆𝐶𝐻𝐷 = (1/2) × 2 × 4 = 4 $c{m}^2$
Base length of the last triangle ∆𝐴𝐷𝐸 = AF + FH + HD = 2 + 4 + 2 = 8 cm
So, area of ∆𝐴𝐷𝐸 = (1/2) × 8 × 3.5 = 4 × 3. 5 = 14 $c{m}^2$

Now, area of trapezium HFBC = (1/2) × (Sum of parallel sides) × (Distance between them)
⇒ (1/2) × (3 + 4) × 4 = (1/2) × 7 × 4
= 7 × 2
= 14 $c{m}^2$

Therefore, total area will be the sum of all areas = 3 + 4 + 14 + 14 = 35 $c{m}^2$