Question

find the area of the following polygons

Answer 1

area of this polygon= area of the triangle EFD+area of the triangle DFC+area of triangle CFB+ area of the triangle BFA

=1/2*5*2+1/2*4*7+1/2*4*7+1/2*6.5*2
=5+14+14+6.5
=39.5 cm^2

Answer 2

As we can see, the polygon ABCDEF is made from 4 triangles i.e.  $\triangle ABF \text{,} \triangle BFC \text{,} \triangle FDC \text{ and } \triangle FDE$.

$\text{Area of a triangle =}$ $\frac{1}{2} \times base \times height$

As per the attached figure:

For $\triangle ABF$, $\text {FB}$ is the base and $\text {AG}$ is the height.

For $\triangle BFC$, $\text {FC}$ is the base and $\text {HB}$ is the height.

For $\triangle FDC$, $\text {FC}$ is the base and $\text {DI}$ is the height.

For $\triangle FDE$, $\text {FD}$ is the base and $\text {EJ}$ is the height.

$\text {The required area = }(\frac{1}{2} \times 6.5 \times 2) + (\frac{1}{2} \times 7 \times 4) + (\frac{1}{2} \times 7 \times 4) + (\frac{1}{2} \times 5 \times 2) \\\Rightarrow 6.5 + 14 + 14 +5\\\Rightarrow 39.5 \text{ cm}^{2}$

Hence the area of polygon is $39.5\text{ cm}^{2}$.