Question

The area of the hexagon, if AB = BC = CD = DE = EF = FA = 13cm and AO = PD = 5cm, is ___ $c{m}^2$

Answer 1

In the given figure, ∆AOB is right-angled triangle right angled at O.

So,

$\Rightarrow$ $A{O}^2$ + $O{B}^2$ = $A{B}^2$.

$\Rightarrow$ $5^2$ + $O{B}^2$ = ${13}^2$.

$\Rightarrow$ $O{B}^2$ = 169 - 25 = 144

$\Rightarrow$ OB = 12cm

So now the total area of the hexagon = 4$\times$ Area of ∆AOB+ 2$\times$ Area of rectangle BCPO

$\Rightarrow$ Area = 4 $\times$ ($\frac{1}{2}$ $\times$ OB $\times$ AO) + 2 $\times$ (BC $\times$ OB)

$\Rightarrow$ Area = 4 $\times$ ($\frac{1}{2}$ $\times$ 12 $\times$ 5) + 2 $\times$ (13 $\times$ 12)

$\Rightarrow$ Area = 4 $\times$ (30) + 2 $\times$ (156)

$\Rightarrow$ Area = 432$c{m}^2$.