Question

Find the number of sides of a polygon if the sum of its interior angles is
(i) ${1440}^{\circ }$
(ii) ${1800}^{\circ }$
(iii) ${540}^{\circ }$

Answer 1

In a polygon $(2n-4)$ right angles $=$ sum of its interior angles

where $n$ is the number of sides of a polygon

$\left(i\right){1440}^o$

$\Rightarrow (2n-4)$ right angles = ${1440}^o$

$\Rightarrow (2n-4)\times 90=1440$

$\Rightarrow 2n-4=16$

$\Rightarrow 2n=20$

$\Rightarrow n=10$ sides

$\left(ii\right){1800}^o$

$\Rightarrow (2n-4)$ right angles = ${1800}^o$

$\Rightarrow (2n-4)\times 90=1800$

$\Rightarrow 2n-4=20$

$\Rightarrow 2n=24$

$\Rightarrow n=12$ sides

$\left(iii\right){540}^o$

$\Rightarrow (2n-4)$ right angles = ${540}^o$

$\Rightarrow (2n-4)\times 90=540$

$\Rightarrow 2n-4=6$

$\Rightarrow 2n=10$

$\Rightarrow n=5$ sides

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