We have

Total length = r + r + r$$\theta $$ = 20

$$ \Rightarrow $$ 2r + r$$\theta $$ = 20

$$ \Rightarrow $$ $$\theta = {{20 - 2r} \over r}$$ .......(1)

A = Area = $${\theta \over {2\pi }} \times \pi {r^2}$$ = $${1 \over 2}{r^2}\theta $$ = $${1 \over 2}{r^2}\left( {{{20 - 2r} \over r}} \right)$$

$$ \Rightarrow $$ A = 10r – r²

For A to be maximum

$${{dA} \over {dr}} = 0$$

$$ \Rightarrow $$ 10 – 2r = 0

$$ \Rightarrow $$ r = 5

$${{{d^2}A} \over {d{r^2}}} = - 2 < 0$$

$$ \therefore $$ For r = 5, A is maximum From (1)

$${\theta = {{20 - 2\left( 5 \right)} \over r}}$$ = 2

A = $${2 \over {2\pi }} \times \pi {\left( 5 \right)^2}$$ = 25 sq. m