A wire of length 20 cm is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is: 

Length of the wire =20 cm
Therefore, Length of wire =2 Radius + Arc length 
i.e. 2 r+l=20 cm
Therefore, $I=2(10-r)=\frac{1}{2}\pi r^2$
Area of Sector $\text{ (A) }=\frac{\theta }{360}\times \pi r^2$
As $\theta =\frac{l}{r}$
$\begin{array}{l}\Rightarrow A=\frac{1}{360}\times \pi r=\frac{20-2r}{360}\times \pi r=\frac{10\pi r-\pi r^2}{180}\\ \Rightarrow \frac{dA}{dr}=\frac{10\pi -2\pi r}{180}=\frac{5\pi -\pi r}{90}\end{array}$
Taking $\frac{dA}{dr}=0\text{, we get, }r=5$
Therefore, $I=20-10=10$
Length of an arc $=\frac{\theta }{360}\times 2\pi r$
$\begin{array}{l}\Rightarrow 10=\frac{\theta }{360}\times 10\pi \\ \Rightarrow \theta =\frac{360}{\pi }\\ \Rightarrow A=\frac{360}{\pi \times 360}\times \pi \times 5^2\\ \Rightarrow A=25\text{ sq cm }\end{array}$