Question

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 ____

• A. $10\, sq. \,cm$
• B. $30\, sq.\, cm$
• C. $20\, sq.\, cm$
• D. $25\, sq.\, cm$

Answer 1

Answer: (D)

Given that of the wire $P=20\, cm$
Then, $P=$ diameter + arc length
$20 =2 r+ S$
$S =20-2 r$
$S =2(10-r)$...(i)
Also, know that area of semicircle
$A=\frac{1}{2} \pi r^{2}$...(ii)
$\Rightarrow A=\frac{1}{2}(\pi r)(r)$
$\because$ Angle $=\frac{\text { Arc }}{\text { Radius }}$
$\Rightarrow \pi=\frac{S}{r}$
$\Rightarrow S=r \pi$ for straight length of wire
$\Rightarrow A=\frac{1}{2} S \cdot r$...(iii)
From Eq. (i)
$A=\frac{1}{2} \cdot 2(10-r) \cdot r$
$A=10 r-r^{2}$...(iv)
Now, $\frac{d A}{d r}=10-2 r$
For max or min area of enclosed by wire
$\Rightarrow \frac{d A}{d r}=0 \Rightarrow 10-2 r=0$
$\Rightarrow r=5$
Then, from Eq. (iv)
$A=10(5)-(5)^{2}$
$A=50-25$
$A=25\, sq\, cm$