Question

If the area and the perimeter of a sector of a circle are $12{\text{cm}}^2$ and $14\text{cm}$ respectively, then possible values of:

• A. $\text{radius of the circle = 6 cm, length of the arc = 8 cm}$
• B. $\text{radius of the circle = 3 cm, length of the arc = 4 cm}$
• C. $\text{radius of the circle = 4 cm, length of the arc = 6 cm}$
• D. $\text{radius of the circle = 3 cm, length of the arc = 8 cm}$

Answer 1

Answer: (C), (D)

$\text{radius of the circle = 3 cm, length of the arc = 8 cm}$

We know that,
$\text{Area of the sector}\left(A\right)=\frac{l\times r}{2}$
$\text{Perimeter of the sector}=2r+l,$
where $r$ is the radius of the circle and $l$ is the length of the arc of sector.

Given, area of the sector is $12{\text{cm}}^2$ and perimeter of the sector is $14\text{cm}.$
$\therefore 12=\frac{l\times r}{2}\dots \left(i\right)$
and $14=2r+l$
$\Rightarrow 14-2r=l$
Substituting $l=14-2r$ in equation $\left(i\right)$
$12=\frac{(14-2r)\times r}{2}$
Taking 2 common from $(14-2r)$
$12=\frac{2(7-r)\times r}{2}$
$12=(7-r)\times r$
Simplifying right hand side of the equation
$12=7\times r-r\times r$
$12=7r-r^2$
$r^2-7r+12=0$
$r^2-3r-4r+12=0(\because 7r=3r+4r)$
$r(r-3)-4(r-3)=0$
Taking $(r-3)$ common
$(r-3)(r-4)=0$
$(r-3)=0\Rightarrow r=3$
$\text{or}(r-4)=0\Rightarrow r=4$

Substituting the values of $r$ in equation $\left(i\right)$

$\text{Case 1: If}r=3\text{, then}$
$12=\frac{l\times 3}{2}$
Multiply both sides by 2
$\Rightarrow 12\times 2=\frac{l\times 3}{2}\times 2$
$\Rightarrow 24=l\times 3$
Divide both sides by 3
$\Rightarrow \frac{24}{3}=\frac{l\times 3}{3}$
$\Rightarrow 8=l$
$\Rightarrow l=8$

$\text{Case 1: If}r=4\text{, then}$
$12=\frac{l\times 4}{2}$
Multiply both sides by 2
$\Rightarrow 12\times 2=\frac{l\times 4}{2}\times 2$
$\Rightarrow 24=l\times 4$
Divide both sides by 4
$\Rightarrow \frac{24}{4}=\frac{l\times 4}{4}$
$\Rightarrow 6=l$
$\Rightarrow l=6$

Hence, possible values of radius of the circle and length of the arc are: $r=3\text{cm},l=8\text{cm}\text{or}r=4\text{cm},l=6\text{cm}.$
Options (c.) and (d.) are correct choices.