Question

The area of sector is 934sq.cm and radius is 42cm then the perimeter of the sector is..?

Answer 1

Given :-

• Area of sector = 934cm² .
• Radius = 42cm .

To Find :-

• Perimeter of Sector ?

Formula used :-

• Area of sector = (Angle at Centre/360°) * π * (radius)²
• Perimeter of sector = 2r + Length of arc Made by angle = 2r + (Angle at centre/360°) * 2 * π * r

Solution :- ( Refer to Image First).

Let Angle at centre is θ ..

Comparing Given Area of Sector we get,

→ (θ/360°) * (22/7) * (42)² = 934

→ (θ/360°) * 22 * 42 * 6 = 934

→ (θ/360°) * 11 * 42 * 6 = 467

→ (θ/360°) = (467) / ( 66 * 42) ---------- Equation (1).

_________________

So,

→ Perimeter of Sector = 2r + (θ/360°) * 2πr

Putting values we get :-

→ P = 2*42 + [(θ/360°) * 2 * (22/7) * 42 ]

Putting Value From Equation (1) Now, we get,

→ P = 84 + [ (467/66*42) * (44 *42)/7 ]

→ P = 84 + [ ( 467/66) * (44/7) ]

→ P = 84 + [(467/3) * (2/7) ]

→ P = 84 + [ (467 * 2) / ( 3 * 7) ]

→ P = 84 + (934/21)

→ P = 84 + 44.47

→ P = 128.47cm. (Ans).

Hence, Perimeter of sector will be 128.47cm.

Answer 2

Let r be the radius of the sector and $\theta$ be the central angle of the sector in radian.

We know the area of the sector is,

$\longrightarrow\sf{A=\pi r^2\cdot\dfrac{\theta}{2\pi}}$

$\longrightarrow\sf{A=\dfrac{r^2\theta}{2}}$

But the perimeter of the sector is,

$\longrightarrow\sf{P=2\pi r\cdot\dfrac{\theta}{2\pi}+2r}$

$\longrightarrow\sf{P=r\theta+2r}$

$\longrightarrow\sf{P=r(\theta+2)}$

Given,

$\longrightarrow\sf{A=934\ cm^2}$

$\longrightarrow\sf{\dfrac{r^2\theta}{2}=934}$

$\longrightarrow\sf{\theta=\dfrac{934\times2}{r^2}}$

$\longrightarrow\sf{\theta=\dfrac{934\times2}{42^2}}$

$\longrightarrow\sf{\theta=\dfrac{467}{441}\ rad}$

Then perimeter,

$\longrightarrow\sf{P=r(\theta+2)}$

$\longrightarrow\sf{P=42\left(\dfrac{467}{441}+2\right)}$

$\longrightarrow\sf{P=42\times\dfrac{1349}{441}}$

$\longrightarrow\sf{\underline{\underline{P=128.48\ cm}}}$