Question

Solve this :-

If the area of a semicircle be 77sq.cm then find its perimeter.​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

$\bold{\boxed{Area \: of \: Semicircle = \frac{1}{2} \pi {r}^{2}}}$

$= > \frac{1}{2} \pi {r}^{2} = 77$

$= > \pi {r}^{2} = 154$

$= > \frac{22}{7} \times {r}^{2} = 154$

$= > {r}^{2} = 154 \times \frac{7}{22} = 7 \times 7 = 49$

$= > {r}^{2} = 49$

$= > r = \sqrt{49} = 7$

Therefore ,radius is 7cm

Now finding perimeter of semi circle:-

$\bold{\boxed{Perimeter \: of \: Semicircle = \pi \: r + 2r}}$

$= \frac{22}{7} \times 7 + 2 \times 7$

$= > 22 + 14 = 36cm$

Therefore, Perimeter of semi-circle is 36cm

Answer 2

$\underline{\underline{\sf{ \color{purple}{\qquad To\:Find\:: \qquad} }}}$

• Perimeter of the circle

$\underline{\underline{\sf{ \color{purple}{\qquad Solution \:: \qquad} }}}$

Given that,

• Area of the semicircle = 77 cm²

$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\purple{Area_{\;(Circle)} = 2 \times Area~ of ~semicircle }}}}$

$\qquad \quad {:} \implies \sf{Area \: of \: circle = \bigg(77 \times 2 \bigg) {cm}^{2} }$

$\qquad \quad {:} \implies \sf{Area \: of \: circle =154 \: {cm}^{2} }$

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Now, find the radius ( r) of the circle :-

$\dag\;{\underline{\frak{We\;know\;that,}}}$

$\star\;{\boxed{\sf{\purple {Area_{\;(Circle)} = \pi \times r² }}}}$

$\qquad \quad {:} \implies \sf{ 154 = \dfrac{22}{7} \times {r}^{2} }$

$\qquad \quad {:} \implies \sf{ 7 \times 7 = {r}^{2} }$

$\qquad \quad {:} \implies \sf{ 49 = {r}^{2} }$

$\qquad \quad {:} \implies \sf{ {r}^{2} = 49 }$

$\qquad \quad {:} \implies \sf{ {r} = \sqrt{ 49} }$

$\qquad \quad {:} \implies \sf{ {r} = { 7} }$

______

$\qquad \quad {:} \implies \sf{ Perimeter = \pi \times r + 2 \times r}$

$\qquad \quad {:} \implies \sf{ Perimeter = 7 \times \dfrac{22}{7} + 2\times 7 }$

$\qquad \quad {:} \implies \sf{ Perimeter = 36 }$

$\sf \: ∴ \: Hence, \: perimeter \: of \: the \: circle \: is \: 36cm$

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