Question

The area of a quadrant of a circle whose circumference is 22 cm, is a. 11 17 cm2 b. 77 cm2 2 7 C. 77 77 cm2 d. 77 cm2 8​

Answer 1

[1].77cm2 8

[2]77upon 8=cm2

Answer 2

$\large{ \sf \underline{Solution - }}$

Here it is given that,

• Circumference of circle = 22 cm

⟹ We have to the area of quadrant.

➢ We know that, one fourth of a circular disc is called a quadrant and the central angle of a quadrant is 90°

• So, Angle (θ) = 90°

Now,

⟶ Circumference of the circle = 2πr

$\sf \therefore 2\pi r=22$

$\sf \implies r=\dfrac{22}{2\pi}$

$\sf \implies r=\dfrac{22}{2 \times \dfrac{22}{7} }$

$\sf \implies r= \dfrac{7}{2} \: cm$

Now,

• θ = 90° [Quadrant]

$\boxed{\sf{Area\:of\:quadrant=\dfrac{\theta}{360^\circ}\times{\pi}{r}^{2}}}$

$\sf \implies \: \dfrac{90^\circ}{360^\circ}\times{ \dfrac{22}{7} } \times \bigg( {\dfrac{7}{2}} \bigg)^{2}$

$\sf \implies \: \dfrac{1}{4}\times{ \dfrac{22 \times 7}{4} }$

$\sf \implies \: \dfrac{77}{8}\: cm^{2}$

Therefore,

$\rm\implies \boxed{ \sf Area \: of \: quadrant = \dfrac{77}{8} \: cm ^{2} }$