Question

- Find the area of a quadrant of a circle whose circumference is 22 cm.​

Answer 1

Answer:

190.2 cm^2 (approx)

Explanation:

1. Circumference of circle = 2πr = 22cm.
2. from 1. we can find r .

r = 22/2π

r = 22 × 7 / 2×22

r = 3.5 cm

Now,

Area of a quadrant = πr^2 /4

= πr×r/2

= 22 × 22× 22 / 2× 7

= 11×11×11/7

= 1331/7

= 190.142 cm^2

= 190.2 cm^2 (approx)

Answer 2

Given,

• $\sf{Circumference\:of\:the\:circle\:is\:22\:cm}$

To find,

• $\sf{Area\:of\:quadrant}$

Solution,

It should be noted that a quadrant of a circle is a sector which is making an angle of 90°

• $\sf{Let\:the\:radios\:of\:the\:circle\:be\:r}$

• As,

$\large{\sf{C=2πr=22}}$

$\large\sf{⇒R=\frac{22}{2π}\:cm}$

$\large\sf{⇒ R=\frac{7}{2}\:cm}$

• So,

$\bf{Area\:of\:the\:quadrant,}$

$\sf{= \frac{θ}{360°} ×πr^2}$

• Here, θ = 90°

• So,

$\sf\large{A=\frac{90°}{360°}×πr^2\:cm^2}$

$\sf\large{=\frac{1}{4}×π (\frac{7}{2})^2\:cm^2}$

$\sf\large{=\frac{1}{4}×π(\frac{49}{4}\:cm^2}$

$\sf\large{=\frac{49}{16}π\:cm^2}$

$\sf\large{=\frac{49}{16}×\frac{22}{7}\:cm^2}$

$\sf\large{=\frac{77}{8}\:cm^2}$

$\sf\large{=9.6\:\:cm^2}$

____________________________

$\large{ \underline{ \overline{ \mid{ \rm{ \red{Answer→9.6\:\:cm^2}} \mid}}}}$

____________________________