Question

find the area of a quadrant of a circle whose circumference is equal to 22 CM​

Answer 1

Answer:

77/8 or 9.625

Step-by-step explanation:

circumference = 2*22/7* r

22= 2*22/7* r

r = 7/2

now .. quadrant os 1/4 of a circle

area of quadrant = 1/4*22/7*(7/2)^2

therefore the answer is 77/8 cm^2

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,

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

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

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

$\sf\large{=\frac{49}{16}π\:cm^2}$

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

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

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

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

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