Question

Find the area of a quadrant of a circle whose circumference is 22 cm. Take PIE= 22/7

Answer 1

circumference
= 2 (pi) r = 22cm
= 2 × 22/7 × r = 22
= 2/7 ×r =1 ( 22 gets cancelled on either side)
so r = 7/2

area of circle = pi r^2
=22/7 × 7/2 ×7/2
=7 ×22/4

area of a quadrant = 1/4 area of a circle
so it is 154/16 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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