Question

find the area of a quadrant of a circle whose circumference is 44cm

Answer 1

Simply, quadrant of a circle is 1/4 of its area.


Area= πr²
=> Area of quadrant = 1/4πr²

Given circumference = 44
Circumference = 2πr = 44
r = 44/2π

$r = \frac{44}{2 \times \frac{22}{7} } \\ = > r = \frac{44}{2 \times 22} \times 7 \\ = > r = 7$
So r = 7


Now area of quadrant = 1/4 πr²
=
$\frac{1}{4} \times \frac{22}{7} \times 49 \\ = \frac{1}{4} \times 22 \times 7 \\ = \frac{77}{2} \\ = 38.5$
hence area of given quadrant is 38.5cm²



Hope it helps dear friend ☺️✌️✌️

Answer 2

Given,

$\sf{Circumference\:of\:the\:circle\:is\:44\:cm}$

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,

$\large{\sf{C=2πr=44}}$

$\large\sf{⇒R=\frac{44}{2π}\:cm}$

$\large\sf{⇒ R=7\:cm}$

• So,

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

$\sf{= \frac{θ}{360°} ×πr^2}$

Here, θ = 90°

• So,

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

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

$\sf\large{=\frac{1}{4}×π×7×7\:cm^2}$

$\sf\large{=\frac{1}{4}×\frac{22}{7}×7×7\:cm^2}$

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

$\sf\large{=38.5\:\:cm^2}$

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$\large{ \underline{ \overline{ \mid{ \rm{ \red{Answer→38.5\:\:cm^2}} \mid}}}}$

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