Question

1. Find the area de a quadrant of a circle whose circumference is 22 cm
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Answer 1

Answer:

Given:

• circumference of circle is 22cm

To find :

•    the area of quadrant

Formula:

Area of circle = πr²

Area of quadrant = 1/4 *πr²

Circumference = 2πr

Solving Question:

  To find the area of quadrant ,we need radius ;we have to find the value of 'r' then substitute in the equation.

Solution:

Circumference = 22

⇒ 2*$\dfrac{22}{7}$ *r = 22

⇒$\dfrac{2r}{7}=1\\\\\implies r=\dfrac{7}{2}=3.5$

$Area\:of\:quadrant=\dfrac{1}{4}\pi r^2\\\\\implies \dfrac{1*22*(3.5)^2}{4*7}\\\\\implies \dfrac{1*22*12.25}{4*7}\\\\\implies \dfrac{269.5}{28}\\\\\implies\boxed{ 9.625cm^2}$

Thus, the area of quadrant is 9.625cm²

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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