Question

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

Answer 1

Answer:

Brainiest me

Step-by-step explanation:

Generally area of a quadrant = 1/4*π*r²,

and circumference of a quadrant = (πr)/2 + 2r,

Where r is a radius,

Assume π = 3.14

Given that circumference of quadrant = 22 cm  

=> πr + 4r = 44 cm,

=> r(π+4) = 44cm

=> r(7.14) = 44

=> r is nearly 6.16 cm, If we can ignore the decimal value r = 6cm,

Remember:  We can find the area of quadrant but we will get a near value,

=> Area = 1/4 * 22/7 * 6 * 6 cm²

=> Area = 28.274 cm² Which is nearly 28 cm² If we ignore the decimal value again,

Here is the answer, It is 28cm² nearly not exactly,

Hope you understand, Have a great day!

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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