Math, asked by blueseas6429, 11 months ago

Find the area of quadrant of a circle whose circumference is 22

Answers

Answered by Aman25130909
3

Answer:

Step-by-step explanation:

1/4×pi×r=circumferences of quadrant

1/4×pi×r=33

1/4×22/7×r=33

r=14cm

Area of quadrant =1/4×pi×r^2

= 1/2×22/7×14×14

=154cm^2

Answered by Anonymous
27

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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