Circumference of a circle is 40 π cm. A quarter of the circle represents the developer layer genuine coupe. Calculate the surface compartment.

We have to find the radius, how can we got it?

\( C=2\pi*r \)

then

\( 40\pi=2\pi*r \)

\( \boxed{r=20~cm} \)

A quarter of the circle will have a surface of a quarter of the circle, there fore

this is the surface of the circle

\( S=\pi*r^2 \)

a quarter of it

\( S_{\frac{1}{4}}=\frac{1}{4}*\pi*r^2 \)

then we can replace the value of r

\( S_{\frac{1}{4}}=\frac{1}{4}*\pi*20^2 \)

\( S_{\frac{1}{4}}=\frac{1}{4}*\pi*400 \)

we can simplify 

\( \boxed{\boxed{S_{\frac{1}{4}}=100\pi~cm^2}} \)

\( 2 \pi r=40 \pi \ cm\ \ \ \Rightarrow\ \ \ r= \frac{40 \pi }{2 \pi } =20\\\\S= \pi r^2\ \ \ \Rightarrow\ \ \ quarter:\ \ A= \frac{1}{4} \cdot S= \frac{1}{4} \cdot \pi \cdot20\cdot20=100 \pi \ [cm^2] \)

<3

Perimeter of the base coupe is 12π
cm and 10 cm is generating. Area buy and surface axial section? Can u pls help me?


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