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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
4. Volumes of Prisms and Cylinders
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Exercise 50 Page 838

Use the formula for the area of a square and for the area of a circle.

Ï€/4

Practice makes perfect

Let's analyze the given diagram.

We are asked to find the ratio of the area of the circle to the area of the square. The radius of the circle is r=2x. Now we can find its area.

\textcolor{darkorange}{A_\text{circle}}=\pi r^2
â–¼
Substitute 2x for r and evaluate
Substitute

r= 2x

\textcolor{darkorange}{A_\text{circle}}=\pi ({\color{#0000FF}{2x}})^2
PowProd

(a * b)^m=a^m* b^m

\textcolor{darkorange}{A_\text{circle}}=\pi (2^2)x^2
CalcPowProd

Calculate power and product

\textcolor{darkorange}{A_\text{circle}}=\textcolor{darkorange}{4\pi x^2}

The side length of the square s is twice the radius of the circle. This tells us that s=2(2x)=4x. Now, let's find the area of the square.

\textcolor{darkviolet}{A_\text{square}} = s^2
â–¼
Substitute 4x for s and evaluate
Substitute

s= 4x

\textcolor{darkviolet}{A_\text{square}} = ({\color{#0000FF}{4x}})^2
PowProd

(a * b)^m=a^m* b^m

\textcolor{darkviolet}{A_\text{square}} = 4^2x^2
CalcPow

Calculate power

\textcolor{darkviolet}{A_\text{square}}=\textcolor{darkviolet}{16x^2}

Finally, let's find the ratio.

\boxed{\text{Ratio}}=\dfrac{\textcolor{darkorange}{A_\text{circle}}}{\textcolor{darkviolet}{A_\text{square}}}
â–¼
Substitute values and evaluate
SubstituteII

\textcolor{darkorange}{A_\text{circle}}={\color{#0000FF}{\textcolor{darkorange}{4\pi x^2}}}, \textcolor{darkviolet}{A_\text{square}}={\color{#009600}{\textcolor{darkviolet}{16x^2}}}

Ratio=4Ï€ x^2/16x^2
ReduceFrac

a/b=.a /4x^2./.b /4x^2.

Ratio=Ï€/4

This tells us that the ratio is π4.

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