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2006 AMC 10B Problems/Problem 8: Difference between revisions

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== Solution ==
== Solution ==
Since the area of the square is <math>40</math> the length of the side is <math>\sqrt{40}=2\sqrt{10}</math>
Since the area of the square is <math>40</math>, the length of the side is <math>\sqrt{40}=2\sqrt{10}</math>.
The distance between the center of the semicircle and one of the bottom vertecies of the square is half the length of the side which is <math>\sqrt{10}</math>
The distance between the center of the semicircle and one of the bottom vertecies of the square is half the length of the side, which is <math>\sqrt{10}</math>.


Using the Pythagorean Theorem to find the square of radius:
Using the Pythagorean Theorem to find the square of radius:
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<math>50=r^2</math>
<math>50=r^2</math>


So the area of the semicircle is <math>\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Rightarrow B </math>
So, the area of the semicircle is <math>\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Rightarrow B </math>


== See Also ==
== See Also ==
*[[2006 AMC 10B Problems]]
*[[2006 AMC 10B Problems]]

Revision as of 12:38, 18 July 2006

Problem

A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?

$\mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi$

Solution

Since the area of the square is $40$, the length of the side is $\sqrt{40}=2\sqrt{10}$. The distance between the center of the semicircle and one of the bottom vertecies of the square is half the length of the side, which is $\sqrt{10}$.

Using the Pythagorean Theorem to find the square of radius:

$(2\sqrt{10})^2 + (\sqrt{10})^2 = r^2$

$50=r^2$

So, the area of the semicircle is $\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Rightarrow B$

See Also

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