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Difference between revisions of "2006 AMC 10B Problems/Problem 8"

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== Solution ==
 
== Solution ==
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Since the area of the square is <math>40</math>, the length of a 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>.
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Since the area of the square is <math>40</math>, the length of a side is <math>\sqrt{40}=2\sqrt{10}</math>. The distance between the center of the semicircle and one of the bottom vertices 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, <math>r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50</math>. So, the area of the semicircle is <math>\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Longrightarrow \boxed{\text{(B)}}</math>.
 
Using the [[Pythagorean Theorem]] to find the square of radius, <math>r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50</math>. So, the area of the semicircle is <math>\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Longrightarrow \boxed{\text{(B)}}</math>.

Revision as of 15:36, 2 June 2021

Problem

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

[asy] defaultpen(linewidth(0.8)); size(100); real r=sqrt(50), s=sqrt(10); draw(Arc(origin, r, 0, 180)); draw((r,0)--(-r,0), dashed); draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle); [/asy]

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

Solution

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

Using the Pythagorean Theorem to find the square of radius, $r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50$. So, the area of the semicircle is $\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Longrightarrow \boxed{\text{(B)}}$.

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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