Difference between revisions of "2006 AMC 10B Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | A [[square]] of area 40 is [[inscribe]]d in a [[semicircle]] as shown. What is the area of the semicircle? | + | A [[square]] of area 40 is [[inscribe]]d in a [[semicircle]] as shown. What is the area of the semicircle? |
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<asy> | <asy> | ||
defaultpen(linewidth(0.8)); size(100); | defaultpen(linewidth(0.8)); size(100); | ||
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draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle); | draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle); | ||
</asy> | </asy> | ||
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<math> \mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi </math> | <math> \mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi </math> | ||
Revision as of 23:24, 18 October 2020
Problem
A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?
Solution
Since the area of the square is
, the length of the side is
. 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
.
Using the Pythagorean Theorem to find the square of radius,
. So, the area of the semicircle is
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.