Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

 
m (minor edit)
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
A [[square]] of area 40 is [[inscribe]]d 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>
 +
 +
<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>
 +
 
== 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>. 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, <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>.
 +
 
== See Also ==
 
== See Also ==
−
*[[2006 AMC 10B Problems]]
+
{{AMC10 box|year=2006|ab=B|num-b=7|num-a=9}}
 +
 
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Problems]]
 +
[[Category:Circle Problems]]
 +
{{MAA Notice}}

Revision as of 00:24, 19 October 2020

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 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, $r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50$. 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_8&oldid=135280"