Art of Problem Solving
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

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

← Older editNewer edit →
Xantos C. Guin (talk | contribs)
editor
580 edits
m added category
mNo edit summary
Line 1: Line 1:
== Problem ==
== Problem ==
A square of area 40 is inscribed 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?  
 
<!-- [[Image:2006amc10b08.gif]] -->
[[Image:2006amc10b08.gif]]
<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>
<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>.
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:


<math>(2\sqrt{10})^2 + (\sqrt{10})^2 = r^2 </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>.
 
<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>


== See Also ==
== See Also ==
*[[2006 AMC 10B Problems]]
{{AMC10 box|year=2006|ab=B|num-b=7|num-a=9}}
 
*[[2006 AMC 10B Problems/Problem 7|Previous Problem]]
 
*[[2006 AMC 10B Problems/Problem 9|Next Problem]]


[[Category:Introductory Geometry Problems]]
[[Category:Introductory Geometry Problems]]
[[Category:Area Problems]]
[[Category:Circle Problems]]

Revision as of 23:23, 20 August 2011

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
Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_8&oldid=41834"