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 "1987 AJHSME Problems/Problem 22"

(New page: ==Problem== <math>\text{ABCD}</math> is a rectangle, <math>\text{D}</math> is the center of the circle, and <math>\text{B}</math> is on the circle. If <math>\text{AD}=4</math> and <math>...)
 
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
<math>\text{ABCD}</math> is a rectangle, <math>\text{D}</math> is the center of the circle, and <math>\text{B}</math> is on the circle.  If <math>\text{AD}=4</math> and <math>\text{CD}=3</math>, then the area of the shaded region is between
+
<math>\text{ABCD}</math> is a [[rectangle]], <math>\text{D}</math> is the center of the [[circle]], and <math>\text{B}</math> is on the circle.  If <math>\text{AD}=4</math> and <math>\text{CD}=3</math>, then the [[area]] of the shaded region is between
  
 
<asy>
 
<asy>
Line 22: Line 22:
 
The area of the shaded region is equal to the area of the quarter circle with the area of the rectangle taken away.  The area of the rectangle is <math>4\cdot 3=12</math>, so we just need the quarter circle.
 
The area of the shaded region is equal to the area of the quarter circle with the area of the rectangle taken away.  The area of the rectangle is <math>4\cdot 3=12</math>, so we just need the quarter circle.
  
−
Applying the Pythagorean theorem to <math>\triangle ADC</math>, we have <cmath>(AC)^2=4^2+3^2\Rightarrow AC=5</cmath> Since <math>ABCD</math> is a rectangle, <cmath>BD=AC=5</cmath>
+
Applying the [[Pythagorean Theorem]] to <math>\triangle ADC</math>, we have <cmath>(AC)^2=4^2+3^2\Rightarrow AC=5</cmath> Since <math>ABCD</math> is a rectangle, <cmath>BD=AC=5</cmath>
  
−
Clearly <math>BD</math> is a radius of the circle, so the area of the whole circle is <math>5^2\pi =25\pi</math> and the area of the quarter circle is <math>\frac{25\pi }{4}</math>.
+
Clearly <math>BD</math> is a [[radius]] of the circle, so the area of the whole circle is <math>5^2\pi =25\pi</math> and the area of the quarter circle is <math>\frac{25\pi }{4}</math>.
  
−
Finally, the shaded region is <cmath>\frac{25\pi }{4}-12 \approx 7.6</cmath>, so the answer is <math>\boxed{\text{D}}</math>
+
Finally, the shaded region is <cmath>\frac{25\pi }{4}-12 \approx 7.6</cmath> so the answer is <math>\boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==
  
−
[[1987 AJHSME Problems]]
+
{{AJHSME box|year=1987|num-b=21|num-a=23}}
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:53, 4 July 2013

Problem

$\text{ABCD}$ is a rectangle, $\text{D}$ is the center of the circle, and $\text{B}$ is on the circle. If $\text{AD}=4$ and $\text{CD}=3$, then the area of the shaded region is between

[asy] pair A,B,C,D; A=(0,4); B=(3,4); C=(3,0); D=origin; draw(circle(D,5)); fill((0,5)..(1.5,4.7697)..B--A--cycle,black); fill(B..(4,3)..(5,0)--C--cycle,black); draw((0,5)--D--(5,0)); label("A",A,NW); label("B",B,NE); label("C",C,S); label("D",D,SW); [/asy]

$\text{(A)}\ 4\text{ and }5 \qquad \text{(B)}\ 5\text{ and }6 \qquad \text{(C)}\ 6\text{ and }7 \qquad \text{(D)}\ 7\text{ and }8 \qquad \text{(E)}\ 8\text{ and }9$

Solution

The area of the shaded region is equal to the area of the quarter circle with the area of the rectangle taken away. The area of the rectangle is $4\cdot 3=12$, so we just need the quarter circle.

Applying the Pythagorean Theorem to $\triangle ADC$, we have \[(AC)^2=4^2+3^2\Rightarrow AC=5\] Since $ABCD$ is a rectangle, \[BD=AC=5\]

Clearly $BD$ is a radius of the circle, so the area of the whole circle is $5^2\pi =25\pi$ and the area of the quarter circle is $\frac{25\pi }{4}$.

Finally, the shaded region is \[\frac{25\pi }{4}-12 \approx 7.6\] so the answer is $\boxed{\text{D}}$

See Also

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 AJHSME/AMC 8 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=1987_AJHSME_Problems/Problem_22&oldid=55537"