Difference between revisions of "1987 AJHSME Problems/Problem 22"
5849206328x (talk | contribs) (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>...) |
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==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> | ||
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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 | + | 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> | + | 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== | ||
| − | [[ | + | {{AJHSME box|year=1987|num-b=21|num-a=23}} |
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 08:59, 31 May 2009
Problem
is a rectangle, 
is the center of the circle, and 
is on the circle. If 
and 
, 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]](http://latex.artofproblemsolving.com/3/4/f/34f83e970acc786d0c8e91464cc6120d79fc07aa.png)
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 
, so we just need the quarter circle.
Applying the Pythagorean Theorem to 
, we have ![\[(AC)^2=4^2+3^2\Rightarrow AC=5\]](http://latex.artofproblemsolving.com/3/6/a/36a32f6d3fdc8120ef6c9b53e3d3c64933c90913.png)
Since 
is a rectangle, ![\[BD=AC=5\]](http://latex.artofproblemsolving.com/2/e/8/2e852f33aab80c1a9710b76854cc3f8dce76af8c.png)
Clearly 
is a radius of the circle, so the area of the whole circle is 
and the area of the quarter circle is 
.
Finally, the shaded region is ![\[\frac{25\pi }{4}-12 \approx 7.6\]](http://latex.artofproblemsolving.com/b/b/5/bb52741db8ea882cb47bc4e70d31d860144b42a2.png)
so the answer is 
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 | ||
