Difference between revisions of "1995 AJHSME Problems/Problem 9"
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Draw a horizontal line <math>PQR</math> that extends to the sides of the rectangle. This line is <math>2</math> diameters long, so it has length <math>4\cdot 2 = 8</math>. It is parallel and congruent to <math>AB</math>, so the width of the rectangle is <math>8</math>. | Draw a horizontal line <math>PQR</math> that extends to the sides of the rectangle. This line is <math>2</math> diameters long, so it has length <math>4\cdot 2 = 8</math>. It is parallel and congruent to <math>AB</math>, so the width of the rectangle is <math>8</math>. | ||
| − | Thus, the area of the rectangle is <math>lw = 4\cdot 8 = 32</math>, and the answer is <math>\boxed{ | + | Thus, the area of the rectangle is <math>lw = 4\cdot 8 = 32</math>, and the answer is <math>\boxed{C}</math>. |
==Solution 2== | ==Solution 2== | ||
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:15, 4 July 2013
Contents
Problem
Three congruent circles with centers 
, 
, and 
are tangent to the sides of rectangle 
as shown. The circle centered at has diameter 
and passes through points and . The area of the rectangle is
![[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,W); label("$Q$",Q,W); label("$R$",R,W); [/asy]](http://latex.artofproblemsolving.com/f/8/c/f8cdb8860743450573eda533b3ba4facf3292fa7.png)
Solution 1
If circle has diameter , then so do congruent circles and . Draw a diameter through parallel to 
. The diameter will be congruent to , and thus 
, which is the height of the rectangle.
Draw a horizontal line 
that extends to the sides of the rectangle. This line is 
diameters long, so it has length 
. It is parallel and congruent to 
, so the width of the rectangle is 
.
Thus, the area of the rectangle is 
, and the answer is 
.
Solution 2
The area of circle and circle almost fill the rectangle. Circle has radius 
, and so does circle . Thus, the sum of their areas is 
. Since 
, the area of the circles is just over 
. The area of the rectangle is greater than the area of the circles. 
would be too high, as the two circles appear to take up much more than half the area of the rectangle. But is too low, as it is less than the area of the two circles. Thus, the only reasonable answer is , which is 
.
See Also
| 1995 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
