Difference between revisions of "2007 AMC 12A Problems/Problem 10"
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==Problem== | ==Problem== | ||
| − | A triangle with side lengths in the | + | A [[triangle]] with side lengths in the [[ratio]] <math>3 : 4 : 5</math> is inscribed in a [[circle]] with [[radius]] 3. What s the area of the triangle? |
| + | |||
| + | <math>\displaystyle \mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18</math> | ||
==Solution== | ==Solution== | ||
| + | [[Image:2007_AMC12A-10.png]] | ||
| − | + | Since 3-4-5 is a [[Pythagorean triple]], the triangle is a [[right triangle]]. Since right triangles can be inscribed in [[semicircle]]s, the hypotenuse is <math>2r = 6</math>. Then the other legs are <math>\frac{24}5=4.8</math> and <math>\frac{18}5=3.6</math>. The area is <math>\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}</math> | |
==See also== | ==See also== | ||
| − | + | {{AMC12 box|year=2007|ab=A|num-b=9|num-a=11}} | |
| − | + | ||
| − | + | [[Category:Introductory Geometry Problems]] | |
Revision as of 14:18, 9 September 2007
Problem
A triangle with side lengths in the ratio 
is inscribed in a circle with radius 3. What s the area of the triangle?
Solution
Since 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since right triangles can be inscribed in semicircles, the hypotenuse is 
. Then the other legs are 
and 
. The area is 
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
