2007 AMC 12A Problems/Problem 10: Difference between revisions

Revision as of 03:49, 1 July 2024

The following problem is from both the 2007 AMC 12A #10 and 2007 AMC 10A #14, so both problems redirect to this page.

Problem

A triangle with side lengths in the ratio $3 : 4 : 5$ is inscribed in a circle with radius 3. What is the area of the triangle?

$\mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18$

Solution

Since 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since the hypotenuse is a diameter of the circumcircle, the hypotenuse is $2r = 6$ . Then the other legs are $\frac{24}5=4.8$ and $\frac{18}5=3.6$ . The area is $\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}$

Solution 2

The hypotenuse of the triangle is a diameter of the circumcircle, so it has length $2 \cdot 3 = 6$ . The triangle is similar to a 3-4-5 triangle with the ratio of their side lengths equal to $\frac{6}{5}$ . The area of a 3-4-5 triangle is $\frac{3\cdot 4}{2} = 6$ . The square of the ratio of their side lengths is equal to the ratio of their areas. Call the area of the triangle $A$ . Therefore, $\left(\frac{6}{5}\right)^2 = \frac{A}{6} \Longrightarrow \frac{36}{25} = \frac{A}{6} \Longrightarrow A = \frac{36\cdot6}{25} = 8.64\ \mathrm{(A)}$

~mobius247

Solution 3 (dimensional analysis)

Using the theorem that the hypotenuse of a right triangle inscribed in a circle is the diameter, one can determine that the diameter of the triangle is 6. As a result the ratio between the triangular ratios and their real values is (6/5). Using dimensional analysis, one can see that because the ratio between lengths is (6/5), the ratio between areas is (6/5)^2. This gives ((3 * 4)(6/5)^2)/2 as the area of the triangle or answer choice (A).

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

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 ==Solution 3 (dimensional analysis)==
-Using the theorem that the hypotenuse of a right triangle inscribed in a circle is the diameter, one can determine that the diameter of the triangle is 6. As a result the ratio between the triangular ratios and their real values is (6/5). Using dimensional analysis, one can see that because the ratio between lengths is (6/5), the ratio between areas is (6/5)^2. This gives ((3 * 4)(6/5)^2)/2 as the area of the triangle or answer choice \mathrm{(A)}$.
+Using the theorem that the hypotenuse of a right triangle inscribed in a circle is the diameter, one can determine that the diameter of the triangle is 6. As a result the ratio between the triangular ratios and their real values is (6/5). Using dimensional analysis, one can see that because the ratio between lengths is (6/5), the ratio between areas is (6/5)^2. This gives ((3 * 4)(6/5)^2)/2 as the area of the triangle or answer choice (A).
 ==See also==

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