2008 iTest Problems/Problem 6

Revision as of 06:32, 12 May 2012 by AlcumusGuy (talk | contribs) (Created page with "==Problem== Let <math>L</math> be the length of the altitude to the hypotenuse of a right triangle with legs 5 and 12. Find the least integer greater than <math>L</math>. ==Solu...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Let $L$ be the length of the altitude to the hypotenuse of a right triangle with legs 5 and 12. Find the least integer greater than $L$.

Solution

By the Pythagorean Theorem, the hypotenuse will be $\sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13$. The area of this triangle can be expressed in 2 ways: half the product of the legs, and half the product of the hypotenuse and its altitude. We can easily find the area using the first method: $\frac{1}{2}\cdot 5 \cdot 12 = 5\cdot 6 = 30$. Therefore, we have $\frac{1}{2}L \cdot 13 = 30$. Multiplying both sides by $2$ and dividing both sides by $13$, we get $L = \frac{60}{13}$. The least integer greater than $\frac{60}{13}$ is $\boxed{5}$