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Difference between revisions of "2008 iTest Problems/Problem 6"

m (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...")
 
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==Solution==
 
==Solution==
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By the Pythagorean Theorem, the hypotenuse will be <math>\sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13</math>. 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: <math>\frac{1}{2}\cdot 5 \cdot 12 = 5\cdot 6 = 30</math>. Therefore, we have <math>\frac{1}{2}L \cdot 13 = 30</math>. Multiplying both sides by <math>2</math> and dividing both sides by <math>13</math>, we get <math>L = \frac{60}{13}</math>. The least integer greater than <math>\frac{60}{13}</math> is <math>\boxed{5}</math>
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By the Pythagorean Theorem, the hypotenuse will be <math>\sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13</math>. 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: <math>\frac{1}{2}\cdot 5 \cdot 12 = 5\cdot 6 = 30</math>. Therefore, we have <math>\frac{1}{2}L \cdot 13 = 30</math>. Multiplying both sides by <math>2</math> and dividing both sides by <math>13</math>, we get <math>L = \frac{60}{13}</math>. The least integer greater than <math>\frac{60}{13}</math> is <math>\boxed{5}</math>.
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==See Also==
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{{2008 iTest box|num-b=5|num-a=7}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 23:52, 21 June 2018

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 $\boxed{5}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 5 		Followed by:
Problem 7
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