Carcass wrote:
In right triangle ABC, the ratio of the lengths of the two legs is 2 to 5. If the area of triangle ABC is 20, what is the length of the hypotenuse?
A. \(7\)
B. \(10\)
C. \(4 \sqrt{5}\)
D. \(\sqrt{29}\)
E. \(2 \sqrt{29}\)
Let the two legs have lengths \(2x\) and \(5x\). Doing so, ensures that the ratio of the lengths of the two legs is \(2\) to \(5\)
The area of triangle ABC is 20Area of a triangle \(= \frac{(base)(height)}{2}\)
Since we're told the area is \(20\), we can write \(\frac{(5x)(2x)}{2}=20\)
Simplify: \(\frac{10x^2}{2}=20\)
Multiply both sides of the equation by \(2\) to get: \(10x^2 = 40\)
Divide both sides by \(10\) to get: \(x^2 = 4\)
Solve: \(x = 2\) OR \(x = -2\)
Since the side lengths can't be negative, we can be certain that \(x = 2\)
Since the two legs have lengths \(2x\) and \(5x\), can now say the lengths of the legs are
\(4\) and
\(10\)What is the length of the hypotenuse?Let h = length of the hypotenuse
Applying the Pythagorean Theorem, we get:
4² +
10² = h²
Simplify: 16 + 100 = h²
Simplify: 116 = h²
So, h = √116 = (√4)(√29) = 2√29
Answer: E
Cheers,
Brent
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Brent Hanneson - founder of Greenlight Test PrepSign up for our GRE Question of the Day emails