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13 Jul 2018, 00:43
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
86% (01:37) correct
14%
(01:29)
wrong
based on 80
sessions
History
Date
Time
Result
Not Attempted Yet
In right triangle ABC, the ratio of the lengths of the two legs (non-hypotenuse sides) 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. \(2 \sqrt{29}\)
A. 7
B. 10
C. \(4 \sqrt{5}\)
D. \(\sqrt{29}\)
E. \(2 \sqrt{29}\)
13 Jul 2018, 00:54
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Bunuel
Let the two legs of the right triangle ABC be 2x and 5x
Now, area = 1/2 * leg1 * leg2
20 = 1/2 * 2x * 5x
x = 2
Putting x's value in respective legs and by pythagoras theorem we can get the hypotenuse
h^2 = 4^2 + 10^2
h^2 = 116
h = 2 root 29
Hence, E.
13 Jul 2018, 00:55
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In right triangle ABC, the ratio of the lengths of the two legs (non-hypotenuse sides) is 2 to 5
=> one leg is 2x and the other leg is 5x
=> These two legs form the base and height for the right triangle
=> Area = \(\frac{1}{2} * base * height = \frac{1}{2} * 2x * 5x = 20\)
=> \(5x^2 = 20\)
=> \(x^2 = 4\)
=> x = 2
The legs of right triangle are 2x = 4 and 5x = 10
length of the hypotenuse = \(\sqrt{(4)^2 + (10)^2}\)
=> \(\sqrt{16 + 100}\)
=> \(2\sqrt{29}\)
Hence option E
=> one leg is 2x and the other leg is 5x
=> These two legs form the base and height for the right triangle
=> Area = \(\frac{1}{2} * base * height = \frac{1}{2} * 2x * 5x = 20\)
=> \(5x^2 = 20\)
=> \(x^2 = 4\)
=> x = 2
The legs of right triangle are 2x = 4 and 5x = 10
length of the hypotenuse = \(\sqrt{(4)^2 + (10)^2}\)
=> \(\sqrt{16 + 100}\)
=> \(2\sqrt{29}\)
Hence option E
