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In right triangle ABC, the ratio of the lengths of the two legs is 2

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In right triangle ABC, the ratio of the lengths of the two legs is 2  [#permalink]

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New post 12 Jul 2018, 23:43
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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. \(2 \sqrt{29}\)

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Re: In right triangle ABC, the ratio of the lengths of the two legs is 2  [#permalink]

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New post 12 Jul 2018, 23:54
Bunuel wrote:
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?


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.
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Re: In right triangle ABC, the ratio of the lengths of the two legs is 2  [#permalink]

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New post 12 Jul 2018, 23:55
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
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In right triangle ABC, the ratio of the lengths of the two legs is 2  [#permalink]

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New post 13 Jul 2018, 16:38
Bunuel wrote:
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. \(2 \sqrt{29}\)

Ratio of legs with multiplier \(x\):
\(\frac{Leg_1}{Leg_1}=\frac{2x}{5x}\)
Area = \(\frac{b*h}{2}=\frac{2x*5x}{2}=20\)
\(5x^2=20\)
\((x^2=4) => x=2\)

Multiplier \(x=2\), so
\(Leg_1=2x=(2*2)=4\)
\(Leg_2=5x=(5*2)=10\)

Hypotenuse
\(4^2+10^2=h^2\)
\(\sqrt{h^2}=\sqrt{116}\)
\(\sqrt{h^2}=\sqrt{2*2*29}\)
\(h=2\sqrt{29}\)

Answer E
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In right triangle ABC, the ratio of the lengths of the two legs is 2 &nbs [#permalink] 13 Jul 2018, 16:38
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