The hypotenuse of a triangle is 26 feet long. One leg of the triangle is 14 feet longer than the other leg. What are the lengths of the legs of the triangle?

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The hypotenuse of a triangle is 26 feet long. One leg of the triangle is 14 feet longer than the other leg. What are the lengths of the legs of the triangle?

2 Answers

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Answer 1 · 2018-03-20T12:36:06

The two lengths are 10 feet and 24 feet.

Explanation:

By Pythagoras Theorem, $a^{2} + b^{2} = c^{2}$ $a^2 + b^2 = c^2$ .
Since the hypotenuse, c is 26 feet long, $26^{2}$ $26^2$ = 676.
$10^{2}$ $10^2$ + $24^{2}$ $24^2$ = 676.
Hence the answer is 10 feet and 24 feet.

Answer 2 · 2018-03-20T12:48:41

Lengths of the legs are $10 and 24$ $10 and 24$ ft

Explanation:

Let length of one leg of the triangle be $x$ $x$ ft , then other leg will

be $x + 14; h = 26$ $x+14 ; h= 26$ ft . We know $h^{2} = x^{2} + {(x + 14)}^{2}$ $h^2 = x^2+ (x+14)^2$ or

$26^{2} = x^{2} + {(x + 14)}^{2} or x^{2} + x^{2} + 28 x + 196 = 676$ $26^2= x^2+ (x+14)^2 or x^2 +x^2 +28x+196 =676$ or

$2 x^{2} + 28 x - 480 = 0 or x^{2} + 14 x - 240 = 0$ $2x^2 +28x-480 =0 or x^2+14x-240=0$ or

$x^{2} + 24 x - 10 x - 240 = 0$ $x^2+24x-10x -240=0$ or

$x (x + 24) - 10 (x + 24) = 0 or (x + 24) (x - 10) = 0$ $x(x+24)-10(x+24)=0 or (x+24)(x-10)=0$

Either $x+24=0:. x=-24 or x-10=0 :.x=10$

leg can not be negative so, $x=10 :.x+14=10+14=24$

Hence lengths of the legs are $10 and 24$ ft . [Ans]

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The hypotenuse of a triangle is 26 feet long. One leg of the triangle is 14 feet longer than the other leg. What are the lengths of the legs of the triangle?

2 Answers

Explanation:

Explanation:

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