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Question
The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle
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Solution
Let the shortest side of the right triangle be x.
∴ Hypotenuse = 6 + 2x
Third side = 2x + 6 – 2
= 2x + 4
In the right triangle ABC,
AC2 = AB2 + BC2
(2x + 6)2 = x2 + (2x + 4)2
4x2 + 36 + 24x = x2 + 4x2 + 16 + 16x
0 = x2 – 24x + 16x – 36 + 16
∴ x2 – 8x – 20 = 0
(x – 10) (x + 2) = 0
x – 10 = 0 or x + 2 = 0
x = 10 or x = – 2 ...(Negative value will be omitted)
The side AB = 10 m
The side BC = 2(10) + 4 = 24 m
Hypotenuse AC = 2(10) + 6 = 26 m
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