The size of one leg of a appropriate triangle is 7 ft longer than the various other leg. The hypotenuse is 17 ft long. Find the lengths the the two legs.

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Hi Jim,

We would usage the pythagorean theorem.

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One leg would be x

the other is x + 7

hypotenuse is 17.

so x^2 + (x +7 )^2 = 17^2

solving x^2 + x^2 + 14 x + 49 = 289

solving ours quadratic would be

2 x ^ 2 + 14 x + -240= 0

factoring x = 8

and x + 7 =15

I expect this helps

Since its a best angled triangle, that will meet the Pythagoras equation.

The amount of the squares of the sides is equal to the square that the hypotenuse.

If one side is x, the various other side is 7+x.

so x2 + (7+x)2 = 172

x2 + x2 + 14x + 49 = 289

2x2 + 14x - 240 = 0.

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x2 + 7x - 120 = 0.

x = (-7 + √(49 + 4*1*120))/2 or (-7 - √(49 + 4*1*120))/2

x = 8, -15

Since negative side that a triangle is not acceptable, the length of various other two legs are 8 ft and 15 ft

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