# Find the short leg, long leg, and hypotenuse of the triangle below

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Find: the short leg, long leg, and hypotenuse of the
triangle below.
Solution
Let the shorter leg be \'x\'
Let the longer leg be1 x/2+11
Hypotenus2x+1
Acc to pythagoras Theorem H^2=P^2+B^2
(2x+1 )^2=x^2+(1 x/2+11)^2
(4x^2+4x+1) =x^2+(x^2/4+11x+121)
3x^2-1x^2/4=7x+120
11x^2/4-7x-120=0
x=3.71
shorter leg be \'x\'=3.71
longer leg be x/2+11=12.855

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Find: the short leg, long leg, and hypotenuse of the triangle below. Solution Let the shorter leg be \'x\' Let the longer leg be1 x/2+11 Hypotenus2x+1 Acc to pythagoras Theorem H^2=P^2+B^2 (2x+1 )^2=x^2+(1 x/2+11)^2 (4x^2+4x+1) =x^2+(x^2/4+11x+121) 3x^2-1x^2/4=7x+120 11x^2/4-7x-120=0 x=3.71 shorter leg be \'x\'=3.71 longer leg be x/2+11=12.855 Hypotenus2 x+1=8.42 Name: Description: ...
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