SOLUTION: The hypotenuse of a right triangle is 1.3 units long. The longer leg is 0.7 units longer than the shorter leg. Find the lengths of the legs of the triangle.
All I am able to figu
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All I am able to figu
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Question 534796: The hypotenuse of a right triangle is 1.3 units long. The longer leg is 0.7 units longer than the shorter leg. Find the lengths of the legs of the triangle.
All I am able to figure out is the formula to use is a^2 + b^2 = c^2. Please help. I am lost from here.
Thank you in advance! Found 2 solutions by stanbon, mananth:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The hypotenuse of a right triangle is 1.3 units long. The longer leg is 0.7 units longer than the shorter leg. Find the lengths of the legs of the triangle.
Let shorter leg be "x"
Then longer leg is "x+0.7"
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use a^2 + b^2 = c^2
----
x^2 + (x+0.7)^2 = 1.3^2
x^2 + x^2 + 1.4x + 0.49 = 1.69
----
2x^2 + 1.4x -1.20 = 0
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x^2 + 0.7x - 0.6 = 0
----
x = [-0.7 +- sqrt(0.49 - 4*1*-0.6)/2
----
x = [-0.7 +- sqrt(2.89)]/2
----
x = [-0.7 +- 1.7]/2
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Positive solution:
x = 1/2 (shorter leg)
x+0.7 = 1.2 (longer leg)
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Cheers,
Stan H.
You can put this solution on YOUR website! let shorter leg be x units
longer leg = x+0.7 units
Hypotenuse = 1.3 units
By Pythagoras theorem
x^2+(x+0.7)^2= 1.3^2
x^2+x^2+1.4x+0.49=1.69
2x^2+1.4x-1.2
/2
x^2+0.7x-0.6=0
multiply by 10 to eliminate the decimals
10x^2+7x-6=0
10x^2+12x-5x-6=0
2x(5x+6)-1(5x+6)=0
(5x+6)(2x-1)=0
x=-6/5 OR 1/2
Ignore negative
x=1/2
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