SOLUTION: a right triangle has hypotenuse of lenght 13 and legs A and B. If the area of the triangle is 14, what is the sum A+B? We do not think this is possible. If the Square of the hyp

Question 261102: a right triangle has hypotenuse of lenght 13 and legs A and B. If the area of the triangle is 14, what is the sum A+B? We do not think this is possible. If the Square of the hypotenuse is 169, and it is an equilateral triangle, the sides would be about 9, and one half of 9*9 (the area of the triangle) is more like 40 than 14
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
a right triangle has hypotenuse of lenght 13 and legs A and B. If the area of the triangle is 14, what is the sum A+B?
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a^2 + b^2 = 169
ab/2 = 14 --> a = 28/b
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(28/b)^2 + b^2 = 169
784 + b^4 = 169b^2
b^4 - 169b^2 + 784 = 0
Sub x for b^2

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=25425 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 164.22609359551, 4.77390640449013. Here's your graph:

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b^2 =~ 164.226
b^2 =~ 4.7739
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b =~ 12.8166, a =~ 2.1849 or vice versa
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a + b = 15
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In any right triange with a specified lenght of the hypotenuse, the max area is
c^2/2.
For a hyp = 13, that's 169/2 = 84.5 sq units. Any area less than that can be made with hyp = 13.

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