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A right triangle has an area of 54 ft2 and a hypotenuse of 25 ft long. What are the lengths of its other two sides?

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By theorem we have the following:


h^2=a^2+b^2

And, we are given:


A=(a\cdot b)/(2)\Rightarrow2A=a\cdot b

Then:


\Rightarrow4A^2=a^2b^2\Rightarrow4A^2=a^2(h^2-a^2)
\Rightarrow a^4-h^2a^2+4A^2=0

Now, we replace h and A:


a^4-(25)^2a^2+4(54)^2=0

And solve for a:


a^4-625a^2+11664=0

Then, the possible values for a are:


a=\begin{cases}a_1=-(29)/(2)-\frac{\sqrt[]{409}}{2} \\ a_2=(29)/(2)-\frac{\sqrt[]{409}}{2} \\ a_3=\frac{\sqrt[]{409}}{2}-(29)/(2) \\ a_4=(29)/(2)+\frac{\sqrt[]{409}}{2} \\ \end{cases}

We can see that a1, and a2 are not solutions, therefore a2 and a4 are.

So, the two possible b sides are then:


b_2=\sqrt[]{25^2-((29)/(2)-\frac{\sqrt[]{409}}{2})^2}\Rightarrow b_2\approx24.99

and:


b_4=\sqrt[^{}]{25^2-((29)/(2)+\frac{\sqrt[]{409}}{2})^2}\Rightarrow b_(4\approx)15.50

So, the lengths of the two sides can be:

a = 4.38 and b = 24.99

or

a = 24.61 and b = 15.50

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