Question

a right triangle with a hypotenuse of sqrt{205} has an area of 39 square inches. find the lengths of the other two sides. question help:
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Explanation:

Step1: Let the two legs be \(a\) and \(b\)

For a right - triangle, by Pythagoras' theorem \(a^{2}+b^{2}=(\sqrt{205})^{2}=205\), and the area formula \(A = \frac{1}{2}ab\). Given \(A = 39\), so \(ab=78\).

Step2: Expand \((a + b)^{2}\) and \((a - b)^{2}\)

We know that \((a + b)^{2}=a^{2}+2ab + b^{2}\) and \((a - b)^{2}=a^{2}-2ab + b^{2}\).
Substitute \(a^{2}+b^{2}=205\) and \(ab = 78\) into the above formulas.
\((a + b)^{2}=205+2\times78=205 + 156=361\), so \(a + b=\sqrt{361}=19\).
\((a - b)^{2}=205-2\times78=205 - 156 = 49\), so \(a - b=\sqrt{49}=7\).

Step3: Solve the system of equations

We have the system of equations \(

$$\begin{cases}a + b=19\\a - b=7\end{cases}$$

\)
Add the two equations: \((a + b)+(a - b)=19 + 7\), \(2a=26\), \(a = 13\).
Substitute \(a = 13\) into \(a + b=19\), we get \(13 + b=19\), \(b=6\).

Answer:

The lengths of the other two sides are \(6\) inches and \(13\) inches.