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In right Delta*A * B * C legAB = legBC The area of the triangle is 12.5. Hypotenuse AC equals to : A. sqrt(5) B. 5sqrt(2) C. 5 D.4 sqrt 5

In right Delta*A * B * C legAB = legBC The area of the triangle is 12.5. Hypotenuse-example-1
User Lschuetze
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1 Answer

15 votes
15 votes

The are of a triangle is:


A=(1)/(2)b\cdot h

b is the base and h is the height

In the given triangle the base is Leg BC and the height is Leg AB:


\begin{gathered} AB=BC \\ A=12.5 \\ \\ 12.5=(1)/(2)AB^2 \end{gathered}

Use the equation above to solve AB:


\begin{gathered} \text{Multiply both sides of the equation by 2:} \\ 12.5*2=2*(1)/(2)AB^2 \\ \\ 25=AB^2 \\ \\ \text{Take the square root of both sides of the equation:} \\ \sqrt[]{25}=√(AB^2) \\ \\ 5=AB \end{gathered}

Then, Legs of the given right triangle have a measure of 5.

To find the measure of the hypotenuse use Pythagorean theorem:


\begin{gathered} \text{hypotenuse}^2=Leg1^2+Leg2^2 \\ \text{hypotenuse}=\sqrt[]{Leg1^2+Leg2^2} \end{gathered}
\begin{gathered} \text{hypotenuse}=\sqrt[]{5^2+5^2} \\ \\ \text{hypotenuse}=\sqrt[]{25+25} \\ \\ \text{hypotenuse}=\sqrt[]{50} \end{gathered}

Find the prime factorization of 50 to simplify it:


50=2*5*5=2*5^2
\begin{gathered} \text{hypotenuse}=\sqrt[]{2*5^2} \\ \\ \text{hypotenuse}=5\sqrt[]{2} \end{gathered}

In right Delta*A * B * C legAB = legBC The area of the triangle is 12.5. Hypotenuse-example-1
User T D Nguyen
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