+0  
 
0
26
2
avatar

The longer leg of a right triangle is three times as long as the shorter leg. The hypotenuse is sqrt(5) What is the area of this triangle?

 
 Oct 13, 2021
 #1
avatar+23 
0

We know that shortleg^2 + longleg^2 = \(\sqrt{5}^2\) or 5

If the shorter leg is x then the longer leg would be 3x

Therefore we have:

\(x^2 + 9x^2 = 5\)

simplify it to

\(10x^2 = 5\)

divide 4 each side

\(x^2 = \frac{5}{10}\)

square root both side

\(x = \sqrt{\frac{5}{10}}\)

So the shorter leg is \(\sqrt{\frac{5}{10}}\) and the longer leg is \(3 \sqrt{\frac{5}{10}}\)

Now we need to find the area

Multiply them:

\(3\sqrt{\frac{5}{10}}\cdot \sqrt{\frac{5}{10}} = 3 \cdot \frac{5}{10} = \frac{15}{10} = \frac{3}{2}\)

And don't forget we also have to divide by 2

\(\frac{3}{2} \div 2 = \frac{3}{4}\)

 
 Oct 14, 2021
 #2
avatar
0

Try this:       x2 + (3x)2 = 5  laugh

 
Guest Oct 14, 2021

28 Online Users

avatar