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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
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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
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Try this:       x2 + (3x)2 = 5

Guest Oct 14, 2021