One leg of a right triangle is 6in more than twice the length of the other leg. the hypotenuse is 9in more than the shorter leg. find the lengths of all sides.

a=2.69 in

b=11.37 in

c=11.69 in

Explanation:

If we mark shorter side as a, we can write all sides expressed through a.

So:  

- short side is a

- longer side is 6 in longer then twice the length of short side b = 2a + 6

- hypotenuse is 9 in longer then short leg c = a + 9

Now we can write the Pythagoras theorem a^2 + b^2 = c^2 as:

a^2 + (2a + 6)^2 = (a + 9)^2

a^2 +4a^2+24a+36 = a^2+18a+81

When we solve this, we get:

4a^2 + 6a - 45 = 0

Now we have a quadratic equation which can be solved as:

x = -b ± √(b^2 - 4ac) / 2a

When we plug in the values, we get that x equals 2.69 or -4.19

It's obvious that length of a side can not be negative, which means that a = 2.69 in.

Side b is 2a + 6, which is 11.37 and c is a + 9 which is 11.69 inches.

These are lengths rounded to two decimals so due to this approximation they may not be completely precise.

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