# The area of a right triangle is 15 cm^2 and the hypotneuse is 9 cm long. Find the lengths of the other two sides.

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Let x and y be the two sides of the right triangle.

To solve, apply the formula of area of triangle.

`A=1/2bh`

So, plug-in b=x and h=y.

`15=1/2xy`

`30=xy`

Then, isolate one of the variable.

`30/x =y` (Let this be EQ1.)

Next, apply the Pythagorean formula.

`a^2+b^2=c^2`

Plug-in a=x and b=y and c=9.

`x^2+y^2=9^2`

`x^2+y^2=81`

Also, substitute EQ1.

`x^2+(30/x)^2=81`

`x^2+900/x^2=81`

Multiply both sides by x^2 to eliminate fraction in the equation.

`x^4 + 900 = 81x^2`

Then, set one side of the equation equal to zero.

`x^4-81x^2+900=0`

To solve using quadratic formula, use another variable.

Let `z=x^2` .

`z^2 - 81z+900=0`

Then, plug-in a=1, b=-81 and c=900 to the quadratic formula.

`z=(-b+-sqrt(b^2-4ac))/(2a)=(-(-81)+-sqrt((-81)^2-4*1*900))/(2*1)=(81+-sqrt(6561-3600))/2`

`z=13.2925` and `67.7075`

Since `z=x^2` , then the values of x are:

`13.2925=x^2` `67.7075=x^2`

`+-sqrt13.2925=x` `+-sqrt67.7075=x`

`+-3.6459=x` `+-8.2285=x`

However, consider only the positive values of x.

So, the values of x are:

`x=3.6459`

`x=8.2285`

Next, plug-in the values of x to EQ1.

`y=30/x`

`y=30/3.6459 =8.2284`

`y=30/8.2285=3.6459`

**Hence, the length of the two sides of the right triangle are 3.6459 cm and 8.2284 cm.**