Question

9. The three squares below join together to form a right triangle. The area and the perimeter for two of the squares is shown. Find the side length of the largest square. Х Area: 81 in 2 Perimeter: 48 in.

Answer 1

We are given the area and perimeter of two squares that form a right triangle. We are asked to find the length of the bigger square. To do that we may use the Pythagorean theorem. Let "a" and "b" be the sides of a triangle and "c" the length of the hypothenuse, we have the following relationship:

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

Now, "a" and "b" are the size of the given squares. For the square which area is given, we can use the following formula for the area of a square:

`A_{\text{square}}=a^2`

We can solve for "a" by taking square root on both sides, like this:

`a=\sqrt[]{A_(square)}`

Replacing the value given for the area, we get:

`\begin{gathered} a=\sqrt[]{81in^2} \\ a=9in \end{gathered}`

Now, for the square which perimeter is given, we can use the fact that the perimeter of a square is the sum of all its sides, like this:

`\begin{gathered} P_{\text{square}}=b+b+b+b \\ P_{\text{square}}=4b \end{gathered}`

solving for "b" we get:

`b=\frac{P_{\text{square}}}{4}`

Replacing the known value for the perimeter, we get:

`b=(48in)/(4)=12in`

Now that we have both sides "a" and "b" we may replace this in the Pythagorean theorem, like this:

`\begin{gathered} c^{2^{}}=a^2+b^2 \\ c^2=9^2+12^2 \end{gathered}`

Solving the operations:

`\begin{gathered} c^2=81+144 \\ c^2=225 \end{gathered}`

Now we solve for "c" by taking square roots on both sides, like this:

`\begin{gathered} c=\sqrt[]{225} \\ c=15 \end{gathered}`

Therefore, the side length of the largest square is 15 in.