Question

solve for x to the nearest tenth.

Explanation:

Step1: Find the length of the diagonal of the rectangle

Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) for the right - angled triangle within the rectangle with sides 3 and 5. Let the diagonal of the rectangle be \(d\). Then \(d^{2}=3^{2}+5^{2}=9 + 25=34\), so \(d=\sqrt{34}\).

Step2: Use the Pythagorean theorem for the large right - angled triangle

In the large right - angled triangle, one leg is \(d=\sqrt{34}\) and the other leg is 9, and the hypotenuse is \(x\). By the Pythagorean theorem \(x^{2}=(\sqrt{34})^{2}+9^{2}\).
Since \((\sqrt{34})^{2}=34\) and \(9^{2}=81\), then \(x^{2}=34 + 81=115\).

Step3: Solve for \(x\)

Taking the square root of both sides, \(x=\sqrt{115}\approx10.7\).

Answer:

\(10.7\)