Question

find the value of x to the nearest tenth.

Explanation:

Step1: Find the length of the vertical line in the right - hand triangle

Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) in the right - hand right - triangle with hypotenuse \(6\) and one side \(3\). Let the other side be \(y\). Then \(y=\sqrt{6^{2}-3^{2}}=\sqrt{36 - 9}=\sqrt{27}=3\sqrt{3}\).

Step2: Find the value of \(x\) in the left - hand triangle

Use the Pythagorean theorem in the left - hand right - triangle with one side \(x\), another side \(3\sqrt{3}\) and hypotenuse \(10\). We have \(x=\sqrt{10^{2}-(3\sqrt{3})^{2}}=\sqrt{100 - 27}=\sqrt{73}\approx 8.5\).

Answer:

\(8.5\)