Question

find the value of x to the nearest tenth.

Explanation:

Step1: Apply Pythagorean theorem to the lower - right triangle

Let's consider the right - triangle at the bottom right of the figure. Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 4\) and \(b = 8\). The length of the hypotenuse of this right - triangle, say \(y\), is \(y=\sqrt{4^{2}+8^{2}}=\sqrt{16 + 64}=\sqrt{80}=4\sqrt{5}\).

Step2: Apply Pythagorean theorem to the upper right - triangle

Now consider the right - triangle in the upper part of the figure. The hypotenuse of this right - triangle is \(10\), and one side is \(y = 4\sqrt{5}\), and the other side is \(x\). Using the Pythagorean theorem \(x^{2}+y^{2}=10^{2}\). Substitute \(y = 4\sqrt{5}\) into the equation: \(x^{2}+(4\sqrt{5})^{2}=100\).

Step3: Solve for \(x\)

Expand the equation: \(x^{2}+80 = 100\). Then \(x^{2}=100 - 80=20\). So \(x=\sqrt{20}=2\sqrt{5}\approx4.5\).

Answer:

\(4.5\)