Question

solve for x to the nearest tenth.

Explanation:

Step1: Find the length of the hypotenuse of the small - right triangle

Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) for the small right - triangle with \(a = 2\) and \(b = 4\).
Let \(c_1\) be the hypotenuse of the small right - triangle. Then \(c_1=\sqrt{2^{2}+4^{2}}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\).

Step2: Consider the large right - triangle

The vertical side of the large right - triangle is \(10\) and the horizontal side (formed by the hypotenuse of the small right - triangle) is \(2\sqrt{5}\).
Use the Pythagorean theorem again to find \(x\). Let the vertical side \(a_2=10\) and the horizontal side \(b_2 = 2\sqrt{5}\). Then \(x=\sqrt{(2\sqrt{5})^{2}+10^{2}}\).
First, calculate \((2\sqrt{5})^{2}=2^{2}\times(\sqrt{5})^{2}=4\times5 = 20\) and \(10^{2}=100\).
So \(x=\sqrt{20 + 100}=\sqrt{120}=2\sqrt{30}\approx 10.954\approx11.0\).

Answer:

\(11.0\)