Question

find the value of x to the nearest tenth.

Explanation:

Step1: Find the base of the rectangle

The bottom right triangle has legs 4 and 2. Using the Pythagorean theorem \( a^2 + b^2 = c^2 \), where \( a = 4 \) and \( b = 2 \). So the base \( b \) of the rectangle is \( \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \).

Step2: Use Pythagorean theorem for the rectangle's diagonal

The diagonal of the rectangle is 9, and one side is \( x \), the other is \( \sqrt{20} \). Using \( x^2 + (\sqrt{20})^2 = 9^2 \). So \( x^2 + 20 = 81 \). Then \( x^2 = 81 - 20 = 61 \). So \( x = \sqrt{61} \approx 7.8 \) (to the nearest tenth).

Answer:

\( 7.8 \)