Question

find the value of x to the nearest tenth.

(image of a rectangle with a diagonal of length 9, and a right triangle attached to the bottom with legs 4 and 2)

answer
attempt 1 out of 2
x ≈ blank submit answer

Explanation:

Step1: Find the base of the rectangle

The lower right triangle has legs 4 and 2. Using the Pythagorean theorem \( c = \sqrt{a^2 + b^2} \), where \( a = 4 \) and \( b = 2 \).
\( \text{Base} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \)

Step2: Use Pythagorean theorem on the rectangle

The rectangle has a diagonal of 9 and one side \( x \), the other side is \( \sqrt{20} \). By Pythagorean theorem \( x^2 + (\sqrt{20})^2 = 9^2 \)
\( x^2 + 20 = 81 \)
\( x^2 = 81 - 20 = 61 \)
\( x = \sqrt{61} \approx 7.8 \)

Answer:

\( x \approx 7.8 \)