Question

In the diagram below of right triangle ACB, altitude CD is drawn to hypotenuse AB.If AB = 45 and AC = 15, what is the length of AD?1) 32) 53) 74) 9

Answer 1

Similarity Shapes.

The correct answer of length AD is 5.

$\begin{gathered} AB^2=AC^2+CB^2 \\ 45^2=15^2+CB^2 \\ \text{Collecting like terms, we have,} \\ 45^2-15^2=AB^2 \\ 2025-225=AB^2 \\ 1800=AB^2 \\ \text{Take the square root of both sides to get,} \\ AB=\sqrt[]{1800}=42.426\approx42.43 \end{gathered}$

By Similarity Theorem,

Triangle ADC is similar to triangle ACB. Hence,

$\frac{AD}{AC}=\frac{AC}{AB}$$\begin{gathered} \frac{AD}{15}=\frac{15}{45} \\ \text{Cross multiply, we get} \\ 45\times AD=15\times15 \\ \text{Dividing both sides by 45, we get,} \\ AD=\frac{15\times15}{45}=5 \end{gathered}$

The correct answer is length AD = 5