Given right triangle $A B C$ with altitude $\overline{B D}$ drawn to hypotenuse $\overline{A C}$ . If $A D=3$ and $A C=30$ , what is the length of $\overline{A B}$ in simplest radical form?

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Q. Given right triangle

A B C

with altitude

\overline{B D}

drawn to hypotenuse

\overline{A C}

. If

A D=3

and

A C=30

, what is the length of

\overline{A B}

in simplest radical form?

Use Geometric Mean Theorem: We are given a right triangle $ABC$ with an altitude $BD$ to the hypotenuse $AC$ . We know that $AD = 3$ and $AC = 30$ . We want to find the length of $AB$ . We can use the geometric mean theorem (also known as the altitude-on-hypotenuse theorem), which states that the altitude drawn to the hypotenuse of a right triangle divides the hypotenuse into two segments, and each segment is the mean proportional between the length of the altitude and the length of the hypotenuse. This gives us two equations: $AD \cdot DC = BD^2$ and $BD \cdot AC = AB^2$ . Since we know $AD$ and $AC$ , we can find $BD$ $0$ using the fact that $BD$ $1$ .
Calculate Length of DC: Calculate the length of DC using the fact that $AC = AD + DC$ . $\newline$ $AC = AD + DC$ $\newline$ $30 = 3 + DC$ $\newline$ $DC = 30 - 3$ $\newline$ $DC = 27$ $\newline$ Now we know that DC is $27$ units long.
Find Length of BD: Apply the geometric mean theorem to find the length of BD. $\newline$ $AD \times DC = BD^2$ $\newline$ $3 \times 27 = BD^2$ $\newline$ $81 = BD^2$ $\newline$ $BD = \sqrt{81}$ $\newline$ $BD = 9$ $\newline$ Now we know that $BD$ is $9$ units long.
Determine Length of AB: Now we can find the length of AB using the second part of the geometric mean theorem. $\newline$ $BD \times AC = AB^2$ $\newline$ $9 \times 30 = AB^2$ $\newline$ $270 = AB^2$ $\newline$ $AB = \sqrt{270}$
Simplify Square Root: Simplify the square root of $270$ to get $AB$ in simplest radical form. $\sqrt{270}$ can be simplified by factoring out perfect squares. $270 = 9 \times 30$ $270 = 9 \times (3 \times 10)$ $270 = (3^2) \times (3 \times 10)$ $\sqrt{270} = \sqrt{(3^2) \times (3 \times 10)}$ $\sqrt{270} = 3 \times \sqrt{3 \times 10}$ $\sqrt{270} = 3 \times \sqrt{30}$ Therefore, $AB = 3 \times \sqrt{30}$ .