# Altitude Of Triangle

### Altitude Of Triangle

##### $$\text{CD}^2 = \text{DA}\times \text{DB}$$

Example 01

Look at the diagram and find the length of BC and AC

Step 1: Find the length of BD

\begin{equation*}
\begin{split}
\text{CD}^2& = \text{DA}\times \text{DB}\\\\
8^2& = 4\times \text{DB}\\\\
64& = 4\times \text{DB}\\\\
\text{DB}& = \frac{64}{4}\\\\
\text{DB}& = 16 \text{ cm}
\end{split}
\end{equation*}

Step 2: Find the length of BC

\begin{equation*}
\begin{split}
\text{BC}^2& = \text{BD}\times \text{BA}\\\\
\text{BC}^2& = 16\times 20\\\\
\text{BC}^2& = 320\\\\
\text{BC}& = \sqrt{320}\\\\
\text{BC}& = \sqrt{64\times 5}\\\\
\text{BC}& = 8\sqrt{5} \text{ cm}
\end{split}
\end{equation*}

Step 3: Find the length of AC

\begin{equation*}
\begin{split}
\text{AC}^2& = 4 \times 20\\\\
\text{AC}^2& = 80\\\\
\text{AC}& = \sqrt{80}\\\\
\text{AC}& = \sqrt{16 \times 5}\\\\
\text{AC}& = 8\sqrt{5} \text{ cm}
\end{split}
\end{equation*}

Example 02

Look at the diagram and find the length of AC and AD!

Step 1: Find the length of BD

Let the length of BD = $$x$$

\begin{equation*}
\begin{split}
\text{AB}^2& = \text{BD}\times \text{BC}\\\\
14^2& = x(x + 21)\\\\
196& = x^2 + 21x\\\\
0& = x^2 + 21x - 196\\\\
0& = (x - 7)(x + 28)\\\\
x - 7 &= 0 \rightarrow x = 7 \text{ cm}\\\\
x + 28& = 0 \rightarrow \cancel{x = -28}\\\\
\text{BD}& = 7 \text{ cm}
\end{split}
\end{equation*}

Step 2: Find the length of AC

\begin{equation*}
\begin{split}
\text{AC}^2& = \text{CD}\times \text{CB}\\\\
\text{AC}^2& = 21 \times 28\\\\
\text{AC}^2& = 588\\\\
\text{AC}& = \sqrt{588}\\\\
\text{AC}& = \sqrt{196 \times 3}\\\\
\text{AC}&= 14\sqrt{3} \text{ cm}
\end{split}
\end{equation*}

Step 3: Find the length of AD

\begin{equation*}
\begin{split}