### $\triangle ABC$ is a triangle such that $\angle C = 90^\circ.$ Suppose $AC = 12 \space cm, AB = 13 \space cm$ and the perpendicular distance from $C$ to $AB$ is $x \space cm.$ Find the value of $x.$

Answer:

$4.62$

Step by Step Explanation:
1. We need to find $x,$ the perpendicular distance from $C$ to $AB,$ if $\angle C = 90^\circ, AC = 12 \space cm ,$ and $AB = 13 \space cm.$ Let $D$ be the point where the perpendicular from $C$ meets $AB.$
Let $D$ be the point where the perpendicular from $C$ meets $AB.$
2. Using Pythagoras' theorem in triangle $ABC$
$BC = \sqrt{ 13^2 - 12^2 } = 5$
3. Since,
\begin{align} & \angle ACB = \angle ADC = 90^\circ \\ & \angle BAC = \angle CAD && [\text{ Common angle }] \\ & \angle CBA = \angle DCA && [\text{ Angle sum property }] \\ & \triangle ACB \text{ and } \triangle ADC \text{ are similar by } AAA \text{ criterion } \\ & \implies \dfrac{ BC }{ AB } = \dfrac{ CD }{ AC } \\ & \implies \dfrac{ 5 }{ 13 } = \frac{ x }{ 12 } \\ & \implies x = 4.62 \end{align}
4. Hence, the value of $x$ is $4.62.$

You can reuse this answer
Creative Commons License