First slide

Properties of triangles

Question

In a $△ A B C, a = 13 cm, b = 12 cm$ and $c = 5 cm .$ The distance of $A$ from $B C,$ is

Moderate

A

$\frac{144}{13}$
B

$\frac{65}{12}$
C

$\frac{60}{13}$
D

$\frac{25}{13}$

Solution

We have,
$a^{2} = b^{2} + c^{2}$
So, $△ A B C$ is right-angled at $A .$
$∴$ Area of $△ A B C = \frac{1}{2} (A B \times A C) = \frac{1}{2} b c = 30 {cm}^{2}$
Also,
Area of $△ A B C = \frac{1}{2} \times B C \times$ (Distance of $A$ from $B C$ )
$\Rightarrow 30 = \frac{1}{2} \times 13 \times$ Distance of $A$ from $B C$
$\Rightarrow$ Distance of $A$ from $B C = \frac{60}{13}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App