Question

In $\Delta ABC,\angle ABC={90}^{\circ }AB=3cm$ and $BC=4cm$. If $BD\perp AC$ where $D$ is a point on $AC$, then find $BD$.

• A. $2.4cm$
• B. $3.6cm$
• C. $4.8cm$
• D. $1.2cm$

Answer 1

The correct option is A $2.4cm$

By pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=9+16=25$

$AC=5cm$

$BD$ is the altitude from $B$ onto $AC$

We have the formula for altitude,

$\frac{1}{h^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Applying the same for $△ABC$

$\frac{1}{B{D}^2}=\frac{1}{A{B}^2}+\frac{1}{B{C}^2}$

$\frac{1}{B{D}^2}=\frac{1}{3^2}+\frac{1}{4^2}$

$B{D}^2=\frac{\left(3^2\right)\left(4^2\right)}{\left(3^2\right)+\left(4^2\right)}$

$BD=\frac{\left(3\right)\left(4\right)}{\left(5\right)}$....... $(\because 3^2+4^2=5^2)$

$BD=\frac{12}{5}=2.4cm$