# ABC is a right angle triangle. AD is drawn perpendicular to BC. If BC = 9cm, BD = 4 cm then find AB?

Jan 7, 2017

$A B = 6$

#### Explanation:

As $A D$ is drawn perpendicular to $B C$ in right angled $\Delta A B C$, it is apparent that $\Delta A B C$ is right angled at $\angle A$ as shown below (not drawn to scale). As can be seen $\angle B$ is common in $\Delta A B C$ as well as $\Delta D B A$ (here we have written two triangles this way as $\angle A = \angle D$, $\angle B = \angle B$ and $\angle C = \angle B A D$) - as both are right angled (obviously third angles too would be equal) and therefore we have

$\Delta A B C \approx \Delta D B A$ and hence

$\frac{B C}{A B} = \frac{A B}{B D} = \frac{A C}{A D}$..............(1)

therefore, we have $\frac{B C}{A B} = \frac{A B}{B D}$ or

$A {B}^{2} = B C \times B D = 9 \times 4 = 36$

Hence $A B = 6$

Jan 7, 2017 Given that $\Delta A B C$ is right angled and $A D$ is drawn perpendicular to $B C$ . So $\angle B A C$ is right angle.

Given $B C = 9 c m \mathmr{and} B D = 4 c m \to C D = B C - B D = 5 c m$

In $\Delta A B C \mathmr{and} \Delta A B D$

• $\angle A B D = \angle A B C$

• $\angle A D B = \angle B A C = \text{right angle}$

• $\angle B A D = \angle A C D \left(\text{remaining}\right)$

So $\Delta A B C \mathmr{and} \Delta A B D$ are similar

Hence

$\frac{A B}{B C} = \frac{B D}{A B}$

$\implies A {B}^{2} = B D \times B C = 4 \times 9 = 36$

$\implies A B = \sqrt{36} c m = 6 c m$