Question

14. A triangle BAC is right angled at A. If BD = 16 cm
and CD = 25 cm then AD =
A. 17cm
C.18cm
B. 7.5cm
D.20cm​

Answer 1

Therefore the answer is $D$

Step-by-step explanation:

Given;

A $\bigtriangleup BAC$ is right angled at $A$ ,$BD=16cm$ and $CD=25cm$

From figure, $AD\perp BC$

In $\bigtriangleup BAC$,

$\angle BAC=90degree$

⇒$\angle BAD+\angle DAC=90$___$1$

In $\bigtriangleup DAC$,

$\angle ADC=90degree$

So, $\angle DAC+\angle ACD=90$___$2$

From equation-$1$ and $2$,

$\angle DAC+\angle ACD=\angle BAD+\angle DAC$

⇒$\angle ACD=\angle BAD$

In $\bigtriangleup BAD$ and $\bigtriangleup ADC$;

• $\angle ACD=\angle BAD$ (Hence proved)
• $\angle BDA = \angle ADC$ (each $90degree$)
• $AD$ common side

∴ $\bigtriangleup BAD$ ≅ $\bigtriangleup ADC$

∴$\frac{BD}{AD }= \frac{AD}{DC }= \frac{AB}{AC}$  (Corresponding sides of similar $\bigtriangleup 's$ are proportional)

⇒$AD^{2} =BD\times CD$

⇒$AD=\sqrt{16\times 25}$

∴$AD=20 cm$

∴ The answer is $D$