Question

The angle bisector of $\angle A$ divides the side BC of $△ABC$ in the ratio 2 :3. The area of $△ABD$ = 20 $c{m}^2$. Area of $△$ ADC =

• A. 30 $c{m}^2$
• B. 20 $c{m}^2$
• C. 10 $c{m}^2$
• D. 12 $c{m}^2$

Answer 1

The correct option is A

30 $c{m}^2$

AD bisects BC in the ratio 2 : 3

Therefore area of $△ABD$ : area of $△ADC$ = 2 : 3

Given, area of $△ABD$ = 20 $c{m}^2$

Therefore area of $△ADC$ = 30 $c{m}^2$