Question

The angle bisector of $\angle A$ divides the side BC of $△ABC$ in the ratio 3 : 5. The area of $△ABD$ = 15 $c{m}^2$. Area of $△$ADC =

• A. 10 $c{m}^2$
• B. 15 $c{m}^2$
• C. 20 $c{m}^2$
• D. 25 $c{m}^2$

Answer 1

The correct option is D 25 $c{m}^2$

AD bisects BC in the ratio 3 : 5

Therefore area of $△ABD$ : area of $△ADC$ = 3 : 5

Given, area of $△ABD$ = 15 $c{m}^2$

area of $△ADC$ = $\frac{5\times 15}{3}$ = 25 $c{m}^2$

Therefore area of $△ADC$ = 25 $c{m}^2$