question_answer

ABC is a triangle with base AB. D is a point on AB such that AB = 5 and DB = 3. What is the ratio of the area of \[\Delta ADC\] to the area of\[\Delta ABC\]?

A)  \[\frac{3}{2}\] 

B)  \[\frac{2}{3}\]

C)  \[\frac{3}{5}\]

D)  \[\frac{2}{5}\]

Correct Answer: D

Solution :

Given, AB = 5 cm, DB = 3 can

\[\therefore \]      \[AD=5-3=2\,\,\text{cm}\]

In the figure we can see that both \[\Delta ADC\]and \[\Delta ABC\]the same height h.

Area of a triangle \[=\frac{1}{2}\times \]Base \[\times \]Height

When, height is constant. Area of triangle \[\propto \]base,

\[\therefore \]      \[\frac{\text{Area}\,\,\text{of}\,\,\Delta ADC}{\text{Area}\,\,\text{of}\,\,\Delta ABC}=\frac{AD}{AB}=\frac{2}{5}\]