Question

In the given figure, BD : DC = 3 : 5, then area of ΔABD : area of ΔACD is:

Options

• 5 : 3

• 3 : 5

• 25 : 9

• 9 : 25

Answer 1

3 : 5

Explanation:

Both triangles ΔABD and ΔACD share the same vertex A and have their bases BD and DC along the same straight line BC. Therefore, they share a common height from vertex A to the base BC.

The area of a triangle is given by `1/2` × base} × height. Taking the ratio of their areas:

`\frac{\text{area of }\triangle ABD}{\text{area of }\triangle ACD} = \frac{\frac{1}{2} \times BD \times h}{\frac{1}{2} \times DC \times h} = \frac{BD}{DC} = \frac{3}{5}`

Thus, the ratio of the area of ΔABD to the area of ΔACD is 3 : 5.