Question

In the figure , BC=7, BD=3.
Then write the ratio
$\frac{ A(∆ABD) }{A(∆ABC)}$
​

Answer 1

Solution:

The bases BD and BC of ∆ABD and  ∆ABC are in the same line and have a common vertex A.

∴ their height is common.

Areas of triangles with the same height are proportional to their corresponding bases.

$\frac{A(∆ABD) }{A(∆ABC) } = \frac{BD}{BC} = \frac{3}{7}$

$Ans. \: \: \: \: \frac{A(∆ABD) }{A(∆ABC) } = \frac{BD}{BC} = \frac{3}{7}$

Answer 2

Step-by-step explanation:

in fig base are same

therefore, the areas proportional to the base

dc = bc-bd

= 7-3

= 4

therefore they have eqale height

areas of triangle with eqale heigh are proportional to the their corresponding base

A(triangle ABD) = 7/3

A(triangle ABC).

therefore the ratio is

7:3