Question

In the figure given, D divides the side BC, (BD:DC) of $△$ABC in the ratio 3:5. What is the area of $△$ABD in terms of area of $△$ABC?

• A. $\frac{2}{5}\times ar\left(\Delta ABC\right)$
• B. $\frac{3}{5}\times ar\left(\Delta ABC\right)$
• C. $\frac{3}{8}\times ar\left(\Delta ABC\right)$
• D. $\frac{2}{9}\times ar\left(\Delta ABC\right)$

Answer 1

The correct option is C

$\frac{3}{8}\times ar\left(\Delta ABC\right)$

Area of Triangle $=\frac{1}{2}\times base\times height$

Drop a perpendicular from A on BC to get the height of both the triangles.

Here, $\frac{BD}{BC}=\frac{3}{3+5}$

$\Rightarrow \frac{\frac{1}{2}\times BD\times AE}{\frac{1}{2}\times BC\times AE}=\frac{3}{8}$

$\Rightarrow \frac{area\left(\Delta ABD\right)}{area\left(\Delta ABC\right)}=\frac{3}{8}$

$(\because \text{Area of Triangle}=\frac{1}{2}\times b\times h)$

$\therefore area\left(\Delta ABD\right)=\frac{3}{8}\times area\left(\Delta ABC\right)$