Question

In the figure give, D divides the side BC of $\Delta ABC$ in the ratio 3:5. What is the area of $\Delta ABD?$

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

Answer 1

The correct option is D $\frac{3}{8}\times ar\left(\Delta ABC\right)$


Let the perpendicular height of $\Delta ABC=h$
$Area\left(\Delta ABC\right)=\left(\frac{1}{2}\right)\times BC\times handarea\left(\Delta ABD\right)=\left(\frac{1}{2}\right)\times BC\times h$
Now, D divides BC in the ratio 3:5
So, BD : DC = 3 : 5
And BD : BC = BD : (BD + DC)
= 3 : (3 + 5)
= 3 : 8
So, $BD=\left(\frac{3}{8}\right)\times BC$
$Area\left(\Delta ABD\right)=\left(\frac{1}{2}\right)\times \left(\frac{3}{8}\right)\times BC\times h\left(\frac{3}{8}\right)\{\left(\frac{1}{2}\right)\times BC\times h\}=(\frac{3}{8}\times Area(\Delta ABC\left)\right)$