Question

State true or false:
The base $BC$ of triangle $ABC$ is divided at $D$ so that $BD=\frac{1}{2}DC$, then the area of $△ABD=\frac{1}{3}$ of the area of $△ABC$

• A. True
• B. False

Answer 1

The correct option is A True

Given, The base BC of triangle ABC is divided at D such that BD$=\frac{1}{2}$DC.
Since, BD$=\frac{1}{2}$ DC, (1)
or, DC$=$ 2BD
Now, BC$=$ BD+DC
or, BC$=$ BD+2BD (As, DC=2BD, From 1)
or, BC$=$ 3BD (2)
Drawing an altitude AE from A to meet BC at E.
Now, Area of $△ABD$ $=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times BD\times AE$
Now, Area of $△ABD$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times BC\times AE$
$=\frac{1}{2}\times 3BD\times AE$ (As,BC=3BD, From 2)
Taking ratio of areas of triangle,
$\frac{Areaof△ABD}{Areaof△ABC}=\frac{\frac{1}{2}\times BD\times AE}{\frac{1}{2}\times 3BC\times AE}=\frac{1}{3}or,Areaof△ABD=\frac{1}{3}\times Areaof△ABC$