Question

CD is the median of $\Delta$ABC. What is the ratio of the area of $\Delta$BCD to the area of $\Delta$ABC?

• A. ∆BDC is an isosceles triangle
• B. ∆ABC is an isosceles triangle
• C. ∆BDC is an equilateral triangle
• D. ∆ABC is an equilateral triangle

Answer 1

The correct option is B

∆ABC is an isosceles triangle

The median bisects the side opposite to the vertex it is drawn from

Here, $AD=DB=\frac{AB}{2}$

Now, the area of triangle ABC = $\frac{1}{2}\times AB\times CF$

Area of Triangle BCD = $\frac{1}{2}\times BD\times CF$

Since $BD=\frac{AB}{2}$,
Area of Triangle CDB = $\frac{1}{2}\times \frac{AB}{2}\times CF$
$=\frac{1}{2}\times \frac{1}{2}\times AB\times CF$
$=\frac{1}{2}\times$ Area of $\Delta$ ABC

Therefore,
$\frac{Areaof\Delta BCD}{Areaof\Delta ABC}=\frac{1}{2}=1:2$