Question

In the figure, D is the mid point of side BC of $\Delta ABC$ and E is the midpoint of AD. Then
the area of $\Delta ABE=$

• A. $\frac{1}{3}$ area $\left(\Delta ABC\right)$
• B. $\frac{1}{2}$ area $\left(\Delta AEC\right)$
• C. $\frac{1}{2}$ area $\left(\Delta BEC\right)$
• D. $\frac{1}{4}$ area $\left(\Delta ABC\right)$

Answer 1

The correct option is D $\frac{1}{4}$ area $\left(\Delta ABC\right)$

We know that, the median of a triangle divides the triangle into two triangles with equal areas

$△ABC$ has median $D$ from vertex $A$

$\therefore Areaof△ADB=Areaof△CAD$

Consider $△ABD$,

$△ABD$ has median $E$ from vertex $B$

$\therefore Areaof△ABE=Areaof△BDE$

$Areaof△ABC=2\times Areaof△ABD$

$Areaof△ABD=2\times Areaof△ABE$

$\therefore Areaof△ABC=4\times Areaof△ABE$

$\therefore Areaof△ABE=\frac{1}{4}\times Areaof△ABC$