Question

In the given figure, $E$ is mid-point of $AB$ and $DE$ meets diagonal $AC$ at point $F$.
If $ABCD$ is a parallelogram and area of $△ADF$ is $60c{m}^2$, then area of $△ADB$ is?

Answer 1

Given, parallelogram ABCD E as a mid-point of AB and DE meets diagonal AC at point F.
In $△AEF$ and $△DFC$
$\angle CAE=\angle ACD$ ( Alternate angles, $\because$ AB$\parallel$CD and AC intersects it. )
$\angle AFE=\angle CFD$ ( Vertically opposite angle )
$\angle AED=\angle EDC$ ( Alternate angles, $\because$ AE$\parallel$CD and ED intersects it. )
$\therefore$ $△AEF\sim △DFC$ ( AAA Similarity property)
Using Similarity Property,
$\frac{EF}{FD}=\frac{AE}{CD}$
or,$\frac{EF}{FD}=\frac{AE}{AB}$ (CD$=$AB, As ABCD is a parallelogram )
or, $\frac{EF}{FD}=\frac{1}{2}$ ( As E is mid point of AB , AB$=$2AE)
EF+FD$=$ED
$\therefore$ FD$=\frac{2}{3}$ ED (1)
Drawing an altitude AG from A to meet DE at G.
Area of $△ADE$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times ED\times AG$
Area of $△ADF$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times DF\times AG$
$=\frac{1}{2}\times \frac{2}{3}\times ED\times AG$ ( FD$=\frac{2}{3}$ ED, From 1)
$\frac{Areaof△ADE}{Areaof△ADF}=\frac{\frac{1}{2}\times ED\times AG}{\frac{1}{2}\times \frac{2}{3}ED\times AG}=\frac{3}{2}$
or, Area of $△ADE=\frac{3}{2}\times$ Area of $△ADF$
$=\frac{3}{2}\times$ 60 ${cm}^2$
$=$ 90 ${cm}^2$
Now, In $△ADB$ , E is mid point. So, DE is median.
Median divides the triangle into two triangles of equal areas.
$\therefore$ Area of $△ADE=$ Area of $△BDE=\frac{1}{2}\times$ Area of $△ADB$
or, Area of $△ADB=$ 2 $\times$ Area of $△ADE$
$=$ 2 $\times$ 90 ${cm}^2$
$=$ 180 ${cm}^2$