Question

$AD$ is the median of the $△ABC$. If the area of $△ABD=10$ sq. units, find the area of $△ABC$

• A. $20$ sq. units
• B. $30$ sq. units
• C. $40$ sq. units
• D. $60$ sq. units

Answer 1

The correct option is A $20$ sq. units

Lets drop a perpendicular from $A$ to the side $BC$

Since, $AD$ is the median, so, $BD=DC$

Area of $△ABD=\frac{1}{2}\times BD\times h$ ... (i)

Area of $△ADC=\frac{1}{2}\times DC\times h$ ... (ii)

Since, $BD=CD\Rightarrow$ (i) $=$ (ii)

$\Rightarrow$ Area of $△ABD=$ Area of $△ADC=10$ sq. units

Now, area of $△ABC=$ area of $△ABD+$ area of $△ADC$

$=10+10=20$ sq. units

Alternatively,
According to the theorem:
The median of a triangle divides the triangle into two triangles of equal areas.

$BD$ divides the $△ABC$ into two triangles, $△ABD$ and $△ADC$.
Area of $△ABD=$ area of $△ADC=10$ sq. units.

Now, area of $△ABC=$ area of $△ABD+$ area of $△ADC$
$=10+10=20$ sq. units