Q28. - Geometry

Question

$\overline{AM}$is a median of  $△ABC$

Prove: Area of$△AMB=\frac{1}{2}$ Area of $△ABC$

Step-by-Step Solution

Answer

Hence proved

$\mathbf{Area}\mathbf{of} △AMB=\frac{1}{2}\mathbf{Area}\mathbf{of}△ABC$ 

Step 1. Modify the given diagram.

Given that base$b=16$ and Height$h=5$

Step 2. Concept used.

The median of a triangle is a line segment joining the vertex of the triangle to the mid-point of its opposite side. It bisects the opposite side, dividing it into two equal parts.

Since$AM$ is the median of$△ABC$

Then,

 $\begin{array}{l}BM=MC\\ BC=2BM\end{array}$

Step 3. Use the concept of area of triangle.

Area of$△ABC$  can be found using the formula 

$\begin{array}{c}A=\frac{1}{2}bh\\ =\frac{1}{2}\left(BC\right)\left(h\right)\\ =\frac{1}{2}\left(2BM\right)\left(h\right)\\ =BM\times h\mathrm{...........}\left(i\right)\end{array}$

Area of $△AMB$ can be found using the formula 

$\begin{array}{l}A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(BM\right)\left(h\right)\\ A\left(AMB\right)=\frac{1}{2}A\left(ABC\right)\left[from\right(i\left)\right]\end{array}$

Therefore, the Area $△AMB=\frac{1}{2}$of Area of $△ABC$