Chapter 11.2, Problem 28WE

To determine

To Prove: Area of $AMB$ = $\frac{1}{2}$ of area $ABC$ .

Explanation of Solution

Given: $BC=15,h=5$ and $AM$ is the median of $ABC$ .

  

  

Since $AM$ is the median of $ABC$ .

Then

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

Now

Area of triangle $ABC$

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

Now

Area of triangle $AMB$

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

Area of $AMB$ = $\frac{1}{2}$ of area $ABC$ .

Hence, proved.