Chapter 11.7, Problem 29WE

To determine

To Find: The ratio of their areas.

Answer to Problem 29WE

  $4:5$

Explanation of Solution

Given: $G$ is the intersection point of the median of $\Delta ABC$ . A line through $G$ parallel to $BC$ divides the triangle into two regions.

  

The median of the triangle intersect at a point is two third of the distance from the vertex to the point of the opposite side.

  $\frac{AG}{AF}=\frac{2}{3}$

Now from the figure

  $\begin{array}{l}\angle AGN=\angle AFB\\ \angle ANG=\angle ABF\\ Thus\\ \Delta AGN\sim \Delta AFB\\ therefore\\ \frac{AN}{AB}=\frac{AG}{AF}=\frac{2}{3}\\ similarly\\ \Delta AMN\sim \Delta ACB\\ i.e.\\ \frac{AN}{AB}=\frac{2}{3}\end{array}$

The ratio of sides is

  $\frac{a}{b}=\frac{2}{3}$

Then ratio of areas

  $\frac{a^2}{b^2}=\frac{2^2}{3^2}=\frac{4}{9}$

We can write it as

  $ar(\Delta ACB)=\frac{9}{4}ar(\Delta AMN)$

Now area of the trapezoid = $ar(ABC)-ar(AMN)$

i.e.

  $\begin{array}{l}ar(MNBC)=\frac{9}{4}ar(AMN)-ar(AMN)\\ ar(MNBC)=\frac{5}{4}ar(AMN)\\ now\\ \frac{ar(AMN)}{\frac{5}{4}ar(AMN)}=\frac{4}{5}\end{array}$

Hence $4:5$ .