Chapter 14.7, Problem 1P

In triangle ADE shown In Flguro 14.92 , the point B Is haltway from A to D and the point C is halfway from A to E. Compare the areas of triangles ABC and ADE. Explain your reasoning clearly. .

Figure 14.92 Triangles

To determine

To compare: The areas of triangles ABC and ADE .

Explanation of Solution

Consider the diagram as shown below.

  

In the above Figure 1 point B is midpoint of AD and C is the midpoint of AE.

  $AB=\frac{1}{2}AD\text{ and }AC=\frac{1}{2}AE$

In the triangle ABC and ADE ,

The angle A is common.

Since $BC\parallel DE$ the angle ABC and ADE is equal.

Similarly, ACB and AED is equal.

Therefore, triangle ABC and ADEis similar triangles.

The ration of the areas of the similar triangle is equal to the square of the ratio of the corresponding side.

Consider $A_1$ and $A_2$ as the areas of triangle ABC and ADE.

Therefore, the ration of areas of the triangle is,

  $\frac{A_1}{A_2}={\left(\frac{AB}{AD}\right)}^2={\left(\frac{AC}{AE}\right)}^2$ .

Substitute, $2AB$ for $AD$ in above equation as,

  $\begin{array}{l}\frac{A_1}{A_2}={\left(\frac{AB}{2AB}\right)}^2\\ \frac{A_1}{A_2}={\left(\frac{1}{2}\right)}^2\\ \frac{A_1}{A_2}=\frac{1}{4}\end{array}$

Therefore, the areas of the triangle are related as in the ratio of $1:4$ .