Question

The vertices of a $∆\mathrm{ABC}$ are $\mathrm{A}(4,6),\mathrm{B}(1,5)$ and $\mathrm{C}(7,2)$ . A line is drawn to intersect sides $\mathrm{AB}$ and $\mathrm{AC}$ at $\mathrm{D}$ and $\mathrm{E}$ respectively, such that . Calculate the area of the $∆\mathrm{ADE}$ and compare it with area of $∆\mathrm{ABC}$. (Recall Theorem $6.2$ and Theorem $6.6$)

Answer 1

Finding the area of $∆\mathrm{ADE}$:

Theorem 6.2: The line must be parallel to the third side if it divides any two sides of a triangle in the same ratio (the opposite of the Basic Proportionality Theorem).

Theorem 6.6: The square of the ratio of the corresponding sides of two comparable triangles is equal to the ratio of their areas.

The vertices of a triangle are given $\mathrm{A}(4,6),\mathrm{B}(1,5)$ and $\mathrm{C}(7,2)$

$\mathrm{AB}$and $\mathrm{AC}$ are divided in the ratio $1:3$ by points $\mathrm{D}$ and $\mathrm{E}$ respectively.

The triangle's area will now be determined:
The formula is as follows:

a triangle's area Triangle's area is = The following is how to calculate $∆\mathrm{ABC}$

$\begin{array}{l}\Rightarrow \frac{1}{2}\left[4\left(5-2\right)+1\left(2-6\right)+7\left(6-5\right)\right]\\ \Rightarrow \frac{1}{2}\left(12-4+7\right)\\ \Rightarrow \frac{15}{2}\text{sq unit}\end{array}$

The following formula can be used to compute the area of a $∆\mathrm{ADE}$:

$\begin{array}{l}\Rightarrow \frac{1}{2}\left[4\left(\frac{23}{4}-5\right)+\frac{13}{4}\left(5-6\right)+\frac{19}{4}\left(6-\frac{23}{4}\right)\right]\\ \Rightarrow \frac{1}{2}\left(3-\frac{13}{4}+\frac{19}{16}\right)\\ \Rightarrow \frac{1}{2}\left(\frac{15}{16}\right)\\ \Rightarrow \frac{15}{32}\mathrm{sq}.\mathrm{unit}\end{array}$

As a result, the ratio of the area of$∆\mathrm{ADE}$ to the area of $∆\mathrm{ABC}$ is $1:16$