Question

$△ABC$ with vertices$A(-2,0),B(2,0)$ and$C(0,2)$ is similar to $△DEF$ with vertices $D(-4,0)E(4,0)$ and $F(0,4)$. State whether the following statement is true or false. Justify your answer.

• A. True
• B. False

Answer 1

The correct option is A True

Step 1: Find the length of sides of $△\mathrm{ABC}$

We know that Distance formula,$d=\sqrt{{(x_2–x_1)}^2+{(y_2–y_1)}^2}$

In triangle $△\mathrm{ABC}$

by applying the distance formula

$\begin{array}{rcl}AB& =& \sqrt{\left(\right(2-{(-2))}^2+0}\\ & =& \sqrt{{(2+2)}^2}\\ & =& \sqrt{4^2}\\ & =& \sqrt{16}\\ & =& 4\end{array}$

$\begin{array}{rcl}\mathrm{BC}& =& \sqrt{{(0-2)}^2+{(2-0)}^2}\\ & =& \sqrt{4+4}\\ & =& 2\sqrt{2}\end{array}$

$$\begin{gathered} CA=\sqrt{{(-2-0)}^2+{(0-2)}^2} \\ =\sqrt{4+4} \\ =2\sqrt{2} \end{gathered}$$

Step 2: Find the length of sides of $△DEF$

Now in $△DEF$

$\begin{array}{rcl}\mathrm{DE}& =& \sqrt{(4-{(-4))}^2+0}\\ & =& \sqrt{{(4+4)}^2}\\ & =& \sqrt{64}\\ & =& 8\end{array}$

$\begin{array}{rcl}EF& =& \sqrt{{(0-4)}^2+{(4-0)}^2}\\ & =& \sqrt{16+16}\\ & =& 4\sqrt{2}\end{array}$

$\begin{array}{rcl}FD& =& \sqrt{{(-4-0)}^2+{(0-4)}^2}\\ & =& \sqrt{16+16}\\ & =& 4\sqrt{2}\end{array}$

Step 2: Find the ratio of corresponding sides

we have the ratio of corresponding sides of the triangle ,

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=\frac{1}{2}$

$\therefore \Delta ABC~\Delta DEF$

Hence, triangle $△ABC$ and$△DEF$ are similar.