Chapter 8.3, Problem 18PSB

To determine

To find: Theorem by which $\Delta ABC$ is similar to $\Delta QPR$

Answer to Problem 18PSB

  $\Delta ABC$ is similar to $\Delta QPR$ by SSS theorem

Explanation of Solution

Given information : Here, we have given that A = $(1,2)$ , B = $(9,8)$ , C = $(1,8)$ , P = $(5,-3)$ , Q = $(-7,6)$ , R = $(-7,-3)$ , AB = $10$ and PQ = $15$ .

Formula used : Distance formula: d = $\sqrt{{\left(x_2-x_1\right)}^2-{(y_2-y_1)}^2}$

Calculation: According to the distance formula, d = $\sqrt{{\left(x_2-x_1\right)}^2-{(y_2-y_1)}^2}$ ,

  

  $\begin{array}{l}\overline{BC}=\sqrt{{(1-9)}^2+{(8-8)}^2}\\ \Rightarrow \overline{BC}=\sqrt{{(-8)}^2}\\ \Rightarrow \overline{BC}=\sqrt{{(8)}^2}\\ \Rightarrow \overline{BC}=8\end{array}$

Similarly,

  $\begin{array}{l}\overline{PR}=\sqrt{{(-7-5)}^2+{(-3+3)}^2}\\ \Rightarrow \overline{PR}=\sqrt{{(-12)}^2}\\ \Rightarrow \overline{PR}=\sqrt{{(12)}^2}\\ \Rightarrow \overline{PR}=12\end{array}$

  $\therefore \frac{BC}{PR}=\frac{8}{12}=\frac{2}{3}$

Now,

  $\begin{array}{l}\overline{QR}=\sqrt{{(-7+7)}^2+{(-3-6)}^2}\\ \Rightarrow \overline{QR}=\sqrt{{(-9)}^2}\\ \Rightarrow \overline{QR}=\sqrt{{(9)}^2}\\ \Rightarrow \overline{QR}=9\end{array}$

And,

  $\begin{array}{l}\overline{AC}=\sqrt{{(1-1)}^2+{(8-2)}^2}\\ \Rightarrow \overline{AC}=\sqrt{{(6)}^2}\\ \Rightarrow \overline{AC}=6\end{array}$

  $\therefore \frac{AC}{QR}=\frac{6}{9}=\frac{2}{3}$

  $\frac{AB}{PQ}=\frac{10}{15}=\frac{2}{3}$ (As, AB = $10$ and PQ = $15$ )

Now, as corresponding sides of $\Delta ABC$ and $\Delta QPR$ are equal,

  $\therefore$$\Delta ABC\sim \Delta QPR$ by SSS rule.