Question

Show that ΔABC, where A(–2, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 2) are similar triangles.

Answer 1

In ΔABC, the coordinates of the vertices are A(–2, 0), B(2, 0), C(0, 2).

$$\begin{gathered} AB=\sqrt{{\left(2+2\right)}^2+{\left(0-0\right)}^2}=4 \\ BC=\sqrt{{\left(0-2\right)}^2+{\left(2-0\right)}^2}=\sqrt{8}=2\sqrt{2} \\ CA=\sqrt{{\left(0+2\right)}^2+{\left(2-0\right)}^2}=\sqrt{8}=2\sqrt{2} \end{gathered}$$

In ΔPQR, the coordinates of the vertices are P(–4, 0), Q(4, 0), R(0, 4).

$$\begin{gathered} PQ=\sqrt{{\left(4+4\right)}^2+{\left(0-0\right)}^2}=8 \\ QR=\sqrt{{\left(0-4\right)}^2+{\left(4-0\right)}^2}=4\sqrt{2} \\ PR=\sqrt{{\left(0+4\right)}^2+{\left(4-0\right)}^2}=4\sqrt{2} \end{gathered}$$

Now, for ΔABC and ΔPQR to be similar, the corresponding sides should be proportional.

$$\begin{gathered} \mathrm{So},\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{PR} \\ \Rightarrow \frac{4}{8}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2} \end{gathered}$$
Thus, ΔABC is similar to ΔPQR.