Question

Show that ΔABC, where A(22, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 4) are similar triangles. [CBSE 2017]

Answer 1

Disclaimer: Instead of 22 it should be $-$2

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

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

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

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

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

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