Question

Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), F(4, 0) and E(0, 4). [CBSE 2017]

Answer 1

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

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

In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).

$$\begin{gathered} \mathrm{DF}=\sqrt{{\left(4+4\right)}^2+{\left(0-0\right)}^2}=8 \\ \mathrm{FE}=\sqrt{{\left(0-4\right)}^2+{\left(4-0\right)}^2}=4\sqrt{2} \\ \mathrm{DE}=\sqrt{{\left(0+4\right)}^2+{\left(4-0\right)}^2}=4\sqrt{2} \end{gathered}$$

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

$$\begin{gathered} \mathrm{So},\frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{FE}}=\frac{\mathrm{AB}}{\mathrm{DE}} \\ \Rightarrow \frac{4}{8}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{2\sqrt{2}}{4\sqrt{2}} \\ \Rightarrow \frac{1}{2}=\frac{1}{2}=\frac{1}{2} \\ \mathrm{Since},\mathrm{the}\mathrm{corresponding}\mathrm{sides}\mathrm{are}\mathrm{proportional} \\ \mathrm{Therefore},\mathrm{given}\mathrm{two}\mathrm{triangles}\mathrm{are}\mathrm{similar}. \end{gathered}$$