△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.

Answer

The given statement is True.

Solution

We know that Distance formula, d = √ ((x2 – x1)2 + (y2 – y1)2)

To determine

\(AB=\sqrt{(2+2)^{2}+ 0}=\sqrt{16}=4\) \(BC=\sqrt{(0-2)^{2}+ (2-0)^{2}}=\sqrt{8}=2\sqrt{2}\) \(CA=\sqrt{(-2-0)^{2}+ (0-2)^{2}}=\sqrt{8}=2\sqrt{2}\) \(DE=\sqrt{(4+4)^{2}+ (0)}=\sqrt{64}=8\) \(EF=\sqrt{(0-4)^{2}+ (4-0)^{2}}=\sqrt{32}=4\sqrt{2}\) \(\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=\frac{1}{2}\)

ΔABC∼ΔDEF

Hence, triangle ABC and DEF are similar.

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