State whether the following statements are true or false. Justify your answer.
$ \triangle \mathrm{ABC} $ with vertices $ \mathrm{A}(-2,0), \mathrm{B}(2,0) $ and $ \mathrm{C}(0,2) $ is similar to $ \triangle \mathrm{DEF} $ with vertices $ D(-4,0) E(4,0) $ and $ F(0,4) $.
Given:
\( \triangle \mathrm{ABC} \) with vertices \( \mathrm{A}(-2,0), \mathrm{B}(2,0) \) and \( \mathrm{C}(0,2) \) is similar to \( \triangle \mathrm{DEF} \) with vertices \( D(-4,0) E(4,0) \) and \( F(0,4) \).
To do:
We have to find whether the given statement is true or false.
Solution:
We know that,
The distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
Therefore,
In \( \triangle A B C \),
The distance between $A(2,0)$ and $B(2,0)$ is,
$A B=\sqrt{[2-(2)]^{2}+(0-0)^{2}}$
$=4$
The distance between $B(2,0)$ and $C(0,2)$ is,
$B C=\sqrt{(0-2)^{2}+(2-0)^{2}}$
$=\sqrt{4+4}$
$=2 \sqrt{2}$
The distance between $C(0,2)$ and $A-(2,0)$ is,
$C A=\sqrt{[0-(2)^{2}]+(2-0)^{2}}$
$=\sqrt{4+4}$
$=2 \sqrt{2}$
In \( \triangle D E F \),
The distance between $F(0,4)$ and $D(-4,0)$ is,
$F D=\sqrt{(0+4)^{2}+(0-4)^{2}}$
$=\sqrt{4^{2}+(-4)^{2}}$
$=4 \sqrt{2}$
The distance between $F(0,4)$ and $E(4,0)$ is,
$F E=\sqrt{(4-0)^{2}+(0-4)^{2}}$
$=\sqrt{4^{2}+4^{2}}$
$=4 \sqrt{2}$
The distance between $E(4,0)$ and $D(-4,0)$ is,
$E D=\sqrt{[4-(-4)]^{2}+(0)^{2}}$
$=\sqrt{8^{2}}$
$=8$
This implies,
$\frac{A B}{D E}=\frac{4}{8}$
$=\frac{1}{2}$
$\frac{A C}{D F}=\frac{2 \sqrt{2}}{4 \sqrt{2}}$
$=\frac{1}{2}$
$\frac{B C}{E F}=\frac{2 \sqrt{2}}{4 \sqrt{2}}$
$=\frac{1}{2}$
Here,
$\frac{A B}{D E}=\frac{A C}{D F}=\frac{B C}{E F}$
Therefore, by SSS similarity,
$\triangle A B C \sim \triangle D E F$
Hence, the given statement is true.
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