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) $.

AcademicMathematicsNCERTClass 10

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.

raja
Updated on 10-Oct-2022 13:28:28

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