question_answer

\[\Delta ABC\]with vertices A(-2, 0), B(2, 0) and C(0,2) is similar to \[\Delta DEF\]with vertices \[D\left( -4,\,0 \right),\,E\left( 4,\,0 \right)\]and F (0, 4).

A) True

B) False

C) Can't say

D) Partially True/False

Correct Answer: A

Solution :

\[\therefore \] Distance between A (-2,0) and B (2, 0),

\[AB=\sqrt{{{\left[ 2-\left( -2 \right) \right]}^{2}}+{{\left( 0-0 \right)}^{2}}}\]

\[=\sqrt{{{\left( 4 \right)}^{2}}+0}=\sqrt{16}=4\]units

[\[\because \]distance between the points \[\left( {{x}_{1}},\,{{y}_{1}} \right)\]and

\[\left( {{x}_{2}},\,{{y}_{2}} \right),\,d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}]\]

Similarly, distance between 5(2, 0) and

\[C\left( 0,\,2 \right),\,BC=\sqrt{{{\left( 0-2 \right)}^{2}}+{{\left( 2-0 \right)}^{2}}}\]

\[=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\] units

In \[\Delta ABC\], distance between C(0, 2) and A (-2, 0),

\[CA=\sqrt{{{\left[ 0-\left( -2 \right) \right]}^{2}}+{{\left( 2-0 \right)}^{2}}}\]

\[=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\]units

Distance between F(0, 4) and -D(-4, 0),

\[FD=\sqrt{{{\left( 0+4 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}=\sqrt{{{4}^{2}}+{{\left( -4 \right)}^{2}}}\]

\[=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\] units

Distance between F(0, 4) and E(4, 0),

\[FE=\sqrt{{{\left( 4-0 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}=\sqrt{{{4}^{2}}+{{\left( -4 \right)}^{2}}}\]

\[=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\]units

and distance between E(4, 0) and D(-4, 0),

\[ED=\sqrt{{{\left[ 4-\left( -4 \right) \right]}^{2}}+{{\left( 0 \right)}^{2}}}\]

\[=\sqrt{{{\left( 4+4 \right)}^{2}}+0}=\sqrt{{{8}^{2}}}=\sqrt{64}=8\] units

Now, \[\frac{AB}{DE}=\frac{4}{8}=\frac{1}{2},\,\frac{AC}{DF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}\],

\[\frac{BC}{EF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}\]

\[\therefore \,\,\,\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}\]

Here, we see that sides of \[\Delta ABC\] and \[\Delta FDE\]are propotional.

Hence, both the triangles are similar.

[by SSS rule]