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  • Traingle ABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to TraingleDEF with vertices D(–4, 0), E(4, 0) and F(0, 4).

\bigtriangleupABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to \bigtriangleupDEF with vertices D(–4, 0), E(4, 0) and F(0, 4).

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Answer [False]
Solution.        
 
 Distance\, Formula= \sqrt{\left ( {x_{2}-x_{1}} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
In  \bigtriangleupABC,
AB= \sqrt{\left ( 2+2 \right )^{2}+\left ( 0-0 \right )^{2}}= \sqrt{\left ( 4 \right )^{2}}= 4
BC= \sqrt{\left ( 0-2 \right )^{2}+\left ( 2-0 \right )^{2}}= \sqrt{\left ( 4+4 \right )}= \sqrt{8}= 2\sqrt{2}
AC= \sqrt{\left ( 0+2 \right )^{2}+\left ( 2-0 \right )^{2}}= \sqrt{\left ( 4+4 \right )}= 2\sqrt{2}
In \bigtriangleupDEF,
DE= \sqrt{\left ( 4+4 \right )^{2}+\left ( 0-0 \right )^{2}}= \sqrt{\left ( 8 \right )^{2}}= 8
EF= \sqrt{\left ( 0-4\right )^{2}+\left ( 4-0 \right )^{2}}= \sqrt{16+16}= \sqrt{32}= 4\sqrt{2}
DF= \sqrt{\left ( 0+4\right )^{2}+\left ( 4-0 \right )^{2}}= \sqrt{16+16}= 4\sqrt{2}
Here,\frac{AB}{DE}= \frac{BC}{EF}= \frac{AC}{DF}
\frac{4}{8}= \frac{2\sqrt{2}}{4\sqrt{2}}= \frac{1}{2},
Hence, \bigtriangleupABC is similar to \bigtriangleupDEF i.e \bigtriangleup ABC\sim \bigtriangleup DEF
 

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