Question

State whether the following statement is true or false.

Enter $1$ for true and $0$ for false

$△$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).

Answer 1

$If\Delta ABC\&\Delta DEFaresimilarthen$

$\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}.$

$Wecanobtaintheabovelengthsbythedistanceformula$

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$andinvestigateiftheratiosareequal.$

$Statement-$

$\Delta ABC\&\Delta DEFaresimilarsinceallthe3sidesofthetriangles$$areinproportion.$

$Justification-$

$Letusobtainthelengthsofthesidesof\Delta ABC\&\Delta DEF$

$bythedistanceformulad=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}.$

$In\Delta ABCtheverticesareA(2,0),B(-2,0),C(0,2).$

$\therefore AB=\sqrt{{(-2-2)}^2+{\left(0\right)}^2}=4units.$

$SimilarlyBC=\sqrt{{(0+2)}^2+{(2-0)}^2}=2\sqrt{2}units$

$andAC=\sqrt{{(0-2)}^2+{(2-0)}^2}=2\sqrt{2}units.$

$Againin\Delta DEFtheverticesareA(4,0),B(-4,0),C(0,4).$

$\therefore DE=\sqrt{{(-4-4)}^2+{\left(0\right)}^2}=8units.$

$SimilarlyEF=\sqrt{{(0+4)}^2+{(4-0)}^2}=4\sqrt{2}units$

$andDF=\sqrt{{(0-4)}^2+{(4-0)}^2}=4\sqrt{2}units.$

$So\frac{AB}{DE}=\frac{4}{8}=\frac{1}{2}.$

$\frac{AC}{DF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}and$

$\frac{BC}{EF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}$

$So\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}.$

$\therefore Thegivenstatementistrue$

.$Ans-TRUE.$