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Coordinate Geometry — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsCoordinate Geometry1 Mark
Question
State whether the following statements are true or false. Justify your answer.
$\triangle\text{ABC}$ with vertices A(-2, 0), B(2, 0) and C(0, 2) is similar to $\triangle\text{DEF}$ with vertices D(-4, 0) and F(0, 4).
✓

Answer

True:
$\triangle\text{ABC}\sim\triangle\text{DEF}$
If $\frac{\text{AB}}{\text{DE}}=\frac{\text{AC}}{\text{DF}}=\frac{\text{BC}}{\text{EF}}=\text{k}$
In $\triangle\text{ABC},$
$\Rightarrow AB^2 = [2 - (-2)]^2 + [0 - (0)]^2$
$\Rightarrow AB^2 = (4)^2 + 0$
$\Rightarrow AB^2 = (4)^2$
$\Rightarrow AB = 4 units$
$\Rightarrow BC^2 = (0 - 2)^2 + (2 - 0)^2$
$\Rightarrow BC^2 = 4 + 4$
$\Rightarrow BC^2 = 8$
$\Rightarrow\text{BC}=2\sqrt{2}\text{ units}$
$\Rightarrow AC^2 = [0 - (-2)]^2 + (-2 - 0)^2$
$\Rightarrow AC^2 = 2^2 + 2^2$
$\Rightarrow AC^2 = 4 + 4$
$\Rightarrow AC^2 = 8$
$\Rightarrow\text{AC}=2\sqrt{2}\text{ units}$
In $\triangle\text{DEF},$
$\Rightarrow DE^2 = [4 - (-4)]^2 + (0 - 0)^2$
$\Rightarrow DE^2 = (8)^2$
$\Rightarrow DE^2 = 8 units$
$\Rightarrow EF^2 = (0 - 4)^2 + (4 - 0)^2$
$\Rightarrow EF^2 = 44 + 44$
$\Rightarrow EF^2 = 16 + 16$
$\Rightarrow EF^2 = 32$
$\Rightarrow\text{EF}=4\sqrt{2}\text{ units}$
$\Rightarrow DF^2= [0 - (-4)]^2+ (4 - 0)^2$
$\Rightarrow DF^2 = 16 + 16$
$\Rightarrow DF^2 = 32$
$\Rightarrow\text{DF}=4\sqrt{2}\text{ units}$
Now, $\frac{\text{AB}}{\text{DE}}=\frac{4}{8}=\frac{1}{2}$
$\frac{\text{BC}}{\text{EF}}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}$
$\frac{\text{AC}}{\text{DF}}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}$
$\therefore\frac{\text{AB}}{\text{DE}}=\frac{\text{AC}}{\text{DF}}=\frac{\text{AB}}{\text{DE}}=\frac{1}{2}$
Hence, $\triangle\text{ ABC}\sim\triangle\text{ DEF}.$

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