Question : If areas of similar triangles $\triangle {ABC}$ and $\triangle {DEF}$ are $ {x}^2 \ \text{cm}^2$ and $ {y}^2 \ \text{cm}^2$, respectively, and EF = a cm, then BC (in cm) is:
Option 1: $\frac{y^2}{a^2 x^2}$
Option 2: $\frac{y}{a x}$
Option 3: $\frac{ax}{y}$
Option 4: $\frac{a^2 x^2}{y^2}$
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Correct Answer: $\frac{ax}{y}$
Solution : Given: $\triangle{ABC}\sim\triangle{DEF}$ $\frac{\text{Area of}\triangle{ABC}}{\text{Area of}\triangle{DEF}}=\frac{{BC}^2}{{EF}^2}$ ⇒ $\frac{ {x}^2 }{{y}^2 }=\frac{{BC}^2}{{a}^2}$ ⇒ $\frac{ {x}}{ {y}}=\frac{{BC}}{{a}}$ $\therefore {BC} = \frac{{ax}}{{y}}$ Hence, the correct answer is $\frac{{ax}}{{y}}$.
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Question : D, E, and F are the midpoints of the sides BC, CA, and AB, respectively of a $\triangle ABC$. Then the ratio of the areas of $\triangle DEF$ and $\triangle ABC$ is:
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