Question : If D and E are the mid-points of AB and AC respectively of $\triangle$ABC then the ratio of the areas of $\triangle$ADE and square BCED is:
Option 1: 1 : 2
Option 2: 1 : 4
Option 3: 2 : 3
Option 4: 1 : 3
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Correct Answer: 1 : 3
Solution : DE || BC DE = $\frac{1}{2}×$BC $\therefore$ BC : DE = 2 : 1 Let BC = 2 units, DE = 1 unit Also, $\triangle$ ABC and $\triangle$ ADE are similar, So, $\frac{\text{Area of triangle ABC}}{\text{Area of triangle ADE}}=(\frac{BC}{DE})^2$ ⇒ $\frac{\text{Area of ABC}}{\text{Area of ADE}}= (\frac{2}{1})^2$ ⇒ $\frac{\text{Area of ABC}}{\text{Area of ADE}}= \frac{4}{1}$ Let the area of $\triangle$ABC = 4 units, area of $\triangle$ADE = 1 unit Area of square BCED = Area of $\triangle$ABC – Area of $\triangle$ADE = 4 – 1 = 3 units Area of $\triangle$ADE : Area of square BCED = 1 : 3 Hence, the correct answer is 1 : 3.
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