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
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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