Question : In $\triangle ABC$, D and E are points on the sides AB and AC, respectively, such that DE || BC and DE : BC = 6 : 7. (Area of $\triangle {ADE}$ ) : (Area of trapezium BCED) = ?
Option 1: 49 : 13
Option 2: 13 : 36
Option 3: 13 : 49
Option 4: 36 : 13
Correct Answer: 36 : 13
Solution :
DE : BC = 6 : 7
Theorem Used:
The ratios of the areas of two similar triangles are equal to the square of the ratio of their corresponding sides
Calculation:
Considering $\triangle$ABC and $\triangle$ADE
$\angle$A = $\angle$A (Common)
As DE || BC
$\angle$ADE = $\angle$ABC (corresponding angles)
$\angle$ACB = $\angle$AED (corresponding angles)
Thus by AAA property, $\triangle$ABC and $\triangle$ADE are similar triangles
DE || BC = 6 : 7
$\frac{\text{Area of} \triangle\text{ADE}}{\text{Area of} \triangle\text{ABC}} = (\frac{\text{DE}}{\text{BC}})^2=(\frac{6}{7})^2=\frac{36}{49}$
Let area of $\text{Area of} \triangle\text{ADE}$ and $\text{Area of} \triangle\text{ABC}$ be 36k and 49k respectively.
Area of Trapezium BCDE = Area of $\triangle$ ABC - Area of $\triangle$ ADE
Area of Trapezium BCDE = 49k – 36k
Area of Trapezium BCDE = 13k
$\frac{\text{Area of} \triangle\text{ADE}}{\text{Area of Trapezium BCDE}} = \frac{36k}{13k}=\frac{36}{13}$
Hence, the correct answer is 36 : 13.
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