Question : ABCD is a trapezium with AD and BC parallel sides and E is a point on BC. The ratio of the area of ABCD to that of AED is:
Option 1: $\mathrm{\frac{AD}{BC}}$
Option 2: $\mathrm{\frac{BE}{EC}}$
Option 3: $\mathrm{\frac{AD+BE}{AD+CE}}$
Option 4: $\mathrm{\frac{AD+BC}{AD }}$
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Correct Answer: $\mathrm{\frac{AD+BC}{AD }}$
Solution :
The area of a trapezium,
$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$\text{Area}_{ABCD} = \mathrm{\frac{1}{2} \times (AD + BC) \times EF}$
The area of triangle AED,
$\text{Area}_{AED} = \mathrm{\frac{1}{2} \times AD \times EF}$
The ratio of the area of ABCD to that of AED,
$\mathrm{\frac{\text{Area}_{ABCD}}{\text{Area}_{AED}} = \frac{\frac{1}{2} \times (AD + BC) \times EF}{ \frac{1}{2} \times AD \times EF} = \frac{AD + BC}{AD}}$
Hence, the correct answer is $ \mathrm{\frac{AD + BC}{AD}}$.
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