Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10 cm and 20 cm, respectively. If the angle \(∠ADC \) is equal to 30° then the area of the parallelogram, in sq. cm, is
\(\frac{25(\sqrt{3}+\sqrt{15})}{2}\)
\(25(\sqrt5+\sqrt{15})\)
\(\frac{25(\sqrt5+\sqrt{15})}{2}\)
\(25(\sqrt3+\sqrt{15})\)
The Correct Option is D
Solution and Explanation
Applying the Cosine Rule in triangle ACD:
Using the cosine rule:
\( AC^2 = AD^2 + CD^2 - 2 \cdot AD \cdot CD \cdot \cos(∠ADC) \)
Given: \( AD = 10 \), \( CD = X \), \( ∠ADC = 30^\circ \), and \( AC = 20 \)
Substituting into the formula:
\( 100 + X^2 - 2 \cdot 10 \cdot X \cdot \cos(30^\circ) = 400 \)
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), we get:
\( X^2 - 10X\sqrt{3} - 300 = 0 \)
Solving the quadratic equation:
\( X = \frac{10\sqrt{3} \pm 10\sqrt{15}}{2} \)
Since \( X \) is the length of a side, it must be positive:
\( \Rightarrow X = \frac{10\sqrt{3} + 10\sqrt{15}}{2} \)
Finding the area of the parallelogram:
Area = base × height = \( 10 \cdot X \cdot \sin(30^\circ) \)
Since \( \sin(30^\circ) = \frac{1}{2} \), we get:
Area = \( 10 \cdot \frac{10\sqrt{3} + 10\sqrt{15}}{2} \cdot \frac{1}{2} \)
\( = \frac{10 \cdot (10\sqrt{3} + 10\sqrt{15})}{4} \)
\( = 25(\sqrt{3} + \sqrt{15}) \)
Final Answer: \( \boxed{25(\sqrt{3} + \sqrt{15})} \) square units
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