Question

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

• A. \(\frac{25(\sqrt{3}+\sqrt{15})}{2}\)
• B. \(25(\sqrt5+\sqrt{15})\)
• C. \(\frac{25(\sqrt5+\sqrt{15})}{2}\)
• D. \(25(\sqrt3+\sqrt{15})\)

Answer 1

The Correct Option is D

Applying the Cosine Rule in triangle ACD:

The cosine rule states: \( AC^2 = AD^2 + CD^2 - 2 \cdot AD \cdot CD \cdot \cos(∠ADC) \)

Given: \( AD = 10 \) cm, \( CD = X \) cm, \( ∠ADC = 30^\circ \), and \( AC = 20 \) cm.
Substituting these values into the formula:
\( 20^2 = 10^2 + X^2 - 2 \cdot 10 \cdot X \cdot \cos(30^\circ) \)
\( 400 = 100 + X^2 - 20X \cdot \frac{\sqrt{3}}{2} \)
\( 400 = 100 + X^2 - 10X\sqrt{3} \)
\( X^2 - 10X\sqrt{3} - 300 = 0 \)

Solving the quadratic equation for X:
\( X = \frac{-(-10\sqrt{3}) \pm \sqrt{(-10\sqrt{3})^2 - 4(1)(-300)}}{2(1)} \)
\( X = \frac{10\sqrt{3} \pm \sqrt{300 + 1200}}{2} \)
\( X = \frac{10\sqrt{3} \pm \sqrt{1500}}{2} \)
\( X = \frac{10\sqrt{3} \pm 10\sqrt{15}}{2} \)

Since \( X \) represents a length, it must be positive.
\( \Rightarrow X = \frac{10\sqrt{3} + 10\sqrt{15}}{2} \)

Calculating the area of the parallelogram:
The area of a parallelogram is given by the formula: Area = base × height. Alternatively, Area = \( AD \cdot CD \cdot \sin(∠ADC) \)
Area = \( 10 \cdot X \cdot \sin(30^\circ) \)
Area = \( 10 \cdot \left(\frac{10\sqrt{3} + 10\sqrt{15}}{2}\right) \cdot \frac{1}{2} \)
\( = \frac{10}{2} \cdot (10\sqrt{3} + 10\sqrt{15}) \cdot \frac{1}{2} \)
\( = 5 \cdot 10(\sqrt{3} + \sqrt{15}) \cdot \frac{1}{2} \)
\( = \frac{50(\sqrt{3} + \sqrt{15})}{2} \)
\( = 25(\sqrt{3} + \sqrt{15}) \)

Final Answer: \( \boxed{25(\sqrt{3} + \sqrt{15})} \) square units