Area of parallelogram ABCD

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 $\angle ADC$ is equal to $30^\circ$ then the area of the parallelogram, in sq. cm, is

Accepted Answer

D. Let $CD = x$. In $	riangle ADC$, using Cosine Rule for $\angle ADC = 30^\circ$: $$AC^2 = AD^2 + CD^2 - 2(AD)(CD)\cos 30^\circ$$ $$20^2 = 10^2 + x^2 - 2(10)(x)\frac{\sqrt{3}}{2}$$ $$400 = 100 + x^2 - 10\sqrt{3}x \implies x^2 - 10\sqrt{3}x - 300 = 0$$ Solving for positive $x$: $$x = \frac{10\sqrt{3} + \sqrt{300 + 1200}}{2} = 5\sqrt{3} + 5\sqrt{15}$$ Area of parallelogram = $AD \cdot CD \cdot \sin 30^\circ = 10 \cdot x \cdot \frac{1}{2} = 5x$: $$	ext{Area} = 5(5\sqrt{3} + 5\sqrt{15}) = 25(\sqrt{3} + \sqrt{15})$$.

Answer

$\frac{25(\sqrt{3}+\sqrt{15})}{2}$

Answer

$25(\sqrt{5}+\sqrt{15})$

Answer

$\frac{25(\sqrt{5}+\sqrt{15})}{2}$

Answer

$25(\sqrt{3}+\sqrt{15})$

Area of parallelogram ABCD

Explanation

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