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Question 65
Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10cm and 20cm, respectively. If the angle $$\angle ADC$$ is equal to $$30^{0}$$ then the area of the parallelogram, in sq.cm is
Solution
Applying cosine rule in triangle ACD,
$$100+X^2-2\times\ 10\times\ X\cos30=400$$
$$X^2-10X\sqrt{\ 3}-300=0$$
Solving, we get X = $$\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)$$
Hence, area = 10Xsin 30 = $$\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$$
= $$25(\sqrt{3}+\sqrt{15})$$
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