CAT 2021 Slot 3 QA Question & Solution
GeometryMedium
Question
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
Options
$\frac{25(\sqrt{5}+\sqrt{15})}{2}$
$25(\sqrt{3}+\sqrt{15})$
$\frac{25(\sqrt{3}+\sqrt{15})}{2}$
${25(\sqrt{5}+\sqrt{15})}$
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 = $\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$
= $25(\sqrt{3}+\sqrt{15})$
