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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
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
Updated On: Sep 13, 2024
\(\frac{25(\sqrt{3}+\sqrt{15})}{2}\)
\(25(\sqrt5+\sqrt{15})\)
\(\frac{25(\sqrt5+\sqrt{15})}{2}\)
\(25(\sqrt3+\sqrt{15})\)
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The Correct Option is D
Solution and Explanation
Applying cosine rule in triangle ACD
\(100+X ^2 −2× 10× Xcos30=400\)
\(X^ 2 −10X \sqrt3−300=0\)
\(⇒ X = ( \frac{10\sqrt3 + 10\sqrt{15}}{ 2} ) \)
X is the length of one of the sides of the parallelogram, hence it can’t be negative.
\(∴ area = 10Xsin 30 =\frac{ (\frac{ 10\sqrt3 + 10\sqrt{15} }{2} )10 }{2}\)
\(= 25 ( \sqrt3 + \sqrt{15} ) \)
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