Question

ABCD is a parallelogram with $\angle ABC = 60^\circ$. If the longer diagonal is of length 7 cm and the area of the parallelogram ABCD is $15\frac{\sqrt{3}}{2}$ sq.cm, then the perimeter of the parallelogram (in cm) is

Accepted Answer

C. Area = $a b \sin 60^\circ = a b \frac{\sqrt{3}}{2} = 15 \frac{\sqrt{3}}{2} \implies a b = 15$. Adjacent angle $\angle DAB = 180^\circ - 60^\circ = 120^\circ$. Longer diagonal $BD = 7	ext{ cm}$. By law of cosines in $	riangle ABD$: $$BD^2 = a^2 + b^2 - 2ab \cos 120^\circ \implies 49 = a^2 + b^2 - 2(15)(-1/2) = a^2 + b^2 + 15$$ $$a^2 + b^2 = 34$$ $$(a + b)^2 = a^2 + b^2 + 2ab = 34 + 30 = 64 \implies a + b = 8$$ Perimeter $= 2(a + b) = 2 	imes 8 = 16	ext{ cm}$.

Parallelogram Perimeter and Area

Explanation

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