Question 85

# ABCD is a parallelogram with $$\angle$$ ABC = 60°. 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

Solution

Area of parallelogram = $$ab sin 60 = 15\frac{\sqrt{3}}{2}$$

=> $$\frac{\sqrt{3}}{2} ab = 15 \frac{\sqrt{3}}{2}$$

=> $$ab = 15$$

Using cosine rule in $$\triangle$$ ABD

=> $$cos 120 = \frac{a^2 + b^2 - 7^2}{2 ab}$$

=> $$\frac{-1}{2} = \frac{a^2 + b^2 - 49}{30}$$

=> $$a^2 + b^2 = 49 - 15 = 34$$

Also, $$(a + b)^2 = a^2 + b^2 + 2ab$$

=> $$(a + b)^2 = 34 + 2(15) = 64$$

=> $$(a + b) = \sqrt{64} = 8$$

$$\therefore$$ Perimeter of parallelogram = $$2 (a + b) = 2 \times 8 = 16$$ cm