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In A B C ,A=(2pi)/3,b-c=3sqrt(3)c m and...

In ` A B C ,A=(2pi)/3,b-c=3sqrt(3)c m` and area of ` A B C=(9sqrt(3))/2c m^2,t h e n (a) `9c m` (b) `18 c m` (c) `27 c m`

A

`6sqrt3 cm`

B

9 cm

C

18 cm

D

27 cm

Text Solution

Verified by Experts

The correct Answer is:
B
`(1)/(2) bc sin.(2pi)/(3) = (9sqrt3)/(2)`
`rArr bc = 18`
Now, we have `b - c = 3sqrt3`
`rArr b^(2) + c^(2) - 36 = 27` [using (1)]
`rArr b^(2) + c^(2) = 63`
Using cosine rule,
`a^(2) = 63 - 2 xx 18 xx cos 120^(@) = 81`
`rArr a = 9 cm`
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