Straight Line Question 220
Question: A regular polygon with equal sides has 9 diagonals. Two of the vertices are at $ A(-1,0) $ and $ B(1,0) $ . Possible areas of polygon is
Options:
A) $ \frac{3\sqrt{3}}{2},2\sqrt{3},6\sqrt{3} $
B) $ 2\sqrt{3},3\sqrt{3},6\sqrt{3} $
C) $ 9\sqrt{3},6\sqrt{3},2\sqrt{3} $
D) $ \frac{3\sqrt{3}}{2},3\sqrt{3},6\sqrt{3} $
Correct Answer: A $ \therefore n=6 $
Now A and B can be adjacent vertices atternate vertices or opposite vertices
If A and B are adjacent then side $ AB=2, $ then
$ area=6\times \Delta OAB $
i.e. area $ =6\times \frac{\sqrt{3}}{4}\times {{(2)}^{2}}=6\sqrt{3} $
If A and B are alternate, then
$ 2\cos 30{}^\circ =a+a\cos 60{}^\circ $ $ \therefore $ Side $ a=\frac{2}{\sqrt{3}} $ $ \therefore $ Area $ =6\times \frac{\sqrt{3}}{4}{{( \frac{2}{\sqrt{3}} )}^{2}}=2\sqrt{3} $
Finally if A and B are opposite vertices then side
$ a=\frac{1}{2}AB=1 $
Then area $ =6\times \frac{\sqrt{3}}{4}{{(1)}^{2}}=\frac{3\sqrt{3}}{2} $Show Answer
Answer:
Solution:
BETA
