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Question 36
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In the given figure, $$PQRSTU$$ is a regular hexagon of side 12 cm. what is the area (in $$cm^2$$) of triangle $$SQU$$?
Solution
Now if we see all t=sides of triangle SQU are the third side of an isosceles triangle with one angle as 120 and other two angles as 30Â
Let us consider triangle PUQÂ
Cos 120 =(12^2+12^2-UQ^2)/2*12*12Â
We get UQ =Â $$12\sqrt{\ 3}$$
Similarly all sides will be $$12\sqrt{\ 3}$$
Now area will be :$$\frac{\sqrt{\ 3}}{4}\times\ \left(12\sqrt{\ 3}\right)^2=108\sqrt{\ 3}\ sq\ cm$$
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