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Conqueror
:
ABCDEF is a regular hexagon of side 12 cm. P. Q and R are the mid points of the sides AB, CD and EF respectively. What is the area (in cm2) of triangle PQR?
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June 16, 2018 at 2:21pm
Kho Yen Hong
:
Let a circle O inscribed in the hexagon. Its diameter can be expressed as [deqn]\sqrt{12^2 + 12 ^2 - 2(12)(12)\cos(120\deg)} = \sqrt{432} cm[/deqn]and so since O is also the circumcircle of traingle PQR, hence the side of traingle can be expressed as[deqn]\sqrt{\frac{432}{4} + \frac{432}{4} - \frac{432}{2}\cos(120\deg)} = 18 cm[/deqn]As such the area of ABC will be equals to[deqn]\frac{1}{2}(18)(18)(\sin [60\deg]) = 81\sqrt{3} cm^2[/deqn]All that I have used is just the cosine rule.
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November 25, 2018 at 11:48am
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