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ricaototti
- Junior | Next Rank: 30 Posts
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- Joined: Wed Feb 27, 2008 4:56 am
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I love Geometry, and in this question I am going to use more weird triangle that one can imagine for this question. See the attached image.
Area of Triangle: (1/2) * Base * Height
Now as shown in the image,
angle OAC = angle OBC (always because OA and OB are equal which make this triangle is isosceles triangle.)
Now let's say this angle OAC = x degree.
So applying trigonometry properties, lets find the height of the given triangle and base of the given triangle.
In triangle OCA,
OC/AO = Sin x
i.e. OC = AO * Sin x (Height of the triangle)
Now as AO = 1 (radius of the circle), therefore OC = Sin x
Now again in triangle OCA,
AC/AO = Cos x
i.e. AC = AO Cos x [1/2 base of the triangle)
Now as AO = 1 (radius of the circle), therefore AC = Cos x
therefore, AB = 2 * AC = 2 Cos x
Now apply the formula of area of the triangle.
area = (1/2) * base * height
area = (1/2) * OC * AB
area = (1/2) * Sin x * 2 * Cos x
area = (1/2) * Sin 2x (by the property Sin 2x = 2 * Sin x * Cos x)
Now finally area = (1/2) of Sin 2x
Now apply Maxima on the both sides,
Max (area) = Max ((1/2) * Sin 2x)
Max (area) = (1/2) * Max (Sin 2x)
(As we know that the maximum value of Sine of any angle = 1, therefore we can replace Max (Sin 2x) = 1)
Hence Max (area) = (1/2) * 1 = (1/2)
Hope this clear your doubt.
