An equilateral triangle PQR is drawn inside a unit square LQST such that Q corresponds to the common vertex of the square and the triangle.
Determine the maximum possible area of the triangle PQR.
The largest possible equilateral triangle will be one that 'splits' the corner of the square, thus leaving equal (15º) angles on each side.
Let's call the length of each side of the triangle,
x. The area of that triangle in terms of
x is:
A = (x²√3)/4.
Consider the triangle QRS where R sits on segment ST. The angles of this triangle are as follows:
Angle QSR = 90º
Angle QRS = 75º
Angle RQS = 15º
By the law of sines, [QS / sin(QRS) = QR / sin(QSR)]
1 / sin(75º) = x / sin(90º)
x = (1/0.9659) = 1.0353
A = 0.4641 square units
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Posted by hoodat
on 2007-10-02 14:27:28 |
Solution
