Maximised area
Of these five figures, which shaded area is the greatest? The large circle in each figure has the same radius.
Problem
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-4-1.png?itok=41nIstiJ)
The large circles in each figure has the same radii.
Which shaded area is the greatest?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
Student Solutions
Answer:
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520answer.png?itok=_rVRGqYD)
The large circles could have any radius and the answer will still be the same.
Choose a nice radius for the large circles, e.g. 1 unit
A:
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520a1.png?itok=VnXYRbLa)
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520a2.png?itok=36yuxUqr)
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520a3.png?itok=77Nb1LKs)
Area = $\pi\times1^2$ $ - $ $2\times\left(\pi\times\left(\frac12\right)^2\right)$
= $\pi$ $-$ $\frac12\pi$ = $\frac12\pi = 1.571$
B:
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520b1.png?itok=58KqFuLa)
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520b2.png?itok=mkJApgbb)
Area of a triangle is $\frac12ab\sin{C} = \frac12\times1\times1\times\sin{120} = \frac{\sqrt3} 4 = 0.433$
Total area is $\frac{3\sqrt3}4 = 1.299$
C:
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520c1.png?itok=3Z7Sh6SY)
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520c2.png?itok=bBIgCbMB)
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520c3.png?itok=DHo_yetr)
D:
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520d1.png?itok=ZWtlSg7p)
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520d2.png?itok=Al5Yyc9d)
E:
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520e1.png?itok=ZTzJKIOC)
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520e2.png?itok=zXogF8l6)
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![Maximised Area Maximised Area](/sites/default/files/styles/large/public/thumbnails/content-04-weekly-prob9-maximised%252520area%252520e3.png?itok=q9FvNEq8)
= $\pi$ $-$ $\frac12(1\times1 )\times4 =\pi-2 = 1.142$