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Area of semicircle (Posted on 2012-12-27) Difficulty: 3 of 5
What is the maximum area of a semicircle inscribed in a square of side s?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Intuitive solution. | Comment 2 of 3 |

With respect, the qbyte solution is unnecessarily abstruse. This is much simpler:

Start with a square, and inscribe a circle inside it. Inscribe another square inside the circle.           
           
By twisting the small square around by half a right angle, we can see that its area is half that of the larger square. (Socrates)           
           
Draw a diagonal of the large square. This splits the circle in half.           
           
We can see at once that if the corners of the small square that are on the diagonal are joined up to the sides of the large one, we will have the desired square.           
           
Using Socrates again, we can see that a square whose side is the radius of the circle will have exactly half the area of the small square. Call this a 'diagonal square'.           
           
Since the large square is twice the size of the small square, the large square is 4 times the size of the diagonal square. Since the circle is larger than the small square, but smaller than the large square, it must be about 3 times the size of the diagonal square.            
           
Call this number (3), as we're not sure it is exactly 3.           
           
So the ratio of the small square to the circle is 2:(3), and the ratio of the small square to half the circle is 4:(3)           
           
Looking at the desired square, we can see that its size is almost exactly half the difference between the small square and the large one. There are just 2 small triangular pieces unaccounted for on the diagonal.           
           
So the ratio of the desired square to the small square is (3/2); not exactly 3/2, but close to it.           
           
Multiplying, the ratio of the desired square to the circle is 4*(3/2):(3).  Since we want the ratio of the circle to the desired square, we simply switch the numbers: (3):4*(3/2)           
           
That is in fact the solution, but we could be more precise with the '(3/2)':            
           
Draw a vertical line through the centre of the diagram.           
           
The side of the desired square is half the side of the small square(ss), plus half the side of the large square(ls).           
           
As we need to square that, we need to know that 1/2*1/2 = 1/4.           
           
But 4 * 1/4 = 1, so we now simply have (3):(ss+ls)^2           
           
Now all we need to know is that (3)='pi' , and ls is 'rt2' times ss: pi/(1+rt2)^2.

Edited on December 28, 2012, 12:17 am
  Posted by broll on 2012-12-28 00:03:14

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