Geometry Forum - Problem of the Week
Solutions - Exploring the 'lunes', Jan. 23-27, 1995
Annie says:The proof of this problem was one that interested folks back in Hippocrates' day, because they thought that if they could have a "curved" figure that had the same area as a "straight" figure, then there might be a way to "square the circle", meaning they might be able to construct a square that had the same area as a given circle. It didn't pan out, but it got people very excited.
Sean Nichols
Hello. This is Sean Nichols, at Burnaby South Secondary School, in Burnaby, BC, Canada. Following is my solution to the Problem of the Week for the week of January 22-27, 1995:If we have the semicircle, with the half square inscribed therein, it can be proved as follows that one quarter of the square (or the triangle ABD) has the _same_area_ as the region between the minor arc AB on the semicircle ABC, and the semicircle AB with diagonal of the side AB:
On the given diagram, we can define the length of the radius of the large semicircle as being x, and the one side of the square as being y. By trigonometry (or any other method), we can find that y = x times sqrt(2).
We can say that the area of the quarter-circle bounded by BD and DA is 1/4*(pi*x^2), or (pi*x^2)/4. The semicircle with edge AB, has an area of 1/2*(pi*(1/2y)^2), or 1/2*(pi*1/4*y^2). But since y = x*sqrt(2), this is the same as 1/2*(pi*2*1/4x^2), or (pi*x^2)/4. Subtracting the region between the Ssemicircle AB and the minor arc AB, we find that the areas of the two regions set out in the problem are the same.
Thank You,
- Sean Nichols
Susan Quan
Annie,Here is the response from Susan Quan Grade 8 Masterman Philadelphia
The area of the triangle and lune are equal.
Let x be the radius of the semicircle ABC.
area of triangle is: 1/2 x^2 /2 = x^2/4
area of segment of the circle: ¼x^2/4 - x^2/4
area of lune: 1/2 ¼(xˆ2/2)^2 - (¼x^2/4 - x^2/4) = ¼x^2 / 4 - ¼x^2/4 + x^2/4 = x^2/4
Eric Wahl
Eric Wahl Grade 11 Masterman Philadelphialet hypotenuse of triangle be x
area of triangle: x^2/4
area of segment of circle: ¼ x^2 / 8 - x^2/4
area of lune: ¼x^2/8 - ¼ x^2 / 8 + x^2/4 = x^2/4
So the area of lune and triangle are equal.
Myron
Cecilia Merediz
Cecilia Merediz, Fairfield HS, grade 9
* * * B
*+++++++* * *
*++++++* *+| * *
*++AL++* *+++| * *
*+++++* A4 *+++++| * *
*++++* *++A2+++| * *
*++* *+++++++++| * *
* * *+++++++++++| * *
A *_____________|_____________* C
r D
A1 is the area of triangle ADB plus the area of 'segment' formed by arc AB and segment AB. A1=1/4 of the area of the circle in which triangle ABC is circumscribed = (pi*r^2)/4.
A2 is the area of triangle ADB. A2 = .5*r*r = (r^2)/2.
A3 is the semicircle with diameter AB. From the Pythagorean Theorem, AB = r*sqrt(2). A3 = .5*pi*(r*sqrt(2)/2)^2 = (pi*r^2)/4.
A4 is the 'segment' formed by arc AB and segment AB. A4 = A1 - area of triangle ADB = (pi*r^2)/4 - (r^2)/2.
AL is the area of the lune. AL = A3 - A4 = (pi*r^2)/4 - ((pi*r^2)/4 - (r^2)/2) = (pi*r^2)/4 - (pi*r^2)/4 + (r^2)/2 = (r^2)/2.
Conclusion: the area of the shaded triangle ABD is equal to the area of the lune.
Ninth Grade Geometry Class
If side AD is 1, then the area of a triangle ABD (shaded triangle) is 1/2. Then the non-shaded segment between semi-circle AB and triangle ABD has an area of (PI/4)-(1/2). The total area of semi-circle AB (radius is (SQUARE-ROOT of 2)/2) is PI/4. Subtract the area of the non-shaded segment from the area of semi-circle AB. The answer comes out to be 1/2, which is the same as the area of triangle ABD.
By:
Ninth-grade geometry class, Wheatland Junior High Class of Mr. Schall
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