Faded-Maximus
New member
- Joined
- Sep 24, 2006
- Messages
- 31
The right angle triangle has legs of length r and a right angle at O. A quarter of a circle is consturcted with the centre O and a radius of r. A semicircle is construted with a diameter of AB. Prove that the area of the shaded lune is equal to the area of triangle AOB.
Area of the triangle: (bh) / 2 = (r^2) / 2
Area of quarter circle: (pi r^2) / 4
Area of semi circle: (pi r^2) / 2
To get the area of the little lune you would subtract the area of the quarter circle by the triangle:
[(pi r^2) /4] - [(r^2) / 2]
. . .= [(pi r^2) / 4] - [(2r^2) / 4]
. . .= (pi - r^2) / 4
To get the area of the big lune (shaded area) you would subtract the area of the semicircle by the area of the little lune.
[(pir^2) / 2] - [(pi - r^2) / 4]
. . .= [2(pir^2) / 4] - [(pi - r^2) / 4]
. . .= (pi + 3r^2) / 4
The area of the trinagle does not equal the area of the shaded region. I have made a mistake somewhere, but am unsure where. Can somebody please help me out? Thanks.
Area of the triangle: (bh) / 2 = (r^2) / 2
Area of quarter circle: (pi r^2) / 4
Area of semi circle: (pi r^2) / 2
To get the area of the little lune you would subtract the area of the quarter circle by the triangle:
[(pi r^2) /4] - [(r^2) / 2]
. . .= [(pi r^2) / 4] - [(2r^2) / 4]
. . .= (pi - r^2) / 4
To get the area of the big lune (shaded area) you would subtract the area of the semicircle by the area of the little lune.
[(pir^2) / 2] - [(pi - r^2) / 4]
. . .= [2(pir^2) / 4] - [(pi - r^2) / 4]
. . .= (pi + 3r^2) / 4
The area of the trinagle does not equal the area of the shaded region. I have made a mistake somewhere, but am unsure where. Can somebody please help me out? Thanks.