Monday, 14 November 2011

Geometry-13

The lower part of the house (see the Figure) in the circle is a square, and the top is an equilateral triangle. Find a relation between the length of the side of the house and the radius of the circle:
Solution follows here:
Solution:
Let ABC be the equilateral triangle and BCDE be square.
Let x be the side of square/triangle. F is the midpoint of the base BC of triangle
=> BF = CF = x/2
Applying Pythagoras to the right triangle AFB, AF2 = AB2 – BF2 = x2 – (x/2)2 = 3x2/4
=> AF = x√3/2
By symmetry, we can say that line joining A and F is diameter. Initially let us assume the centre to be at F.
DF2= x2 + (x/2)2 = 5x2/4 => DF = x√5/2 => AF ≠ DF => F is not the centre and as AF < DF,
Centre will be somewhere away from F down the line AF.
Let O be centre => r = AO = OD
Let OF=h
r = x√3/2 + h => h = r - x√3/2
r2 = (x/2)2+(x-h) 2 => r2 = (x/2)2 + {x-(r - x√3/2)} 2
=> r2 = (x/2)2 + (x - r + x√3/2)2
=> r2 = x2/4 + x2 + r2 + 3x2/4 - 2rx - √3rx + √3x2
=> r2 = (2+√3)x2+r2-(2+√3)rx => (2+√3)x2 = (2+√3)rx
=> x2-rx = 0 => x(x-r) = 0 => x-r = 0 => x = r
=> “Length of the side of the house and the radius of the circle are equal”