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”