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, AF

^{2}= AB^{2}– BF^{2}= x^{2}– (x/2)^{2}= 3x^{2}/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.
DF

^{2}= x^{2 }+ (x/2)^{2 }= 5x^{2}/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

r

^{2}= (x/2)^{2}+(x-h)^{ 2}=> r^{2}= (x/2)^{2 }+ {x-(r - x√3/2)}^{ 2}
=> r

^{2}= (x/2)^{2 }+ (x - r + x√3/2)^{2}
=> r

^{2}= x^{2}/4 + x^{2 }+ r^{2 }+ 3x^{2}/4 - 2rx - √3rx + √3x^{2}
=> r

^{2}= (2+√3)x^{2}+r^{2}-(2+√3)rx => (2+√3)x^{2 }= (2+√3)rx^{}
=> x

=> “Length of the
side of the house and the radius of the circle are equal”^{2}-rx = 0 => x(x-r) = 0 => x-r = 0 =>**x = r**
## No comments:

## Post a Comment