Question 262981: There are two circles, one circle is inscribed and another circle is circumscribed over a square. What is the ratio of area of inner to outer circle?
Answer by ajmisra(4)   (Show Source):
You can put this solution on YOUR website! It quite easy to solve this one..
Let each of the sides of the square be 'x'
Then, by pythagoras theorem we know the diagonal of the spuare would be
square root of (x^2 + x^2) i.e. (sq root of 2)times 'x'
Now the inscribed circle is within the bounds of the sqaure, hence has a radius of half the sqaure's sides; i.e radius of inscribed circle is 'x/2'
Thus area of inscribed circle is 'pie' times (x/2)^2
For the circumscribed circle the diameter of the circle would be the diagonal of the square which we know to be (sq root of 2)times 'x'
Hence radius of this circumscribed circle would be 1/2 the diameter
Thus area of this circumscribed circle is 'pie' times {1/2 [(sq root of 2)times 'x']}^2
Taking the required ratio we have,
Inscribed Circle Area/Circumscribed Circle Area =
{pie' times (x/2)^2} /{'pie' times {1/2 [(sq root of 2)times 'x']}^2}
i.e. cancelling 'pie' in numerator and denominator we have
{(x^2)/4} / [(sq root of 2)^2 times x^2]/4
further removing the common '4' denominators we have
x^2 / (sq root of 2)^2 times x^2
Further reducing would give us
x^2 / 2 times x^2 (since (sq root of 2)^2 is simply 2)
Again reducing getting rid of x^2 from numerator and denominator gives us:
1/2
Thus,
Inscribed Circle Area/Circumscribed Circle Area = 1/2 as required!
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