This question was uploaded on 22/09/21 on social media accounts.
Two concentric circles are arranged such that one chord passing through both circles making equal intercepts (One in the inner circle and the remaining on the outer circle). Find the area between the circles if intercepts having length 'a'.
Solution 1:
Both triangles (x, a/2, r and x, 3a/2, R) are right-angled triangles.
In triangle with side x, a/2, r:
\[x^2=r^2-\left(\frac{a}{2}\right)^2\]
In triangle with side x, 3a/2, R:
\[x^2=R^2-\left(\frac{3a}{2}\right)^2\]
\[\Rightarrow r^2-\left(\frac{a}{2}\right)^2=R^2-\left(\frac{3a}{2}\right)^2\]
\[\Rightarrow R^2-r^2=2a^2\]
Blue Area:
\[A = \pi(R^2-r^2)=2\pi a^2\]
Solution 2:
Using Chord Theorem:
At the point of intersection of diameter and chord:
\[(R-r)(R+r)=(2a)(a)\]\[\Rightarrow R^2-r^2=2a^2\]
Blue Area:
\[A = \pi(R^2-r^2)=2\pi a^2\]
Puzzle related to Geometry, Circle, Area.
