This question was uploaded on 22/10/21 on social media accounts.
A rectangle and a semicircle are drawn such that one vertex is the same and one side of the square is passing through the top of the semicircle. The length of the chord from the common vertex to the common point is given. Find the area of the rectangle.
Solution 1:
By two right-angled triangles in the figure:
\[x^2=r^2-a^2=2^2-(r+a)^2\]
\[\Rightarrow r^2-a^2=4-r^2-a^2-2ar\]
\[\Rightarrow r(a+r)=2\]
\[Area=r(a+r)=2\]
Solution 2:
\[r=\frac{a+b}2\]
By right-angled triangle:
\[x^2=2^2-b^2\]
By intersecting chord theorem:
\[x^2=ab\]
\[\Rightarrow 4-b^2=ab\]
\[\Rightarrow (a+b)b=4\]
\[Area = rb = \left(\frac{a+b}2\right)b=2\]Puzzle related to Geometry, Circle, Rectangle, Area.
