A diameter PQ is drawn to a circle whose diameter length is 1 m. A square is drawn using the diameter PQ as one of its sides. Assuming that $\pi$ is 22/7, what is the area of the part of the square lying outside the circle ?
3/28 sq. m
11/28 sq. m
15/28 sq. m
17/28 sq. m
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2020
– A square is drawn using the diameter PQ as one of its sides. This means the side length of the square is equal to the length of the diameter, which is 1 m.
– The area of the square is side * side = $1^2 = 1$ sq. m.
– The area of the circle is $\pi r^2 = \pi (1/2)^2 = \pi/4$ sq. m.
– Using $\pi = 22/7$, the area of the circle is $(22/7) / 4 = 22/28 = 11/14$ sq. m.
– Let’s interpret “using the diameter PQ as one of its sides” to mean that the line segment PQ forms one boundary of the square. Let PQ lie on the x-axis from (0,0) to (1,0). The square would then occupy the region $0 \le x \le 1$ and $0 \le y \le 1$ (assuming it’s drawn above PQ). The circle with diameter PQ is centered at the midpoint of PQ, which is (0.5, 0), and has radius 0.5. Its equation is $(x-0.5)^2 + y^2 = 0.5^2 = 0.25$.
– The area of the part of the square lying outside the circle is the Area of the Square minus the Area of the region common to both the square and the circle.
– The square is defined by $0 \le x \le 1$ and $0 \le y \le 1$. The circle is defined by $(x-0.5)^2 + y^2 \le 0.25$.
– The part of the circle within the square is where $y \ge 0$ and $(x-0.5)^2 + y^2 \le 0.25$. This describes the upper semi-circle bounded by the diameter PQ (the base of the square).
– The area of this upper semi-circle is half the area of the full circle = (Area of circle) / 2 = $(\pi/4) / 2 = \pi/8$.
– Using $\pi = 22/7$, the area of the semi-circle is $(22/7) / 8 = 22/56 = 11/28$ sq. m.
– The area of the part of the square lying outside this semi-circular region is Area of Square – Area of the semi-circle.
– Area outside = 1 sq. m – 11/28 sq. m = $(28/28) – (11/28) = (28-11)/28 = 17/28$ sq. m.