11 11 votes A square with sides of length $6 \mathrm{~cm} $ is given. The boundary of the shaded region is defined by two semi-circles whose diameters are the sides of the square, as shown. The area of the shaded region is_______ $\mathrm{cm}^2.$ $6 \pi$ $18$ $20$ $9 \pi$ Quantitative Aptitude gate2023-ee quantitative-aptitude geometry circle + – admin 1.6k views answer comment Share Follow 0 reply Please log in or register to add a comment.
10 10 votes This is the answer Krishna Reddy kyp answered Apr 4, 2024 • moved May 12 by Arjun Krishna Reddy kyp comment Share Follow See 1 comment 1 1 comment reply anujs commented Sep 7, 2024 i moved by Arjun May 13 reply Follow flag Imp. point to note: After calculating the area of the non-shaded common region b/w both semicircles. you have to substract it 2 times because it got counted 2 times when calculating areas of both semicircles. 3 3 replyShare Please log in or register to add a comment.
5 5 votes Imagine replicating the semicircle on remaining sides of the squares.Add all four semicircular area (Each overlapping part is added twice) -$= 2*\pi*r^2 = 18\pi$ .... (i)Removing the area of square from (i), gives us the 4 overlapping areas.$18\pi - 36 = 18(\pi-2)$ .... (ii)1 overlapping area $= \frac{18}{4}(\pi-2) = \frac{9}{2}(\pi-2)$Required shaded area $=$ Area of both semicircle $- 2*$Overlapping area $=9\pi - 9\pi + 18 = 18$ chetan-naik answered Oct 30, 2024 • moved May 12 by Arjun chetan-naik comment Share Follow 0 reply Please log in or register to add a comment.
4 4 votes The best way, said by my friend, sreenadh annaluru answered Dec 23, 2024 • moved May 12 by Arjun sreenadh annaluru comment Share Follow See all 2 Comments 2 2 Comments reply ruchirjain commented Jan 24 i moved by Arjun May 13 reply Follow flag WOWWWW! 0 0 replyShare Vasu_Sapehia commented Jan 25 i moved by Arjun May 13 reply Follow flag WoW,the easiest explanation 0 0 replyShare Please log in or register to add a comment.
