Advertisements
Advertisements
प्रश्न
If three circles of radius a each, are drawn such that each touches the other two, prove that the area included between them is equal to `4/25 a^2`. [Take `sqrt(3) = 1.73` and π = 3.14.]
सिद्धांत
Advertisements
उत्तर
When three circles touch each other, their centres form an equilateral triangle, with each side being 2a.
Area of the triangle`=sqrt(3)/4xx2"a"xx2"a" = sqrt(3)"a"^2`
Total area of the three sectors of circles `=3xx60/360xx22/7xx"a"^2 = 1/2xx22/7 "a"^2 = 11/7"a" ^2`
Area of the region between the circles = Area of the triangle - Area of three sectors
`=(sqrt(3)-11/7)"a"^2`
= (1.73 - 1.57)a2
= 0.16 a2
= 0.16 a2
`=4/25"a"^2 `
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
