Home
Class 12
MATHS
The area of the smaller segment cut off ...

The area of the smaller segment cut off from the circle `x^(2)+y^(2)=9` by x = 1 is

Text Solution

Verified by Experts

The correct Answer is:
`= 9 cos^(-1)frac(1)(3) -2sqrt(2) = 9sec^(-1)3-sqrt(8)` sq. units
Promotional Banner

Topper's Solved these Questions

  • AREA UNDER THE CURVE

    MOTION|Exercise EXERCISE - 1|17 Videos
  • AREA UNDER THE CURVE

    MOTION|Exercise EXERCISE - 2 (LEVEL-I)|25 Videos
  • BASIC MATHEMATIC & LOGARITHM

    MOTION|Exercise Exercise - 4|4 Videos

Similar Questions

Explore conceptually related problems

The area of smaller segment cut off from the circle x^(2)+y^(2)=9 by x=1 is

The area of the smaller part of the circle x^(2)+y^(2)=2 cut off by the line x=1 is

The area of the smaller region bounded by the circle x^(2) + y^(2) =1 and the lines |y| = x +1 is

" The area of the smaller portion of the circle "x^(2)+y^(2)=4" cut-off by the line "x=1" is "((n pi-3sqrt(3))/(3))m^(2)" .Find the value of "n" ."

The area of the smaller portion of the circle x^(2)+y^(2)=4 cut off the line x+y=2 is

Find the area of the smaller part of the circle x^(2)+y^(2)=a^(2) cut off by the line x=(a)/(sqrt(2))

Find the ara of the smaller part of the circle x ^(2) +y ^(2) =a ^(2) cut off by the line x = (a)/(sqrt2).

The area of smaller part between the circle x^(2)+y^(2)=4 and the line x=1 is