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helpless06
Nov 12, 2007, 11:11 AM
FInd the area of the figure given on the coordinate plane (XOY) as a system of inequalities:
x(x + y - sqrt2) < 0
x^2 + y^2 < 2
CaptainRich
Nov 12, 2007, 11:13 AM
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nycfunction
Nov 12, 2007, 05:58 PM
FInd the area of the figure given on the coordinate plane (XOY) as a system of inequalities:
x(x + y - sqrt2) < 0
x^2 + y^2 < 2
Where's the figure? Your question states "figure given" but you forgot to upload the picture of the figure.
terryg752
Nov 13, 2007, 06:42 PM
FInd the area of the figure given on the coordinate plane (XOY) as a system of inequalities:
x(x + y - sqrt2) < 0
x^2 + y^2 < 2
x^2 + y^2 < 2
The above figure consists of all points within the circle x^2 + y^2 = 2 i.e. circle whose centre is the Origin and whose radius is sqrt 2
Regarding second condition: x(x + y - sqrt2) < 0
This is possible when
1. x is negative but x + y - sqrt2 is positive.
Note: x + y - sqrt2 = 0 is a straight line as follows:
Suppose the circle cuts positive side of X axis at A and positive side of Y axis at B
then the line is AB
So x + y - sqrt2 is positive only for points lying above AB
None of the points in the circle meet the above TWO conditions
2. x is positive but x + y - sqrt2 is negative.
All points within the circle, on right of Y-axis, but below the straight line AB
meet the above 2 conditions.
The figure consists of:
Triangle OAB + a quarter of the circle
AREA = 1/2(OA times OB) + 1/4 pi r^2
OA = OB = sqrt 2
r = sqrt 2
AREA = 1/2 2 + 1/4 pi 2
= 1 + 1/2 pi
= Approx 1 + 1/2 3.14
= 2.57