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I first draw a line through the centers of the two circles, the length of that line is d. If we consider that line to go along the y-axis of a tilted coordinate system which has its origin in the center of the small circle, then the equations for the two circles become:

x²+(y-d)²=25

and

x²+y²=6.25

If we subtract the lower from the upper, we get

-2dy+d²=18.75

By the Pythagorean theorem we get d

d=sqrt(31.25)

So we put that into our equation

-2*sqrt(31.25)*y+31.25=18.75    

y²=1.25

So we put this y² into

x²+y²=6.25

and we get    

x₁=sqrt(5)

x₂=-sqrt(5)

Which is where the two circles cross eachother and which are the end points of the chord of length c drawn in the figure (notice that the y-value is the same for both).

So

c=2*sqrt(5)    

The area A of a segment cut off from a circle with radius R by the chord of length c is:

A = arcsin(c/(2R))*R² - (1/2) * sqrt(R² - (c/2)²) * c

A = arcsin(sqrt(5)/R)*R² - sqrt(R² - 5) * sqrt(5)

We have two circles to find the segment area of, we call the segment areas a and b. One of the circles has radius 5 cm.

R₁=5

a = arcsin(sqrt(5)/R₁)*R₁² - sqrt(R₁² - 5) * sqrt(5)

a = arcsin(sqrt(5)/5)*25 - sqrt(25 - 5) * sqrt(5)    

a = 1.5912

The other circle one has radius 2.5 cm.

R₂=2.5

b = arcsin(sqrt(5)/R₂)*R₂² - sqrt(R₂² - 5) * sqrt(5)

b = arcsin(sqrt(5)/2.5)*6.25 - sqrt(6.25 - 5) * sqrt(5)    

b = 4.4197

qb = quarter of big circle area

hs = half of small circle area

qb = Pi*5²/4 = 19.6350

hs = Pi*2.5²/2 = 9.8175

sa = square area

sa = 25

fa = fish area

fa = sa -qb -hs +2*(a+b)    

fa = 7.5692 (cm²)