Two circles of radii 5 cm and 12 cm are drawn, partly overlapping. Their centers are 13cm apart. Find the area common to the two circles.
I assume you made a sketch
On mine, I labeled the centre of the smaller circle P and the centre of the larger circle Q
There is a common chord, I called that AB
Adding the areas of the two circles would count the overlap twice, so we have to subtract the overlap.
The overlap consists of a segment in the smaller and a segment in the larger circle with the common chord AB
We need some angles.
Consider triangle APQ
We know we have sides 5, 13 and 12
AHHH, but 5^2 + 12^2 = 13^2, so we have a right angle at A !
tan(APQ) = 13/5
angle APQ = 68.96..°
so central angle P = 137.925°
find the area of sector APQ:
area APQ/137.925.. = 25π/360
area of sector APQ = 30.09...
area of triangle APB
= (1/2)(5)(5)sin 137.925
= 8.376..
area of segment in smaller circle
= 30.09... - 8.376.. = 21.714.... ***
Now find the segment in the larger circle and you have the overlap.
We already know that the central angle at Q
= 2(90-68.96) = 42.075..°
So proceed from there.
Once you have the segment area of ABQ, add that to ***
and finally subtract that sum from the addition of the two circle areas.
To find the area common to the two circles, we need to determine the area of the intersection region.
Let's label the centers of the circles as A and B. We are given that the distance between the centers of the circles (AB) is 13 cm.
Next, we can draw the line segment connecting the centers of the circles and extend it until it intersects the outer boundary of both circles. This line segment will divide the area common to the two circles into two regions: a triangle and a segment of a circle.
Let's label the point where the extended line segment intersects the outer boundary of the smaller circle as C, and the point where it intersects the outer boundary of the larger circle as D.
Since we know the radii of the circles are 5 cm and 12 cm, respectively, the length of line segment AC will be 5 cm, and the length of line segment BD will be 12 cm.
Now, we can calculate the length of the line segment CD by subtracting AC (5 cm) from AB (13 cm).
CD = AB - AC = 13 cm - 5 cm = 8 cm
The next step is to calculate the length of the line segment AD, which is the hypotenuse of the right triangle ACD. We can use the Pythagorean theorem to find this length.
AD^2 = AC^2 + CD^2
AD^2 = 5 cm^2 + 8 cm^2
AD^2 = 25 cm^2 + 64 cm^2
AD^2 = 89 cm^2
Taking the square root of both sides, we find:
AD = √89 cm
Now, we can calculate the angle formed by line segments AD and AC using the cosine rule:
cos(∠DAC) = (AC^2 + AD^2 - CD^2) / (2 * AC * AD)
cos(∠DAC) = (5 cm^2 + (√89 cm)^2 - 8 cm^2) / (2 * 5 cm * √89 cm)
cos(∠DAC) = (25 cm^2 + 89 cm - 64 cm^2) / (10 cm * √89 cm)
cos(∠DAC) = (114 cm - 39 cm^2) / (10 cm * √89 cm)
cos(∠DAC) = (114 cm - 39 cm^2) / (10 √89 cm)
Using a calculator, we get:
cos(∠DAC) ≈ 0.4706
Now, we can calculate the area of the sector formed by the two circles using the formula:
Area of sector = (θ/360°) * π * r^2
The angle of the sector is given by 2∠DAC (as there are two circles):
θ = 2 * (∠DAC) = 2 * cos^(-1)(0.4706)
Using a calculator, we get:
θ ≈ 2 * 63.29° = 126.58°
The radius of the smaller circle is 5 cm, so the area of the sector formed by the smaller circle is:
Area_sector_small = (126.58°/360°) * π * (5 cm)^2 = (126.58/360) * π * 25 cm^2
Similarly, the radius of the larger circle is 12 cm, so the area of the sector formed by the larger circle is:
Area_sector_large = (126.58°/360°) * π * (12 cm)^2 = (126.58/360) * π * 144 cm^2
Finally, to calculate the area common to the two circles, we subtract the area of the triangle formed by the line segments AD, AC, and CD from the sum of the areas of the two sectors:
Area_common = Area_sector_small + Area_sector_large - Area_triangle
To find the area of the triangle, we can use the formula for the area of a triangle given two sides and the included angle:
Area_triangle = (1/2) * AD * AC * sin(∠DAC)
Using a calculator, we get:
Area_triangle ≈ (1/2) * (√89 cm) * (5 cm) * sin(cos^(-1)(0.4706))
Now, calculate the Area_common using the above formulas.
To find the area common to the two circles, we need to first determine the points where the two circles intersect or overlap. Let's call the centers of the circles A and B, with A being the circle of radius 5 cm and B being the circle of radius 12 cm.
Since the distance between the centers A and B is 13 cm, we can form a right triangle with sides of length 5 cm, 12 cm, and 13 cm. This is a Pythagorean triple, which indicates that the two circles will intersect at two points.
Now, let's find the coordinates of the two intersection points. Assume that the center A is located at the origin (0, 0) of a coordinate system. The coordinates of the center B would then be (13, 0).
Using the formula for a circle, we can write the equation for each circle:
Circle A: x^2 + y^2 = 5^2
Circle B: (x - 13)^2 + y^2 = 12^2
Now, we can solve these two equations simultaneously to find the coordinates of the intersection points.
Substituting y^2 = 5^2 - x^2 into the equation for Circle B:
(x - 13)^2 + (5^2 - x^2) = 12^2
Expanding and simplifying the equation:
x^2 - 26x + 169 + 25 - x^2 = 144
-26x + 194 = 144
-26x = -50
x = 50/26
Substituting this value back into the equation for Circle A to solve for y:
(50/26)^2 + y^2 = 5^2
y^2 = 25 - (50/26)^2
Now, we have the coordinates of one intersection point (50/26, y) and can use these coordinates to find the other intersection point.
The area common to the two circles can be found by calculating the sum of the areas of the two segments cut by the intersecting chord.
The area of a segment of a circle can be found using the formula:
Area = r^2 * theta/2 - (1/2) * r^2 * sin(theta)
First, let's find the angle theta of each segment. This can be calculated using the formula:
theta = 2 * arccos((d^2 + r1^2 - r2^2)/(2 * d * r1))
Where r1 is the radius of the smaller circle (5 cm), r2 is the radius of the larger circle (12 cm), and d is the distance between the centers of the circles (13 cm).
Using this formula with the known values:
theta = 2 * arccos((13^2 + 5^2 - 12^2)/(2 * 13 * 5))
Now that we have the angle theta, we can calculate the area of each segment and then sum them to find the total area common to the two circles.
Area of segment 1 = (5^2 * theta)/2 - (1/2) * 5^2 * sin(theta)
Area of segment 2 = (12^2 * theta)/2 - (1/2) * 12^2 * sin(theta)
Finally, the total area common to the two circles is the sum of these two segment areas.
Area common to the two circles = Area of segment 1 + Area of segment 2