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The area of the disc and its parts
When determining the area of a disc, the goal is to map to a number in accordance with the concept of area. This requires methods of a specific area of mathematics, the analysis. The proofs of the propositions are more advanced than the topics studied in high school, thus we will omit the details.
It is clear that if we inscribe a polygon in a circle, then we add a new vertex, the circumference of the polygon will be closer and closer to that of the circle. This approximation leads to the well-known formula for the circumference of the circle, that is, the circumference of a circle with radius of length \latex{ r } is
\latex{c=2r\times\pi}.
The area of the disc
When inscribed and circumscribed regular polygons are drawn to a disc, one can see that the area of the inscribed polygons are smaller, while that of the circumscribed ones are larger than the area of the disc. If the number of the sides the two polygons is increased, then the area of the inscribed one increases, while the area of the circumscribed one decreases. It can be shown that when the number of sides is increased, their areas flank one single number, in other words, they converge to a specific number. This number is the area of the disc.
If the number of sides is denoted by \latex{ n }, then the following table shows the areas for some cases with the radius of the a circle being \latex{ 1 }. (Figure 34)
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ r }
\latex{ n=4 }
\latex{ n=6 }
\latex{ n=8 }
Figure 34
Number of sides (n) | Inscribed area | Circumscribed area | |
\latex{ 4 } | \latex{ 2 } | < area of the disc < | \latex{ 4 } |
\latex{ 6 } | \latex{ 2.59807... } | < area of the disc < | \latex{ 3.46410… } |
\latex{ 8 } | \latex{ 2.82842... } | < area of the disc < | \latex{ 3.31370… } |
\latex{ 3 } \latex{ 1 } \latex{ 4 }
One: A Poem
\latex{ 1 } \latex{ 5 }
A Raven
\latex{ 9 } \latex{ 2 } \latex{ 6 } \latex{ 5 } \latex{ 3 } \latex{ 5 }
Midnights so dreary, tired and weary,
\latex{ 8 } \latex{ 9 } \latex{ 7 } \latex{ 9 }
Silently pondering volumes extolling
\latex{ 3 } \latex{ 2 } \latex{ 3 } \latex{ 8 } \latex{ 4 }
all by-now obsolete lore.
\latex{ 6 } \latex{ 2 } \latex{ 6 } \latex{ 4 } \latex{ 3 }
During my rather long nap
\latex{ 3 } \latex{ 8 } \latex{ 3 }
– the weirdest tap!
\latex{ 2 } \latex{ 7 } \latex{ 9 } \latex{ 5 }
An ominous vibrating sound
\latex{ 0 } \latex{ 2 } \latex{ 8 } \latex{ 8 }
disturbing my chamber’s antedoor.
\latex{ 4 } \latex{ 1 } \latex{ 9 } \latex{ 7 } \latex{ 1 } \latex{ 6 }
“This”, I whispered quietly, “I ignore”.
One: A Poem
\latex{ 1 } \latex{ 5 }
A Raven
\latex{ 9 } \latex{ 2 } \latex{ 6 } \latex{ 5 } \latex{ 3 } \latex{ 5 }
Midnights so dreary, tired and weary,
\latex{ 8 } \latex{ 9 } \latex{ 7 } \latex{ 9 }
Silently pondering volumes extolling
\latex{ 3 } \latex{ 2 } \latex{ 3 } \latex{ 8 } \latex{ 4 }
all by-now obsolete lore.
\latex{ 6 } \latex{ 2 } \latex{ 6 } \latex{ 4 } \latex{ 3 }
During my rather long nap
\latex{ 3 } \latex{ 8 } \latex{ 3 }
– the weirdest tap!
\latex{ 2 } \latex{ 7 } \latex{ 9 } \latex{ 5 }
An ominous vibrating sound
\latex{ 0 } \latex{ 2 } \latex{ 8 } \latex{ 8 }
disturbing my chamber’s antedoor.
\latex{ 4 } \latex{ 1 } \latex{ 9 } \latex{ 7 } \latex{ 1 } \latex{ 6 }
“This”, I whispered quietly, “I ignore”.
Let us call the resulting number of this approximation the area of the disc. In case of the unit circle, this is exactly the number \latex{\pi} introduced for the circumference of the circle.
Since the unit disc can be transformed into a disc with radius of length \latex{ r } using a similarity with ratio \latex{ r }, the following holds for the area of the disc with radius \latex{ r }:
\latex{t=r^2\times\pi}.
The area of the sector
The plane figure bounded by two radii of a circle and the arc between them is called sector. Its area is proportional to the corresponding central angle of size \latex{\alpha}. (Figure 35)
If we let \latex{ t } to denote the area of the sector belonging to a central angle \latex{\alpha}, then, keeping in mind that the central angle corresponding to the whole disc is \latex{2\pi}, the following ratio can be written:
\latex{i=r\times\alpha}
\latex{\alpha}
\latex{ r }
\latex{ r }
\latex{ t }
\latex{ i }
\latex{ B }
\latex{ A }
\latex{ O }
Figure 35
\latex{\frac{\alpha }{2\pi } =\frac{t}{r^2\times \pi }}.
It follows that the area of the sector is
\latex{t=\frac{r^2\times \alpha }{2} ,}
where \latex{\alpha} is the size of the central angle in radians.
If we recall the definition about the size of angles in radians, that \latex{\alpha =\frac{i}{r} }, where i is the length of the arc belonging to a central angle of size \latex{\alpha}, then we can express the area in a different form:
\latex{t=\frac{r^2\times \alpha }{2} =\frac{r\times r\times \alpha }{2}}, from which
\latex{t=\frac{r^2\times i }{2} .}
If the angles are measured in degrees then the following form can be stated:
\latex{\frac{\alpha }{360°} =\frac{t}{r^2\times \pi }}, from which
\latex{t=\frac{r^2\times \pi \times \alpha }{360°} .}
Example 1
Draw half-discs above the sides of a right angled triangle as seen in Figure 36. Prove that the area of the shaded parts equals to the area of the right angled triangle. (Lunes of Hippocrates)
\latex{ C }
\latex{ B }
\latex{ A }
\latex{ c }
\latex{ b }
\latex{ a }
Figure 36
Solution
Let us denote the sides of the triangle by \latex{ a }, \latex{ b } and \latex{ c }, as usual. The radii of the semi-discs drawn above the sides are \latex{\frac{a}{2},\frac{b}{2}} and \latex{\frac{c}{2},} thus their areas are, respectively, \latex{\frac{a^2\times\pi}{8},\frac{b^2\times\pi}{2}} and \latex{\frac{c^2\times\pi}{8}}.
We can get the area of the lunes if we subtract the area of the semi-disc above the hypotenuse from the sum of the area of the triangle (\latex{ t }) and the area of the semi-discs above the legs:
\latex{t+\frac{a^2\times \pi }{8}+ \frac{b^2\times \pi }{8}-\frac{c^2\times \pi }{8}=t+\frac{\pi }{8}\times (a^2+b^2-c^2)=t,}
since it follows from the Pythagorean theorem the expression in the brackets is \latex{a^2 + b^2 - c^2 = 0}. Therefore the total area of the lunes equals the area of the triangle.
The area of the segment
The area of the segment can be determined using the area of the sector and the triangle. It is obvious that this area is nothing else than the difference (or sum) of a sector and the appropriate triangle. (Figure 37)
In a disc with radius r, the area t of the segment with central angle \latex{\alpha} is
In a disc with radius r, the area t of the segment with central angle \latex{\alpha} is
\latex{t=\frac{r^2\times \alpha }{2}-\frac{r^2\times \sin \alpha }{2},} from which
\latex{t=\frac{r^2}{2}\times (\alpha -\sin \alpha )}
\latex{\alpha}
\latex{\alpha}
\latex{ O }
\latex{ B }
\latex{ A }
\latex{ O }
\latex{ r }
\latex{ r }
\latex{ A }
\latex{ B }
Figure 37
The area of the annulus
Two concentric circles with radii being \latex{ R } and \latex{ r } determine an annulus. Its area can be obtained by subtracting the area of the disc with smaller radius from the one with larger radius. (Figure 38)
\latex{t=R^2\times \pi -r^2\times \pi =\pi \times (R-r)\times (R+r),} from which
\latex{t=2\times \frac{R+r}{2} \times (R-r)\times \pi }
\latex{\rho}
\latex{\rho =\frac{R+r}{2} }
\latex{m=R-r}
\latex{ R }
\latex{ m }
\latex{ r }
Figure 38
The circle which is equally distant from both bounding circle of an annulus is called its midcircle.
Its radius is actually \latex{\rho =\frac{R+r}{2} } and the width of the annulus is \latex{m = R - r.}
Thus the area of the annulus can be expressed using these quantities:
Thus the area of the annulus can be expressed using these quantities:
\latex{t=2\times\rho\times m\times\pi }
The part of an annulus between two radii is called an annulus sector. Let us denote the length of the bounding arcs by \latex{ i_1 } and \latex{ i_2 }, its width by \latex{ m }. (Figure 39)
Since both arcs belong to the same central angle, it follows that
\latex{\frac{i_1}{R}=\frac{i_2}{r},} that is, \latex{i_1\times r=i_2\times R}.
The area of the annulus sector is equal to the difference of two sectors:
\latex{t=\frac{i_1\times R}{2} -\frac{i_2\times r}{2},} from which
\latex{m=R-r}
\latex{ R }
\latex{ R }
\latex{ r }
\latex{ r }
\latex{ i_2 }
\latex{ m}
\latex{ i_1}
Figure 39
\latex{t=\frac{1}{2}\times (i_1\times R-i_2\times r) .}
Since it follows from the equality based on the central angles that \latex{i_2 \times R - i_1\times r = 0}, we can further transform the last formula using this expression:
\latex{t=\frac{1}{2}\times (i_1\times R-i_2\times r+i_2\times R-i_1\times r)=\frac{1}{2} \times (i_1+i_2)\times (R-r) ,} from which
\latex{t=\frac{i_1+i_2}{2}\times m}.
Note: It worth mentioning that the formula for the area of the sector resembles the formula for the area of the triangle, while the formula for the area of the annulus sector resembles the formula for the area of the trapezoid.
Exercises
{{exercise_number}}. The radius of a circle is increased by a factor of \latex{ 3 }. By what factor do the circumference and the area of the circle increase?
{{exercise_number}}. The area of a circle is increased by a factor of \latex{ 2 }. By what factor does its circumference increase?
{{exercise_number}}. The area of a circle is increased by a factor of \latex{ 2 }. By what factor does its circumference increase?
{{exercise_number}}. If we think of the Earth as a perfect sphere, then the length of the Equator would be \latex{ 40,000\, km }. How much more wire would be needed to create a huge circle \latex{ 1\, km } above the equator?
{{exercise_number}}. Determine the area of the sector with the following data known (where r is the radius of the circle, i is the length of the arc and a is the central angle of the sector):
- \latex{r=2\; \text{cm},\;\alpha=60°};
- \latex{r=2\; \text{cm},\;i=3\; \text{cm}};
- \latex{i=2\; \text{cm},\;\alpha=60°}.
{{exercise_number}}. What is the area of the segment whose central angle is \latex{ 70º } and the radius of the circle is \latex{ 2\, m }?
{{exercise_number}}. Determine the area of the annulus sector with the following data given.
- \latex{r=2\; \text{cm},\;R=5\; \text{cm},\;\alpha=30°};
- \latex{i_1=5\; \text{cm},\;i_2=3\; \text{cm},\;\alpha=30°};
- \latex{r=2\; \text{cm},\;\rho=5\; \text{cm},\;\alpha=30°};
- \latex{r=2\; \text{cm},\;\rho=5\; \text{cm},\;i_1=3\;\text{cm}}.
{{exercise_number}}. Determine the size of the dark area in each figure compared to the area of the square.
P, Q, R, S are midpoints
\latex{ R }
\latex{ R }
\latex{ P }
\latex{ P }
\latex{ Q }
\latex{ Q }
\latex{ S }
\latex{ S }
\latex{ a })
\latex{ b })
{{exercise_number}}. Prove that the area of the yellow and deep blue regions in each of the figures below are equal.
\latex{ OP } is an angle bisector
\latex{ O }
\latex{ O }
\latex{ P }
\latex{ b })
\latex{ a })
{{exercise_number}}. In the figure, we have marked the twelve vertices of a regular \latex{ 12 }-gon on a circle. Which one of the regions with different colours have the largest area?
{{exercise_number}}. Two discs with radius of length \latex{ r = 10\, cm } intersect such that their centres are \latex{ 15\, cm } far from each other. Determine the area of the intersection of the two discs.
{{exercise_number}}. The area of a regular \latex{ 12 }-gon is \latex{ 100\, cm^2 }. Find the area of the annulus determined by the inscribed and circumscribed circle of the polygon.
{{exercise_number}}. Three coins with radius being \latex{ r_1= 3\, cm }, \latex{ r_2= 4\, cm } and \latex{ r_3= 5\, cm }, respectively, are placed on a surface such that any two touch. Determine the area of the triangle circumscribed around the circles.