5.8: Circular arcs and circular sectors
- Page ID
- 23459
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Length is defined for straight line segments, and area is defined in terms of rectangles; neither measure is defined for shapes with curved boundaries - unless, that is, they can be cunningly dissected and the pieces rearranged to make a straight line, or a rectangle.
Figure 5: Dumbbell.
Problem 201 Four identical semicircles of radius 1 fit together to make the dumbbell shape shown in Figure 5. Find the exact area enclosed without using the formula for the area of a circle.
In general, making sense of length and area for shapes with curved boundaries requires us to combine a little imagination with what we know about straight line segments and polygons. Our goal here is to lead up to the familiar results for the length of circular arcs and the area of circular sectors. But first we need to explore the perimeter and area of regular polygons, and the surface area of prisms and pyramids.
As so often in mathematics, to make sense of the perimeter and area of regular polygons we need to look beyond their actual values (which will vary according to the size of the polygon), and instead interpret these values as a function of some normalizing parameter - such as the radius. The calculations will be simpler if you first prove a general result.
Problem 202
(a) A regular n-gon and a regular 2n-gon are inscribed in a circle of radius 1. The regular n-gon has edges of length sn = s, while the regular 2 n-gon has edges of length s2n = t. Prove that
(b) A regular “2-gon” inscribed in the unit circle is just a diameter (repeated twice), so has two identical edges of length s2 = 2. Use the result in part (a) to calculate the edge length s4 of a regular 4-gon, and the edge length s8 of a regular 8-gon inscribed in the same circle.
(c) A regular 6-gon inscribed in the unit circle has edge length s6 equal to the radius 1. Use the result in part (a) to calculate the edge length s3 of a regular 3-gon inscribed in the unit circle, and the edge length s12 of a regular 12-gon inscribed in the unit circle.
(d) In Problem 185 we saw that a regular 5-gon inscribed in the unit circle has edge length
Use the result in part (a) to calculate the edge length s10 of a regular 10-gon inscribed in the same circle.
Problem 203
(a) A regular n-gon is inscribed in a circle of radius r.
(i) Find the exact perimeter pn (in surd form): when n = 3; when n = 4; when n = 5; when n = 6; when n = 8; when n = 10; when n = 12.
(ii) Check that, for each n:
for some constant cn, where
(b) A regular n-gon is circumscribed about a circle of radius r.
(i) Find the exact perimeter Pn (in surd form): when n = 3; when n = 4; when n = 5; when n = 6; when n = 8; when n = 10; when n = 12.
(ii) Check that, for each n:
for some constant Cn, where
(c) Explain why c12 < C12.
It follows from Problem 203 that
Hence
- the perimeters pn and Pn of regular n-gons inscribed in, or circumscribed about, a circle of radius r all have the same form:
- The perimeters of inscribed regular n-gons all increase with n, but remain less than the perimeter of the circle, while
- the perimeters of the circumscribed regular n-gons all decrease with n, but remain greater than the perimeter of the circle.
- the perimeter P of the circle appears to have the form P = K × r, where the ratio
satisfies
In particular, the value of the constant K lies somewhere between c42 = 6.21 . . . and C12 = 6.43 . . . . If we now define the quotient K to be equal to “2π”, we see that
2 π r , where π denotes some constant lying between 3.1 and 3.22
In this spirit one might reinterpret the first two bullet points as defining two sequences of constants "πn" and "Πn" for , such that
- (perimeter of a regular n-gon with circumradius r) = 2πnr, where
etc.,
and
- (perimeter of a regular n-gon with inradius r)= 2Πnr, where
etc.,
Moreover
Problem 204 Find the exact length (in terms of π)
(i) of a semicircle of radius r;
(ii) of a quarter circle of radius r;
(iii) of the length of an arc of a circle of radius r that subtends an angle θ radians at the centre.
In the next problem we follow a similar sequence of steps to conclude that the quotient
is also constant. The surprise lies in the fact that this different constant is so closely related to the previous constant K.
Problem 205
(a) A regular n-gon is inscribed in a circle of radius r.
(i) Find the exact area an (in surd form): when n = 3; when n = 4; when n = 5; when n = 6; when n = 8; when n = 10; when n = 12.
(ii) Check that, for each n:
for some constant dn, where
(b) A regular n-gon is circumscribed about a circle of radius r.
(i) Find the exact area An (in surd form): when n = 3; when n = 4; when n = 5; when n = 6; when n = 8; when n = 10; when n = 12.
(ii) Check that, for each n:
for some constant Dn, where
(c) Explain why d12 < D12.
It follows from Problem 205 that
- the areas an and An of regular n-gons inscribed in, or circumscribed about, a circle of radius r all have the same form:
- The areas of inscribed regular n-gons all increase with n, but remain less than the area of the circle, while
- the areas of the circumscribed regular n-gons all decrease with n, but remain greater than the area of the circle, whence
- the area A of the circle appears to have the form A = L × r2, where the ratio
satisfies
In particular, the value of L lies somewhere between d12 = 3 and . The surprise lies in the fact that the constant L is exactly half of the constant K - that is, L = π, so
The next problem offers a heuristic explanation for this surprise.
Figure 6: Circle cut into 8 slices.
Problem 206 A regular 2n-gon ABCDE . . . is inscribed in a circle of radius r. The 2n radii OA, OB,... joining the centre O to the 2n vertices cut the circle into 2n sectors, each with angle (Figure 6).
These 2n sectors can be re-arranged to form an “almost rectangle”, by orienting them alternately to point “up” and “down”. In what sense does this “almost rectangle” have “height = r” and πr?
Problem 207
(a) Find a formula for the surface area of a right cylinder with height h and with circular base of radius r.
(b) Find a similar formula for the surface area of a right prism with height h, whose base is a regular n-gon with inradius r.
Problem 208
(a) Find the exact area (in terms of π)
(i) of a semicircle of radius r;
(ii) of a quarter circle of radius r;
(iii) of a sector of a circle of radius r that subtends an angle θ radians at the centre.
(b) Find the area of a sector of a circle of radius 1, whose total perimeter (including the two radii) is exactly half that of the circle itself.
Problem 209
(a) Find a formula for the surface area of a right circular cone with base of radius r and slant height l.
(b) Find a similar formula for the surface area of a right pyramid with apex A whose base BCDE. . . is a regular n-gon with inradius r.
Problem 210
(a) Find an expression involving “ ” for the ratio
(b) Find an expression involving “ ” for the ratio
Problem 211
(a) Find an expression involving “ ” for the ratio
(b) Find an expression involving “ ” for the ratio