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Circle Measurement and Constructions: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two circular sizes

A circle has exact diameter $23/4$ cm. A second circle is its dilation by the positive factor $8/5$ .

(a) Find the first circle's circumference and disk area exactly.

(b) Find the second circle's circumference and disk area exactly.

Only the first diameter is supplied.

Text description of this figure

The diameter crosses center O and is 23/4 centimeters long. No recovered length or area is shown.
1. Hint 1
  Circumference is a length, while disk area responds to a scale factor in two dimensions.
2. Hint 2
  Halve the first diameter before finding area. The second circumference has factor $8/5$ and the second area has factor $(8/5)^2$ .
Answer

(a)
$\begin{gathered}C_1=\frac{23\pi}{4}\text{ cm}\\A_1=\frac{529\pi}{64}\text{ cm}^2\end{gathered}$
(b)
$\begin{gathered}C_2=\frac{46\pi}{5}\text{ cm}\\A_2=\frac{529\pi}{25}\text{ cm}^2\end{gathered}$

Full solution

(a) A diameter gives circumference directly.
$C_1=\frac{23\pi}{4}\text{ cm}.$

The positive radius is half that diameter.
$r_1=\frac{23}{8}\text{ cm}.$
Square it for disk area.
$A_1=\frac{529\pi}{64}\text{ cm}^2.$

(b) A dilation scales circumference by its length factor.
$\begin{aligned}C_2&=\frac85\cdot\frac{23\pi}{4}\\C_2&=\frac{46\pi}{5}\text{ cm}.\end{aligned}$

Area scales by the square of the length factor.
$\begin{aligned}A_2&=\frac{64}{25}\cdot\frac{529\pi}{64}\\A_2&=\frac{529\pi}{25}\text{ cm}^2.\end{aligned}$

Answer

(a)
$\begin{gathered}C_1=\frac{23\pi}{4}\text{ cm}\\A_1=\frac{529\pi}{64}\text{ cm}^2\end{gathered}$
(b)
$\begin{gathered}C_2=\frac{46\pi}{5}\text{ cm}\\A_2=\frac{529\pi}{25}\text{ cm}^2\end{gathered}$

Key idea

A circle dilation changes boundary length by its scale factor and disk area by its square.
Problem 2 One turn on two guides

A selected minor arc is $43.2$ cm long on a circular guide of radius $0.18$ m.

(a) Find its central turn exactly in radians and in degrees. Round the degree measure to the nearest hundredth as well.

(b) The same turn is used on a second circular guide of radius $35$ mm. Find the second selected arc length in millimeters. Explain why its arc-to-radius ratio agrees with the first.
1. Hint 1
  Compare arc length with radius only after giving them matching length units.
2. Hint 2
  Find the radian ratio on the first guide, then express that same turn in the other angular unit.
3. Hint 3
  The corresponding arc and radius scale by the same factor, so their quotient is preserved.
Answer

(a)
$\begin{gathered}\theta=\frac{12}{5}\text{ radians}\\\alpha=\frac{432}{\pi}^\circ\\\alpha\approx137.51^\circ\end{gathered}$
(b) $84$ mm; both arc-to-radius ratios equal $12/5$ .

Full solution

(a) The first radius is $18$ cm.
Using matching lengths gives
$\begin{aligned}\theta&=\frac{43.2}{18}\\\theta&=\frac{12}{5}\text{ radians}.\end{aligned}$

A half-turn is both pi radians and 180 degrees.
Convert the same turn.
$\begin{aligned}\alpha&=\frac{12}{5}\cdot\frac{180}{\pi}\\\alpha&=\frac{432}{\pi}^\circ\\\alpha&\approx137.51^\circ.\end{aligned}$
The turn is smaller than a half-turn, consistent with the selected minor arc.

(b) Use the same radian measure with the second radius.
$\begin{aligned}s_2&=35\cdot\frac{12}{5}\\s_2&=84\text{ mm}.\end{aligned}$

The second quotient is $84/35$ , which reduces to $12/5$ .
A dilation scales each chord of a polygonal arc approximation and its radius by the same factor.
Arbitrarily close approximations give that factor for the curved arc too, so the factor cancels from the quotient.

Answer

(a)
$\begin{gathered}\theta=\frac{12}{5}\text{ radians}\\\alpha=\frac{432}{\pi}^\circ\\\alpha\approx137.51^\circ\end{gathered}$
(b) $84$ mm; both arc-to-radius ratios equal $12/5$ .

Key idea

The arc-to-radius ratio measures the turn independently of circle size or the shared length unit.
Problem 3 A polygon on the given circle

A circle has known center $O$ and radius $9\sqrt7/2$ cm. Copy it on paper at any convenient scale.

Construct its inscribed square with a compass and unmarked straightedge. Retain the helper arcs, state their centers and opening, and justify both the equal sides and the right angles.

Find the square's exact side length, apothem, perimeter and area, using the stated radius.

Construct the polygon on paper from the circle.

Text description of this figure

The given circle has center O, boundary point A and radius OA of nine square root seven over two centimeters. No square or helper crossing is drawn.
1. Hint 1
  A square can be located by two perpendicular diameters of the original circle.
2. Hint 2
  Use the two endpoints of one diameter as centers of equal helper circles. An opening equal to the whole diameter gives two distinct crossings.
3. Hint 3
  Join the helper crossings, then use its intersections with the original circle. A right center triangle supplies the side, and its perpendicular half gives the apothem.
Answer

Helpers at $A,C$ use opening $9\sqrt7$ cm. Lengths in cm:
$\begin{gathered}s=\frac{9\sqrt{14}}2\\a=\frac{9\sqrt{14}}4\\P=18\sqrt{14}\end{gathered}$
Area $567/2$ square cm.

Full solution

Draw a diameter $AC$ through $O$ .
Its length is $9\sqrt7$ cm.
With center $A$ and then center $C$ , retain equal compass arcs with the full opening $AC$ .
The opening exceeds half the center separation, so the helper circles meet at two distinct points $U,V$ .

Each crossing is equally distant from $A,C$ .
The perpendicular-bisector converse puts both on the unique perpendicular bisector of $AC$ .
Join $U,V$ ; their line passes through its midpoint $O$ .
Mark its original-circle intersections $B,D$ .
Join $A,B,C,D$ in circle order.

The four center angles are right angles.
Their center triangles have equal pairs of radii and equal included angles, so SAS gives four equal chords.
Each interior angle subtends the diameter joining its neighboring vertices, so the semicircle theorem makes each a right angle.
The constructed quadrilateral is a square.

A center triangle has perpendicular radius legs, so its side is radius times square root 2.
The apothem is half that square side.
$\begin{gathered}s=\frac{9\sqrt{14}}2\text{ cm}\\a=\frac{9\sqrt{14}}4\text{ cm}.\end{gathered}$

Count four straight sides for perimeter, and square one side for area.
$\begin{gathered}P=18\sqrt{14}\text{ cm}\\A=\frac{567}{2}\text{ cm}^2.\end{gathered}$
The helper opening is a diameter, not a square side.

The retained compass arcs establish the second diameter.

Answer

Helpers at $A,C$ use opening $9\sqrt7$ cm. Lengths in cm:
$\begin{gathered}s=\frac{9\sqrt{14}}2\\a=\frac{9\sqrt{14}}4\\P=18\sqrt{14}\end{gathered}$
Area $567/2$ square cm.

Key idea

Equal-distance compass crossings justify the perpendicular diameters before right triangles supply square measurements.
Problem 4 A chord and the region beside it

A circle centered at $O$ has chord $AB$ of length $12\sqrt6$ cm. The perpendicular distance from $O$ to $AB$ is $6\sqrt2$ cm.

(a) Find the circle radius and the smaller central angle $AOB$ .

(b) Find the exact area and complete perimeter of the minor circular segment cut off by $AB$ .
1. Hint 1
  A perpendicular from a circle center to a chord also reaches its midpoint.
2. Hint 2
  The radius is the hypotenuse of a right triangle with half the chord and the given center distance as legs.
3. Hint 3
  Identify the special right triangle, then distinguish the sector area from the cap area and the arc from the full cap boundary.
Answer

Lengths are in centimeters and area is in square centimeters.
$\begin{gathered}r=12\sqrt2\\\alpha=120^\circ\\A=96\pi-72\sqrt3\\P=8\pi\sqrt2+12\sqrt6\end{gathered}$

Full solution

(a) Let $M$ be the perpendicular foot.
The chord-bisection theorem gives half-chord $AM$ of $6\sqrt6$ cm.
Pythagoras in right triangle $OMA$ gives
$\begin{aligned}r^2&=(6\sqrt6)^2+(6\sqrt2)^2\\r^2&=288\\r&=12\sqrt2\text{ cm}.\end{aligned}$

The center-distance leg is half the hypotenuse, so the right triangle has a 30-degree angle at $A$ and a 60-degree angle at $O$ .
The equal halves of the isosceles center triangle therefore give
$\alpha=120^\circ.$

(b) The minor sector takes one third of the disk area.
$A_{\rm sector}=96\pi\text{ cm}^2.$
The central triangle uses the chord as base and its perpendicular center distance as height.
$\begin{aligned}A_{\triangle}&=\frac12(12\sqrt6)(6\sqrt2)\\A_{\triangle}&=72\sqrt3\text{ cm}^2.\end{aligned}$

Subtract the triangle from the sector.
$A=(96\pi-72\sqrt3)\text{ cm}^2.$

The selected arc takes one third of the circumference.
The segment closes with its chord, not with the two radius guides.
$\begin{gathered}s=8\pi\sqrt2\text{ cm}\\P=(8\pi\sqrt2+12\sqrt6)\text{ cm}.\end{gathered}$

The perpendicular height and chord determine the center triangle.

Answer

Lengths are in centimeters and area is in square centimeters.
$\begin{gathered}r=12\sqrt2\\\alpha=120^\circ\\A=96\pi-72\sqrt3\\P=8\pi\sqrt2+12\sqrt6\end{gathered}$

Key idea

A perpendicular chord-distance triangle can determine the central angle before a segment area and its entire boundary are measured.
Problem 5 What the measurements certify

Polygon bounds establish the following circumference enclosure, with $C$ in centimeters:
$69.1156<C<69.1164.$
Certified pi bounds are
$3.14159<\pi<3.14160.$
(a) Give the midpoint circumference estimate and its guaranteed strict error bound. Certify the circumference rounded to the nearest hundredth.

(b) Find a guaranteed interval for the positive radius and certify its value rounded to the nearest hundredth of a centimeter. Explain whether either rounded decimal is an exact measurement.
1. Hint 1
  A finite interval controls the possible error and can settle rounding without giving an exact decimal.
2. Hint 2
  The midpoint error is at most half the interval width, with strictness inherited from these open bounds.
3. Hint 3
  For the radius lower bound use the smaller circumference and the larger positive pi value. Reverse those choices for the upper bound.
Answer

(a) Midpoint $69.116$ cm; error less than $0.0004$ cm; rounded circumference $69.12$ cm. (b)
$\begin{gathered}r>69.1156/6.28320\\r<69.1164/6.28318\end{gathered}$
Rounded radius $11.00$ cm. Both decimals are approximate.

Full solution

(a) Average the two endpoints.
$\frac{69.1156+69.1164}{2}=69.116.$
Half the width is $0.0004$ cm.
Every allowed circumference is strictly less than that distance from the midpoint.

The whole circumference interval lies strictly between $69.115$ cm and $69.125$ cm, the rounding boundaries for $69.12$ cm.
Thus the rounded circumference is certified.

(b) The positive radius is circumference divided by twice pi.
The numerator and denominator bounds must be chosen in opposite directions.
$\begin{gathered}r>\frac{69.1156}{6.28320}\\r<\frac{69.1164}{6.28318}.\end{gathered}$

The lower endpoint exceeds $11$ because $11(6.28320)$ is $69.1152$ .
The upper endpoint is less than $11.001$ because $11.001(6.28318)$ is $69.12126318$ .
Therefore
$11<r<11.001.$
Every radius in that interval rounds to $11.00$ cm.

These are certified finite-precision conclusions.
The radius and circumference remain unknown within their enclosures, so the rounded decimals are approximate.

Answer

(a) Midpoint $69.116$ cm; error less than $0.0004$ cm; rounded circumference $69.12$ cm. (b)
$\begin{gathered}r>69.1156/6.28320\\r<69.1164/6.28318\end{gathered}$
Rounded radius $11.00$ cm. Both decimals are approximate.

Key idea

Certified rounding depends on the whole allowed interval, and inverse bounds pair a smaller numerator with a larger denominator.
Problem 6 Recover the sector

A proper sector has area $54\text{ cm}^2$ and complete perimeter $36$ cm. Its positive radius and selected central turn are unknown. The turn must be strictly between zero and $2\pi$ radians.

Determine every sector satisfying these data. Give its exact radius, selected arc length and central turn in radians, and explain why any other algebraic candidate fails the geometry.

Justify how the selected turn determines the sector area relationship used in your reconstruction.
1. Hint 1
  Separate the two radius edges from the curved part of the perimeter.
2. Hint 2
  The same arc length also connects area with radius. Eliminate that arc length to obtain an equation in the radius.
3. Hint 3
  The area and boundary relations are
  $\begin{gathered}54=rs/2\\36=2r+s.\end{gathered}$
  Check both positive radius candidates against the turn domain.
4. Hint 4
  For each positive radius candidate, divide its positive arc length by the radius and compare that turn with a full turn.
Answer

One sector:
$\begin{gathered}r=9+3\sqrt3\text{ cm}\\s=18-6\sqrt3\text{ cm}\\\theta=4-2\sqrt3\text{ radians}\end{gathered}$
The other radius candidate gives a turn greater than $2\pi$ . The area relation is $A=r^2\theta/2$ .

Full solution

For a proper radian turn, compare its measure with a full turn.
Congruent sectors have equal areas, so the selected fraction of the disk gives
$A=\frac{\theta}{2\pi}(\pi r^2).$
Cancellation yields
$A=\frac12r^2\theta.$
Since the selected arc length is radius times radian measure, the same relationship is
$A=\frac12rs.$

Let $s$ be the selected arc length.
The complete perimeter includes both radii.
$s=36-2r.$

The area is half radius times selected arc length.
Substitution gives
$\begin{aligned}54&=\frac12r(36-2r)\\r^2-18r+54&=0\\(r-9)^2&=27.\end{aligned}$

Both algebraic radii are positive.
$r=9\pm3\sqrt3\text{ cm}.$
Their corresponding arc lengths come from the same boundary equation.

For the larger radius, the selected length and radian turn are
$\begin{gathered}s=18-6\sqrt3\text{ cm}\\\theta=4-2\sqrt3\text{ radians}.\end{gathered}$
The turn is positive and less than 1 radian, so it meets the proper-sector domain.

For the smaller radius, the arc length is $18+6\sqrt3$ cm and the turn is $4+2\sqrt3$ radians.
The finite hexagon enclosure gives pi less than $2\sqrt3$ .
Also, square root 3 is less than 2.
Hence
$\begin{gathered}2\pi<4\sqrt3\\4\sqrt3<4+2\sqrt3.\end{gathered}$
This turn exceeds a full turn and cannot bound the stated proper sector.

The retained radius and arc give perimeter $36$ cm.
Their product is $108\text{ cm}^2$ , so half that product is the stated area $54\text{ cm}^2$ .
There is exactly one admissible sector.

Only the admissible sector is drawn.

Answer

One sector:
$\begin{gathered}r=9+3\sqrt3\text{ cm}\\s=18-6\sqrt3\text{ cm}\\\theta=4-2\sqrt3\text{ radians}\end{gathered}$
The other radius candidate gives a turn greater than $2\pi$ . The area relation is $A=r^2\theta/2$ .

Key idea

Inverse area and boundary equations must be checked against the allowed angle domain.
Problem 7 Marks that close around a circle

A circle has known center $O$ , radius $\sqrt{37}$ cm and boundary point $A$ . Copy it on paper at any convenient scale.

(a) Construct regular hexagon $ABCDEF$ with a compass and unmarked straightedge, keeping every compass arc. Name the center, opening and excluded intersection at every forward step, including the return to $A$ . Justify closure and regularity.

(b) Construct an equilateral triangle from alternate hexagon vertices and justify its equal sides. Find the exact hexagon apothem and area, and the exact triangle area.

Carry out both linked constructions on paper.

Text description of this figure

Only the given circle, center O, point A and radius OA of square root 37 centimeters are drawn. No other vertices or construction arcs are present.
1. Hint 1
  The compass can carry the given radius around the original circle as a straight chord length.
2. Hint 2
  At each new vertex, discard the crossing that would return to the preceding vertex. Equal-radius center triangles determine how much each forward step turns.
3. Hint 3
  Join alternate justified vertices. Bisect one hexagon center triangle to obtain the perpendicular apothem, then use the polygon perimeter-apothem area relationship.
Answer

Construct hexagon $ABCDEF$ with opening $OA$ and triangle $ACE$ .
$\begin{gathered}a_H=\frac{\sqrt{111}}2\text{ cm}\\A_H=\frac{111\sqrt3}{2}\text{ cm}^2\\A_T=\frac{111\sqrt3}{4}\text{ cm}^2\end{gathered}$

Full solution

(a) Set the opening to $OA$ .
With center $A$ , choose one of its two crossings with the original circle as $B$ .
This fixes a travel direction.
Keep the opening unchanged for every later arc.

Center B: mark C at the original-circle crossing other than A.

Center C: mark D at the original-circle crossing other than B.

Center D: mark E at the original-circle crossing other than C.

Center E: mark F at the original-circle crossing other than D.

Center F: the crossing other than E returns to A.
Retain the arcs from all six centers and join A, B, C, D, E, F, A in order.

Every step's chord equals the compass opening.
Its two radii belong to the original circle and have that same length.
Each center triangle is equilateral, so each forward turn is $60^\circ$ .
Six turns make $360^\circ$ ; the first five are smaller than a full turn and locate distinct vertices.
Adjacent center triangles give equal 120-degree interior angles.
Thus the closed polygon has equal sides and equal angles.

(b) Join $AC$ , $CE$ and $EA$ .
Each spans two consecutive 60-degree center turns.
The three center triangles have equal radii and equal included 120-degree angles, so SAS gives three equal chords.
Triangle $ACE$ is equilateral.

The hexagon side is $\sqrt{37}$ cm.
Its right half-center triangle has hypotenuse $\sqrt{37}$ cm and half-side $\sqrt{37}/2$ cm.
Its perpendicular apothem is therefore
$a_H=\frac{\sqrt{111}}2\text{ cm}.$

The perimeter is $6\sqrt{37}$ cm.
The apothem provides the height in the area sum.
$\begin{aligned}A_H&=\frac12(6\sqrt{37})\frac{\sqrt{111}}2\\A_H&=\frac{111\sqrt3}{2}\text{ cm}^2.\end{aligned}$

The triangle side is $\sqrt{111}$ cm and its apothem is $\sqrt{37}/2$ cm, from the right halves of a 120-degree center triangle.
Use its own perimeter and apothem.
$\begin{aligned}A_T&=\frac12(3\sqrt{111})\frac{\sqrt{37}}2\\A_T&=\frac{111\sqrt3}{4}\text{ cm}^2.\end{aligned}$

The completed construction retains the marks used in the proof.

Answer

Construct hexagon $ABCDEF$ with opening $OA$ and triangle $ACE$ .
$\begin{gathered}a_H=\frac{\sqrt{111}}2\text{ cm}\\A_H=\frac{111\sqrt3}{2}\text{ cm}^2\\A_T=\frac{111\sqrt3}{4}\text{ cm}^2\end{gathered}$

Key idea

Retained radius-opening arcs prove the hexagon, and alternate vertices transfer that justification to an equilateral triangle.
Problem 8 A plate with two curved cuts

A rectangular plate is $30$ cm by $18$ cm. A semicircular region of diameter $11$ cm is removed inward from the middle of one 30 cm side. A second semicircular region of diameter $7$ cm is removed inward from the middle of the opposite side. The removed regions do not overlap.

Find the exact remaining area and the complete perimeter of the plate after both cuts.

Trace the solid boundary after both inward cuts.

Text description of this figure

A rectangle measures 30 centimeters horizontally and 18 centimeters vertically. A semicircle of diameter 11 centimeters is cut inward from the middle of the top edge. A semicircle of diameter 7 centimeters is cut inward from the middle of the bottom edge. The two removed regions do not overlap. The kept material is shaded and its complete boundary is solid. Dashed diameters identify the cut openings.
1. Hint 1
  Area is removed, while the cuts replace some straight boundary with curved edges.
2. Hint 2
  Halve each cut diameter for its radius. The two semicircle areas must both be removed.
3. Hint 3
  Subtract the two opening diameters from the original rectangle perimeter, then add each exposed semicircular arc.
Answer

$\begin{gathered}A=\left(540-\frac{85\pi}{4}\right)\text{ cm}^2\\P=(78+9\pi)\text{ cm}\end{gathered}$

Full solution

The original rectangle area is $540\text{ cm}^2$ .
The cut radii are $11/2$ cm and $7/2$ cm.
Their semicircle areas are $121\pi/8\text{ cm}^2$ and $49\pi/8\text{ cm}^2$ respectively.

Both cuts remove material.
Their combined area is $85\pi/4\text{ cm}^2$ , so the remaining area is
$A=\left(540-\frac{85\pi}{4}\right)\text{ cm}^2.$

The original perimeter is $96$ cm.
The two cuts remove straight openings totaling $18$ cm, leaving $78$ cm of straight boundary.

The first new arc is $11\pi/2$ cm and the second is $7\pi/2$ cm.
Add their combined length to the retained straight edges.
$P=(78+9\pi)\text{ cm}.$
The dashed opening diameters are not edges of the remaining plate.

Answer

$\begin{gathered}A=\left(540-\frac{85\pi}{4}\right)\text{ cm}^2\\P=(78+9\pi)\text{ cm}\end{gathered}$

Key idea

A cut subtracts region area and replaces boundary pieces; its opening diameter does not remain exposed.
Problem 9 How tight can the enclosure become?

A circle has radius $7$ cm and circumference $C$ . Start with a regular inscribed hexagon and repeatedly bisect every corresponding arc. After $n$ refinements, where $n$ is a positive integer, let the new side length be $s$ , inner perimeter $P$ , inner apothem $h$ and parallel tangent-polygon perimeter $Q$ , all in centimeters.

At every refinement the numerical centimeter measures satisfy
$\begin{gathered}P<50\\h>2.\end{gathered}$
Each arc bisection makes the squared side length less than half its preceding value.

(a) Express $7-h$ in terms of $s,h$ from a right half-center triangle. Obtain a bound for $Q-P$ in terms of $n$ , and find the fewest refinements for which your bound certifies a perimeter gap less than $0.01$ cm.

(b) Derive the exact disk area in terms of $C$ from the polygon enclosures. Explain why a positive discrepancy is impossible, and state the corresponding formula for any positive radius $r$ .

(c) Does rearranging this disk into a row of $64$ equal curved sectors make an exact parallelogram? Explain the finite figure's role.
1. Hint 1
  The half-side and apothem are perpendicular legs, while the circle radius is their hypotenuse.
2. Hint 2
  Factor the difference of the squared radius and squared apothem. Then use the given side-shrinking rate.
3. Hint 3
  The parallel tangent polygon has perimeter $Q=7P/h$ . The bounds on $P,h$ control the factor multiplying the radial gap.
4. Hint 4
  For area, compare the two polygon areas and the value they approach. A nonzero fixed difference cannot fit inside every sufficiently narrow enclosure.
Answer

(a) $7-h=s^2/[4(7+h)]$ . Gap less than $175/(4\cdot2^n)$ cm; $13$ refinements suffice by this bound. (b)
$A=7C/2.$
In general $A=Cr/2$ . (c) No; a finite row retains curved edges.

Full solution

(a) The right half-center triangle gives
$49=h^2+\frac{s^2}{4}.$
Factor the difference of squares and divide by the positive sum.
$7-h=\frac{s^2}{4(7+h)}.$
Since $h>0$ , this is less than $s^2/28$ .

The original hexagon side is $7$ cm.
After $n$ positive refinements the squared side satisfies
$s^2<\frac{49}{2^n}.$

The tangent polygon is the $7/h$ dilation of the inner polygon.
Its perimeter difference is
$Q-P=\frac{P(7-h)}h.$
Use $P<50$ , $h>2$ and the radial-gap bound.
$\begin{aligned}Q-P&<\frac{25s^2}{28}\\Q-P&<\frac{175}{4\cdot2^n}\text{ cm}.\end{aligned}$

For this bound to certify the requested accuracy, its denominator must satisfy
$2^n>4375.$
Twelve refinements give $4096$ , which is too small for this bound.
Thirteen give $8192$ , so thirteen is the fewest refinements this bound certifies.
This is not a claim that the actual gap first crosses the threshold at that stage.

(b) Let $I$ and $U$ be the inner and outer polygon areas.
Their perpendicular heights give
$\begin{gathered}I=Ph/2\\U=7Q/2.\end{gathered}$
The disk area lies between them.
So does $7C/2$ , because the circumference lies between the two perimeters and $h<7$ .

Both the perimeter gap and radial gap become smaller than any chosen positive tolerance.
Their area contributions are
$U-I=\frac72(Q-P)+\frac P2(7-h).$
The perimeter P stays bounded, so the area gap also becomes as small as requested.

If the disk area and $7C/2$ differed by a fixed positive amount, a narrower common enclosure would be impossible.
Therefore
$A=7C/2.$
Nothing in that reasoning depends on radius 7.
For any positive radius the same argument gives
$A=Cr/2.$
Substituting the circumference expression gives $A=\pi r^2.$

(c) Rigidly rearranging the $64$ sectors preserves their total area exactly.
Their circular edges remain curved, so this finite row is not a parallelogram.
Finer rearrangements support the limiting comparison; an exact area conclusion needs control of the remaining discrepancy, as in the polygon enclosure argument.

Answer

(a) $7-h=s^2/[4(7+h)]$ . Gap less than $175/(4\cdot2^n)$ cm; $13$ refinements suffice by this bound. (b)
$A=7C/2.$
In general $A=Cr/2$ . (c) No; a finite row retains curved edges.

Key idea

Exhaustion establishes equality by ruling out every positive discrepancy, while each finite approximation still has an error.
Problem 10 One side becomes a cut

An equilateral triangle $ABC$ is inscribed in a circle centered at $O$ . Its perimeter is $9\sqrt{21}$ cm.

Remove the minor circular segment between side $AB$ and the minor arc with those endpoints. Keep the rest of the disk.

Find the exact kept area and complete perimeter. Explain which perpendicular triangle height your area calculation needs, and why the full altitude from $C$ is a different length.

The kept region lies on the major-arc side of chord AB.

Text description of this figure

Equilateral triangle ABC is inscribed in circle O. The minor cap on side AB is removed. The major segment containing C is shaded, with chord AB and the major arc as its solid boundary. The triangle perimeter is nine square root twenty-one centimeters.
1. Hint 1
  The triangle perimeter determines its side, which in turn determines the supporting circle radius.
2. Hint 2
  Bisect the 120-degree center triangle on the cut side. Its perpendicular center-to-side distance is the apothem.
3. Hint 3
  The kept region is the complement of the minor cap. Its complete boundary has the same chord and the longer arc through the third triangle vertex.
Answer

$\begin{gathered}A=\left(42\pi+\frac{63\sqrt3}{4}\right)\text{ cm}^2\\P=(4\pi\sqrt7+3\sqrt{21})\text{ cm}\end{gathered}$
The needed center-triangle height is $3\sqrt7/2$ cm; the full altitude is $9\sqrt7/2$ cm.

Full solution

The triangle side is one third of its perimeter, or $3\sqrt{21}$ cm.
An inscribed equilateral triangle side is square root 3 times its circle radius.
Thus
$r=3\sqrt7\text{ cm}.$

Side $AB$ subtends one of three equal 120-degree center angles.
The right halves of triangle $OAB$ have hypotenuse r and shorter leg equal to half r.
The perpendicular center-to-side height is therefore
$OM=\frac{3\sqrt7}{2}\text{ cm}.$
This is the apothem.
The full altitude from C contains the radius from C to O as well, giving $9\sqrt7/2$ cm.

The minor sector is one third of the disk.
The central triangle uses side AB as base and OM as height.
$\begin{gathered}A_{\rm sector}=21\pi\text{ cm}^2\\A_{OAB}=\frac{63\sqrt3}{4}\text{ cm}^2.\end{gathered}$
Using the full altitude would instead find the area of the entire equilateral triangle ABC.

Subtract the central triangle from the minor sector to get the removed cap.
Then subtract that cap from the whole disk.
$\begin{aligned}A_{\text{cap}}&=21\pi-\frac{63\sqrt3}{4}\\A_{\rm kept}&=63\pi-A_{\text{cap}}\\A_{\rm kept}&=42\pi+\frac{63\sqrt3}{4}.\end{aligned}$
These areas are in square centimeters.

The kept major arc uses two thirds of the circumference.
$s_{\rm major}=4\pi\sqrt7\text{ cm}.$
Its straight boundary is the chord AB, of length $3\sqrt{21}$ cm.
Hence
$P=(4\pi\sqrt7+3\sqrt{21})\text{ cm}.$
The other two triangle sides are interior guides and add no exposed boundary.

The solution adds the center-triangle height.

Answer

$\begin{gathered}A=\left(42\pi+\frac{63\sqrt3}{4}\right)\text{ cm}^2\\P=(4\pi\sqrt7+3\sqrt{21})\text{ cm}\end{gathered}$
The needed center-triangle height is $3\sqrt7/2$ cm; the full altitude is $9\sqrt7/2$ cm.

Key idea

Transfer through the supporting radius, then use the selected segment and the correct perpendicular height.

Multiple choice

Core practice