HSG.GMD.A.1: Informal Arguments for Circle, Cylinder, Pyramid and Cone Formulas
In plain English: HSG.GMD.A.1 is the Common Core geometry standard that asks students to give informal arguments for the formulas for the circumference and area of a circle and the volumes of a cylinder, pyramid and cone. Students explain the formulas with dissection, Cavalieri's principle and limits of polygons with more and more sides. It is usually taught in high school Geometry.
Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.
Common Core State Standards for Mathematics · Domain: Geometric Measurement and Dimension (GMD) · Cluster: Explain volume formulas and use them to solve problems Also written as HSG-GMD.A.1 or G-GMD.1 · Official standard
Students already know the formulas C = 2πr, A = πr², V = πr²h, V = (1/3)Bh and V = (1/3)πr²h from middle school. This lesson asks why they are true. Students give an informal argument for each formula using the three methods the standard names: dissection (cut a figure into pieces and rearrange them), Cavalieri's principle (compare two solids slice by slice) and informal limit arguments (polygons with more and more sides get closer to a circle).
The arguments build on one another. Similar circles give a constant ratio C/d = π. Sectors of a circle rearrange into a shape close to a parallelogram with base πr and height r. Stacks of thin disks give the cylinder. A cube splits into three congruent pyramids, which explains the one-third, and a cone behaves like a pyramid whose base polygon has more and more sides. Students finish by writing one argument in their own words.
Learning Objectives
By the end of this lesson, students will be able to:
Explain why C = 2πr using similar circles and inscribed polygons whose perimeters approach the circumference
Explain why A = πr² by cutting a circle into sectors and rearranging them into a shape that approaches a parallelogram
Use Cavalieri's principle to explain why V = πr²h for right and oblique cylinders
Explain the one-third in V = (1/3)Bh for a pyramid by dissecting a cube into three congruent pyramids
Explain why a cone has V = (1/3)πr²h by comparing it with pyramids of the same base area and height
Prior Knowledge Required
Students should already be comfortable with:
Area and circumference formulas for circles 7.G.B.4
Volume formulas for cylinders and cones 8.G.C.9
Volume of a right rectangular prism as base area times height 6.G.A.2
Similarity and scale factors, including that all circles are similar HSG.C.A.1
Area of a triangle as one half of base times height
Groups receive three or four round objects, such as a jar lid, a roll of tape and a paper plate. They wrap string around each one to measure the circumference, measure the diameter, and compute C ÷ d to two decimal places.
Warm-Up Prompt
"Your ratios C ÷ d are all a little above 3. Why should a jar lid and a paper plate give the same ratio, when one is much bigger than the other?"
Collect ratios on the board; string measurements usually give values between 3.0 and 3.3. Lead students to the reason: all circles are similar, so a dilation that maps one circle onto another multiplies the circumference and the diameter by the same scale factor, and their ratio stays the same. That constant is called π, so C = πd = 2πr.
Direct Instruction25 minutes
Three methods. A dissection argument cuts a figure into pieces and rearranges them into a figure whose area or volume is known. Cavalieri's principle: if two solids have the same height and every plane parallel to their bases cuts them in cross-sections of equal area, the solids have equal volumes. An informal limit argument looks at a sequence of approximations that get as close as we like to the figure. Build the five formulas in this order:
Circumference (similarity and limit): C/d is the same for all circles, and π is its value. Inscribe regular polygons with 6, 12, 24, ... sides. Each side is a chord slightly shorter than its arc, so each perimeter is a little less than C, and the table shows the perimeters climbing toward 2πr ≈ 6.2832r.
Area (dissection and limit): cut the circle into many equal sectors and place them tip up, tip down in a row (Diagram 1). The curved edges add up to the circumference, half on top and half on the bottom, so the base is close to πr and the height is close to r. With more sectors, the shape gets closer to a parallelogram, so A = πr · r = πr².
Cylinder (Cavalieri): a prism has volume Bh because it is a stack of congruent slices of area B. A cylinder with radius r and height h has every slice equal to a disk of area πr², so it matches a prism with base area πr² at every level: V = πr²h. The same holds for an oblique cylinder (Diagram 2).
Pyramid (dissection and Cavalieri): a cube with edge s splits into three congruent square pyramids that share one vertex of the cube. Each has base s², height s and volume s³/3 = (1/3)Bh. Cross-sections of any pyramid shrink by the same scale factor at the same fraction of the height, so pyramids with equal base areas and heights have equal slices and equal volumes: V = (1/3)Bh for every pyramid.
Cone (limit and Cavalieri): inscribe regular polygons with more and more sides in the base of a cone and build pyramids with the same apex. Each pyramid has V = (1/3)Bnh, and the base areas Bn approach πr², so V = (1/3)πr²h. Equivalently, a cone and a pyramid with the same base area and height have equal slices at every level.
Regular polygons inscribed in a circle of radius r
Number of sides
Perimeter
Area
6
6.0000r
2.5981r²
12
6.2117r
3.0000r²
24
6.2653r
3.1058r²
96
6.2821r
3.1394r²
Circle
2πr ≈ 6.2832r
πr² ≈ 3.1416r²
Work through the examples, asking before each one: "Which method is this: dissection, Cavalieri or a limit?"
Circumference by an informal limit
A regular hexagon and a regular 12-gon are inscribed in a circle of radius 10 cm. Their perimeters are 60 cm and about 62.12 cm. What do the perimeters approach as the number of sides grows, and why?
Equation: each side is a chord a little shorter than its arc, so the perimeters increase toward the circumference 2π(10) = 20π ≈ 62.83 cm
Area by dissection
A circle with a radius of 6 cm is cut into 16 equal sectors and rearranged with the tips alternating. Estimate the base and height of the new shape and its area.
Equation: base ≈ half the circumference = 6π cm, height ≈ 6 cm, area ≈ 6π · 6 = 36π ≈ 113.1 cm²
Cylinder by Cavalieri's principle
A right cylinder and an oblique cylinder both have a radius of 3 cm and a height of 10 cm. Explain why their volumes are equal and find them.
Equation: every horizontal slice of both is a disk of area 9π cm² and the heights match, so V = 9π · 10 = 90π ≈ 282.7 cm³ for each
Pyramid by dissection
A cube with an edge of 6 cm is cut into three congruent pyramids that share one vertex of the cube. Find the volume of each pyramid and compare it with (1/3)Bh.
Equation: cube 6³ = 216 cm³, so each pyramid is 216 ÷ 3 = 72 cm³, and (1/3)(6²)(6) = 72 cm³
Cone as a limit of pyramids
A cone has a radius of 4 cm and a height of 9 cm. Pyramids with the same apex have regular 6-, 12- and 24-gon bases inscribed in the base of the cone. What do their volumes approach?
Equation: about 124.7, 144.0 and 149.1 cm³, approaching (1/3)(16π)(9) = 48π ≈ 150.8 cm³
Stress what makes these arguments informal: we do not prove that the sector shape becomes exactly a parallelogram or that the polygon areas reach πr², but the error can be made as small as we like. A full proof uses limits, which students meet again in calculus.
Guided Practice10-15 minutes
Pairs answer four prompts on whiteboards and name the method each time. (a) A regular 12-gon inscribed in a circle of radius 1 is made of 12 triangles, each with area ½ · 1 · 1 · sin 30° = ¼. Find its area (3) and compare it with π. (b) A straight stack of 40 coins is pushed into a slanted stack. Did the volume change? (No: same height and the same disk at every level.) (c) A cube with an edge of 3 cm is cut into three congruent pyramids. Find the volume of each (9 cm³). (d) A hollow cone and a hollow cylinder have radius 5 cm and height 12 cm. How many full cones of water fill the cylinder? (Three: 100π cm³ versus 300π cm³.) Listen for students who give only the formula; ask them for the picture behind it.
Independent Practice10 minutes
Students work alone. (a) A regular 24-gon is inscribed in a circle of radius 5. Its perimeter is 240 · sin 7.5° ≈ 31.33. Compare it with the circumference and explain the difference. (b) Explain why an oblique cone and a right cone with the same base and height have the same volume. (c) Write a paragraph that argues for one formula of your choice from this lesson. Name the method, describe the picture, and say where the formula comes from.
Closure5 minutes
Exit ticket, choose one: (1) "Explain why the area of a circle is πr² and not 2πr, using the sector picture." (2) "Explain where the one-third in the cone formula comes from." Sort the tickets by method (dissection, Cavalieri, limit) and start the next lesson with two strong examples.
Differentiation Strategies
For Struggling Students
Give pre-cut paper circles in 8 and 16 sectors so students can focus on the rearrangement instead of cutting
Use physical stacks (coins, index cards, sticky-note pads) to show Cavalieri's principle before any formulas
Provide a sentence frame: "I cut the figure into ___, rearranged them into ___, so the area or volume is ___"
For Advanced Students
Show that a regular n-gon inscribed in a circle has area (1/2) · perimeter · apothem, and explain why this approaches (1/2)(2πr)(r)
Use a spreadsheet to compute n · sin(180°/n) for n = 6, 12, 24, 48 and 96 and compare with π, as Archimedes did by hand
Explain why every pyramid can be cut into triangular pyramids, so the argument for the square pyramid covers all bases
Assessment Guidance
What to Look For
A complete informal argument names the method, describes the picture and connects it to each part of the formula: where the π comes from, why r appears twice in πr², and why the pyramid and cone have a one-third. Watch for students who only restate the formula or plug in numbers; ask them to point to the base and the height in the rearranged sectors or the three pyramids in the cube. In Cavalieri arguments, check that students mention both conditions: equal heights and equal cross-sectional areas at every level.
02
Classroom Activities
3 Activities
1
Sector Rearrangement
20 minPairs
Pairs cut a paper circle into sectors and rearrange them into a near-parallelogram, first with 8 sectors and then with 16. The activity is the dissection and limit argument for A = πr², done by hand.
Procedure
Draw two circles with a radius of 8 cm. Fold and cut one into 8 equal sectors and the other into 16
Arrange each set in a row with tips alternating up and down, and glue it on paper. Cut one end sector in half so both ends are straight
Measure the base and height of each shape. Expected values: base close to 8π ≈ 25.1 cm, height close to 8 cm, so area close to 64π ≈ 201.1 cm²
Compare the 8-sector and 16-sector shapes: which looks more like a parallelogram, and why?
Discussion Questions
Why is the base of the new shape half of the circumference and not all of it?
Which length in the new shape is the radius?
What would the shape look like with 1,000 sectors?
Modification for Distance Learning
Use a dynamic geometry file with a slider for the number of sectors. Students record the base and height for 8, 16 and 32 sectors in a shared table and write the argument under a screenshot.
2
Three Pyramids in a Cube
20 minGroups of 3
Each student folds one pyramid from a cardstock net. The group fits its three congruent pyramids together into a cube, then checks the one-third with a cone, a cylinder and rice.
Procedure
Each net has a 5 cm by 5 cm square base, two right-triangle faces with legs 5 cm and 5 cm, and two right-triangle faces with legs 5 cm and 5√2 ≈ 7.1 cm. The apex sits directly above one corner of the base, 5 cm up
Fold and tape the three pyramids, then fit them together into a cube with an edge of 5 cm
Record: cube volume 125 cm³, so each pyramid is 125 ÷ 3 ≈ 41.7 cm³, which equals (1/3)(5²)(5)
Fill a hollow cone with rice and pour it into a cylinder with the same base and height. Count the full cones needed to fill the cylinder
Discussion Questions
How do you know the three pyramids have the same volume?
Where is the base and where is the height of each pyramid in the cube?
How does the rice experiment support the cone formula, and why is it not a proof?
3
Cavalieri Stacks
15 minGroups of 3-4
Groups build stacks of identical cards and coins, slant and twist them, and decide what changes and what stays the same. They then use the idea to compare solids that look very different.
Stations
Station 1: stack 30 identical index cards straight, then push them into a slant and a twist. Measure the height each time and describe each card as a cross-section
Station 2: stack 30 identical coins as a right cylinder, then as an oblique cylinder. Explain why both have volume πr²h
Station 3: two pyramids have the same height, one with a 6 cm by 6 cm square base and one with a triangular base of area 36 cm². Explain why their slices at every level have equal areas
Station 4: write Cavalieri's principle in your own words, including both conditions
Challenge Variation
Archimedes' polygons: use the areas of inscribed regular 6-, 12- and 24-gons (2.598r², 3r² and 3.106r²) to argue that the area of a circle is a little more than 3.1r², and explain what happens as the number of sides doubles again.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Rearranging Sectors to Find the Area of a Circle
The circle is cut into 12 equal sectors. Placed with their tips alternating, six curved edges form the top and six form the bottom, so the base is close to half the circumference, πr, and the height is close to r. With more sectors the shape approaches a parallelogram with area πr · r = πr². Sectors are drawn to scale: 30° each, radius 80 units.
Diagram 2: Cavalieri's Principle for Cylinders
A right cylinder and an oblique cylinder with the same radius r and height h. A plane at any level cuts both in a disk of area πr², just as a slanted stack of coins has the same coins as a straight one. Equal heights and equal cross-sections at every level give equal volumes, so both have V = πr²h.
04
Homework Assignment
~30 min
HSG.GMD.A.1 Homework: Informal Arguments for Area and Volume Formulas
Directions: For each problem, name the method you use (dissection, Cavalieri's principle or an informal limit), include a labeled sketch, and write your argument in complete sentences. Give exact answers in terms of π and decimals to the nearest tenth.
Part 1: Circles (Problems 1-3)
A circle has a radius of 15 cm. The perimeter of an inscribed regular hexagon is 90 cm, and the perimeter of an inscribed regular 12-gon is about 93.2 cm. Compare both with the circumference and explain what these numbers suggest about the formula C = 2πr.
A round pizza with a radius of 7 in is cut into 20 equal slices, which are laid in a row with the points alternating. Estimate the base and height of the shape, use them to estimate the area, and explain why the estimate gets better with more slices.
A regular 12-gon is inscribed in a circle with a radius of 2 cm. It is made of 12 congruent triangles, each with area ½ · 2 · 2 · sin 30°. Find the area of the 12-gon and compare it with the area of the circle. Then explain why the polygon areas approach πr² as the number of sides grows.
Part 2: Solids (Problems 4-6)
A straight stack of coins forms a cylinder with a radius of 1.2 cm and a height of 5 cm. The stack is pushed so it leans. Use Cavalieri's principle to explain why the volume does not change, and find the volume.
A wooden cube with an edge of 9 in is cut into three congruent pyramids that share one vertex of the cube. Describe the base and the height of each pyramid, find the volume of each, and show that it equals (1/3)Bh.
A cone and a cylinder both have a radius of 6 cm and a height of 5 cm. Find both volumes. Then explain, using pyramids with polygon bases or Cavalieri's principle, why the cone holds one third as much as the cylinder.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Method
Correct method named for every problem
Method named for some problems
No method named
Argument
Picture connected to every part of the formula
Argument partly explained
Formula restated with no reasoning
Calculations
All values correct, exact and rounded
One or two calculation errors
Many errors
Cavalieri Conditions
States equal heights and equal cross-sections
States only one condition
Conditions missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Why is the ratio of circumference to diameter the same for every circle?
Answer: B
A dilation maps any circle onto any other and multiplies all lengths by the scale factor, so C/d does not change; this constant is π. Choice C is true but says nothing about the circumference. Choice D is false: 22/7 is only an approximation of π.
Question 2 of 20 · Multiple Choice
A regular hexagon is inscribed in a circle with a radius of 5 cm, so its perimeter is 30 cm. What can you say about the circumference?
Answer: C
Each of the six sides cuts across an arc, and a chord is shorter than its arc, so C > 30 cm; in fact C = 10π ≈ 31.4 cm. Choice A reverses the comparison. Choice D doubles the perimeter with no reason.
Question 3 of 20 · Multiple Choice
Regular polygons with 6, 12, 24, 48, ... sides are inscribed in a circle of radius r. What do their perimeters approach?
Answer: A
The perimeters increase (6r, about 6.21r, about 6.27r, ...) and get as close as we like to the circumference, 2πr ≈ 6.28r. Choice B is an area, not a length. Choice C is only the hexagon, the first polygon in the list.
Question 4 of 20 · Multiple Choice
A circle with a radius of 9 cm is cut into many equal sectors, which are arranged in a row with the tips alternating. What length does the base of the new shape approach?
Answer: D
Half of the curved edges lie along the bottom, so the base is half the circumference: ½ · 2π(9) = 9π cm. Choice A is the whole circumference. Choice B is the height of the shape, which is the radius.
Question 5 of 20 · Multiple Choice
In the same rearranged shape, what does the height approach?
Answer: A
Each sector stands on its tip, and the distance from the tip to the arc is the radius, so the height approaches r and the area approaches πr · r = πr². Choice B would double the area.
Question 6 of 20 · Multiple Choice
A regular polygon inscribed in a circle of radius r has perimeter P and apothem a, and its area is ½Pa. As the number of sides grows, what does ½Pa approach?
Answer: B
The perimeter approaches 2πr and the apothem approaches r, so ½Pa approaches ½ · 2πr · r = πr². Choice A is the limit of P alone. Choice D forgets that P approaches the whole circumference, not half of it.
Question 7 of 20 · Multiple Choice
Which statement is Cavalieri's principle?
Answer: C
Cavalieri's principle needs both conditions: equal heights and equal cross-sectional areas at every level. Choice B checks only the bases, so a cone and a cylinder with the same base would wrongly have the same volume. Choice D is false: a slanted stack of coins has the same volume as a straight one.
Question 8 of 20 · Multiple Choice
An oblique cylinder has a radius of 2 m and a height of 7 m. What is its volume?
Answer: A
Every horizontal slice is a disk of area π(2²) = 4π, just like a right cylinder with the same height, so V = 4π · 7 = 28π m³. Choice C uses the one-third from the cone formula. Choice B uses the radius instead of its square.
Question 9 of 20 · Multiple Choice
A cube with an edge of 12 in is cut into three congruent pyramids that share one vertex of the cube. What is the volume of each pyramid?
Answer: D
The cube has volume 12³ = 1,728 in³, and the three congruent pyramids share it equally: 1,728 ÷ 3 = 576 in³, which is (1/3)(12²)(12). Choice B divides by 4 instead of 3. Choice C is the area of the base, not a volume.
Question 10 of 20 · Multiple Choice
Which argument explains the factor 1/3 in the pyramid formula V = (1/3)Bh?
Answer: B
Each of the three pyramids has the cube's face as its base and the cube's edge as its height, so each has one third of Bh. Choice A counts faces, which has nothing to do with volume, and a square pyramid has four faces at the apex anyway. Choice C is false.
Question 11 of 20 · Multiple Choice
A hollow cone and a hollow cylinder have the same base and the same height. How many full cones of water are needed to fill the cylinder?
Answer: B
The cone has V = (1/3)πr²h and the cylinder has V = πr²h, so three full cones fill the cylinder. Choice A is a common guess from the picture. Choice D confuses the ratio with the number π, which is in both formulas.
Question 12 of 20 · Multiple Choice
Pyramids with regular n-gon bases inscribed in the base of a cone share the cone's apex. What do their volumes approach as n grows?
Answer: C
Each pyramid has volume (1/3)Bnh, and the base areas Bn approach πr², so the volumes approach (1/3)πr²h. Choice A leaves out the one-third that every pyramid has. Choice B leaves out the π that comes from the circular base.
Question 13 of 20 · Multiple Choice
Two pyramids have a height of 10 cm. One has a square base with an area of 36 cm², and the other has a triangular base with an area of 36 cm². Which is true?
Answer: D
At a fraction t of the way down from the apex, each slice has area t² · 36, so the slices match at every level and Cavalieri's principle gives equal volumes: (1/3)(36)(10) = 120 cm³. Choice C is false because the heights are equal. Choice B misses that only the base area matters.
Question 14 of 20 · Multiple Choice
A pyramid has a base area of 64 cm². What is the area of the cross-section halfway between the apex and the base?
Answer: B
Halfway up, every length in the slice is ½ of the matching length in the base, so the area is (½)² · 64 = 16 cm². Choice A scales the area by ½ instead of ¼. This scaling is why pyramids with equal base areas and heights have equal slices.
Question 15 of 20 · Short Answer
Explain why C = 2πr for every circle, using similarity and the definition of π.
All circles are similar: a dilation with scale factor k maps a circle of diameter d onto one of diameter kd and multiplies its circumference by k too. So C/d is the same for every circle, and π is defined as that ratio. Then C = πd, and since d = 2r, C = 2πr. Inscribed polygons whose perimeters approach this value support the argument.
Question 16 of 20 · Short Answer
A circle with a radius of 10 m is cut into a very large number of equal sectors, which are rearranged with alternating tips. Give the base, the height and the area that the shape approaches, and explain each.
The base approaches half the circumference, 10π m, because half of the arcs lie on the bottom. The height approaches the radius, 10 m, the distance from each tip to its arc. The area approaches 10π · 10 = 100π ≈ 314.2 m², which is πr² for r = 10.
Question 17 of 20 · Short Answer
A regular 12-gon is inscribed in a circle with a radius of 3 cm. It is made of 12 triangles with two sides of 3 cm and a 30° angle between them. Find its area and compare it with the area of the circle.
Each triangle has area ½ · 3 · 3 · sin 30° = 9/4 cm², so the 12-gon has area 12 · 9/4 = 27 cm². The circle has area 9π ≈ 28.3 cm². The 12-gon is a little smaller because it leaves out 12 thin slivers between the chords and the arcs; with more sides the slivers shrink and the areas approach 9π.
Question 18 of 20 · Short Answer
A stack of poker chips with a radius of 2 cm is 18 cm tall. It is pushed so that it leans to one side. Find its volume and explain why leaning the stack does not change it.
Each chip is still a disk of area π(2²) = 4π cm², and the height is still 18 cm. Every horizontal slice of the leaning stack has the same area as the matching slice of the straight stack, so by Cavalieri's principle the volumes are equal: V = 4π · 18 = 72π ≈ 226.2 cm³.
Question 19 of 20 · Short Answer
A cube has an edge of 4 cm. It is cut into three congruent pyramids that share one vertex. Find the volume of each pyramid and explain how the cube shows the formula V = (1/3)Bh.
The cube has volume 4³ = 64 cm³, so each pyramid has 64/3 ≈ 21.3 cm³. Each pyramid has one face of the cube as its base (B = 16 cm²) and an edge of the cube as its height (h = 4 cm), so Bh is the cube's volume and each pyramid is one third of Bh: (1/3)(16)(4) = 64/3.
Question 20 of 20 · Short Answer
A cone has a radius of 3 in and a height of 7 in. Use a pyramid with the same base area and height and Cavalieri's principle to explain its volume formula, and find its volume.
Take a pyramid with base area 9π in² and height 7 in. At the same fraction of the height, both slices are scaled copies of their bases with the same scale factor, so they have equal areas. By Cavalieri's principle the cone has the pyramid's volume, (1/3)Bh = (1/3)πr²h. Here V = (1/3)(9π)(7) = 21π ≈ 66.0 in³.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GMD.A.1 mean?
HSG.GMD.A.1 asks students to explain, informally, why the formulas for the circumference and area of a circle and the volumes of a cylinder, pyramid and cone are true. The official text names the methods: dissection arguments, Cavalieri's principle and informal limit arguments.
Is HSG.GMD.A.1 taught in Geometry?
Yes, it is usually taught in high school Geometry, in the unit on area and volume. Students learn and use the formulas in Grades 7 and 8 (7.G.B.4 and 8.G.C.9); HSG.GMD.A.1 adds the reasons behind them.
What counts as an informal argument?
An informal argument explains why a formula makes sense with a picture, a model or a pattern, without a complete formal proof. For example, rearranging sectors of a circle into a near-parallelogram shows why the area is πr · r, even though the shape is never exactly a parallelogram.
What is Cavalieri's principle?
Cavalieri's principle says that two solids with the same height have the same volume if every plane parallel to the base cuts them in cross-sections of equal area. A straight stack of coins and a slanted stack of the same coins show the idea.
Why is there a one-third in the pyramid and cone formulas?
Because three congruent pyramids fill a cube, each with the cube's face as its base and the cube's edge as its height. So a pyramid holds one third of the prism with the same base and height. A cone is a limit of pyramids with polygon bases, so it keeps the one-third.
How can you show the area of a circle is πr² without calculus?
Cut the circle into many equal sectors and arrange them in a row with the tips alternating. The shape is close to a parallelogram with base πr, half the circumference, and height r, so its area is close to πr². The more sectors, the closer the fit.
What is an informal limit argument in HSG.GMD.A.1?
It is an argument that looks at approximations that get better and better. Inscribed polygons with 6, 12, 24, ... sides have perimeters and areas that get as close as we like to the circumference and area of the circle, and pyramids with polygon bases get as close as we like to a cone.
What mistakes do students make with these formulas?
Common errors include:
mixing up 2πr and πr², or using the diameter where the radius belongs
thinking an oblique cylinder or cone holds more or less than a right one with the same base and height
forgetting the one-third for pyramids and cones
checking only the bases in Cavalieri's principle and forgetting the equal heights
Does HSG.GMD.A.1 include the volume of a sphere?
No. The sphere is in HSG.GMD.A.2, an advanced (+) standard that uses Cavalieri's principle to argue for the volume of a sphere. Using the sphere formula to solve problems belongs to HSG.GMD.A.3.
How does HSG.GMD.A.1 connect to later math?
The same arguments return in HSG.GMD.A.3, where students use the volume formulas in problems, and in HSG.GMD.A.2, where Cavalieri's principle gives the volume of a sphere. The slicing and limit ideas also prepare students for integrals in calculus, where volumes are found by adding thin slices.
07
Related Standards
6 standards
These standards connect to HSG.GMD.A.1: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.G.B.4Prerequisite
Know circle area and circumference formulas and relate the two informally
