Math II · G-GMD.1

Explaining Circumference, Area, and Volume Formulas Instead of Just Memorizing Them

This objective teaches students that geometry formulas are not magic sayings. They are compressed explanations about how space is built, sliced, rearranged, and measured.

Concept Geometry
Domain Geometric Measurement and Dimension
Read time 10 minutes

What this learning objective is really asking you to learn

This objective asks students to move from formula users to formula explainers. Many students know that the circumference of a circle is 2πr, the area of a circle is πr², the volume of a cylinder is πr²h, and the volume of a cone or pyramid is one third of a matching cylinder or prism. But knowing a formula by memory is not the same thing as understanding what the formula represents. A student who only memorizes formulas is carrying a bag of isolated commands. A student who understands the formulas is carrying a map of space.

The standard focuses on informal arguments, which means students do not need a fully formal calculus proof. They do need a serious explanation. “Because the teacher said so” is not enough. “Because it is on the formula sheet” is not enough. Students should be able to say why a cylinder’s volume is the area of its base multiplied by its height, why a cone’s volume is one third of the cylinder with the same base and height, why a circle’s area depends on the square of its radius, and why circle circumference is proportional to radius.

The objective names three powerful kinds of reasoning: dissection, Cavalieri’s principle, and informal limits. Dissection means cutting a shape apart and rearranging the pieces into a shape whose area or volume is easier to understand. A rectangle is easy because its area is rows times columns. A parallelogram can be cut and rearranged into a rectangle. A circle can be cut into many sectors and rearranged into an almost-parallelogram. Dissection tells students that area can remain the same even when the shape changes, as long as the pieces are not stretched or overlapped.

Cavalieri’s principle is a slicing idea. It says, in a student-friendly form, that if two solids have equal cross-sectional areas at every height and the same height, then they have the same volume. Imagine two stacks of paper. One stack is perfectly vertical. Another stack is slanted or shifted sideways. If every sheet has the same area and the same thickness, the total volume is the same. The shape may lean, but the amount of material has not changed. This idea explains why the volume of an oblique cylinder is still base area times height and why many volume comparisons can be made by slicing.

Informal limits are about approaching an exact result through increasingly fine approximations. A circle can be approximated by regular polygons with more and more sides. A curved boundary can be approximated by many tiny straight pieces. A cone can be approximated by a stack of thinner and thinner circular disks. The more slices or pieces you use, the closer the approximation gets to the real curved object. This is the doorway to calculus, but students can understand the intuition before they learn formal limits.

Why students should learn this math

Students should learn this math because formulas are only useful when you know what they measure, when they apply, and why they behave the way they do. In real life, people do not only plug numbers into formulas. They choose the right formula, check whether the assumptions fit, estimate whether the answer is reasonable, and adapt the formula when the object is not perfectly textbook-shaped.

A construction worker estimating concrete, an engineer designing a tank, a designer sizing packaging, a baker scaling a cylindrical cake, a city planner estimating pipe capacity, a mechanic thinking about a piston, and a scientist modeling flow all need measurement reasoning. If a student understands that volume is built from cross-sectional area times height, then a formula becomes flexible. The student can reason about unusual containers, layered materials, slanted solids, and approximations. If the student only memorized a list, the first nonstandard situation breaks the method.

This objective also answers one of the biggest student complaints: “Why do I need to know where formulas come from?” Because derivations build transfer. If you know that cylinder volume is base area times height, you can remember the formula even when you forget the letters. If you know that a cone is one third of a cylinder with the same base and height, you can compare volumes without redoing everything. If you know that circle area is related to rearranging many tiny sectors into a nearly rectangular shape, you can see why πr² has square units and why doubling the radius quadruples the area.

There is a practical honesty here. Many students can survive a quiz by memorizing formulas the night before. But they cannot solve unfamiliar problems that way. Real problems often ask, “What matters here?” “What can be ignored?” “What shape does this resemble?” “What would happen if the dimension doubled?” “What is a reasonable estimate?” A student who understands the machinery of measurement can answer those questions. A memorizer often cannot.

This math also teaches students how humans discovered reliable measurement. The formulas did not drop from the sky. They came from people needing to measure land, build structures, store grain, design containers, navigate, and study the heavens. Measurement formulas are part of civilization’s operating system. They allowed people to compare areas, plan materials, understand capacity, and design at scale. When students learn why the formulas work, they are not just learning school math; they are entering a long human project of making space intelligible.

The historical machinery behind these formulas

The history of measurement is older than formal algebra. Ancient societies needed to measure fields, build canals, tax land, store food, and construct buildings. Rectangular area and box volume were natural early ideas because they connected to counting equal units. Circles and curved solids were harder. A circle does not break naturally into square tiles without gaps, and a cone does not stack into equal rectangular layers. To understand curved shapes, mathematicians needed approximation, comparison, and clever rearrangement.

The constant π emerged from the observation that every circle has the same ratio between circumference and diameter. That is a remarkable fact. A small coin, a bicycle wheel, and a circular stadium all share the same circumference-to-diameter ratio. The size changes, but the ratio stays fixed. Ancient mathematicians estimated this ratio by comparing circles to polygons. If you inscribe a many-sided polygon inside a circle and circumscribe another outside it, the circle’s circumference lies between their perimeters. Increasing the number of sides tightens the estimate. This is a classic informal limit idea: the polygon becomes a better and better stand-in for the circle.

Circle area also developed through comparison. One famous student-friendly dissection argument cuts a circle into many equal sectors, like pizza slices, then rearranges the slices alternating up and down. The result looks increasingly like a parallelogram or rectangle. The height is approximately the radius. The base is approximately half the circumference, because half the arc lengths lie along one side and half along the other. So the area is approximately \((1/2)(2πr)(r) = πr²\). As the slices become thinner, the rearranged shape approaches a true rectangle. This is not a formal proof by modern standards, but it captures the essential mechanism.

Volume formulas also have a long history. The cylinder formula follows from stacking congruent circular layers. If every layer has area \(B\) and the height is \(h\), then the volume is \(Bh\). Pyramids and cones are subtler. Ancient mathematicians discovered that a pyramid has one third the volume of a prism with the same base area and height, and a cone has one third the volume of a cylinder with the same base area and height. These facts can be explained by dissection in special cases, by comparison of cross-sections, or by limiting stacks of slices.

Cavalieri’s principle came much later in formal historical development, associated with Bonaventura Cavalieri in the seventeenth century, but the slicing idea is accessible earlier. It says volume can be understood by adding up cross-sectional areas. If two solids have matching slices at every height, they have the same volume. This principle helped bridge classical geometry and calculus. It gave mathematicians a powerful way to handle curved solids without having to cut them into obvious finite pieces.

The technical machinery: circumference and area of a circle

The circumference formula begins with the definition of π. For every circle, \(π = C/d\), where \(C\) is circumference and \(d\) is diameter. Since diameter is twice radius, \(d = 2r\). Rearranging gives \(C = πd = 2πr\). This formula is not arbitrary. It says circumference grows linearly with radius. If the radius doubles, the circumference doubles. This makes sense because all circles are similar; enlarging every length by the same scale factor doubles every length.

The area formula works differently because area is two-dimensional. If the radius doubles, the area does not double; it quadruples. That is why the formula uses . The dissection argument gives a visual explanation. Cut the circle into many sectors. Each sector has two radius sides and a small arc. Rearranged, the sectors form a shape whose height is close to \(r\) and whose base is close to half the circumference, πr. Multiplying base by height gives \(πr \cdot r = πr²\).

Another explanation uses rings. Imagine a circle as many very thin circular rings nested inside each other. A ring at radius \(x\) has circumference about 2πx and tiny thickness. The total area is like adding the circumferences of all rings from radius 0 to radius \(r\). The rings near the center are short; the rings near the outside are longer. This is an informal calculus-style argument. It helps students see why circle area is connected to accumulating circumferences.

Students should be careful with units. Circumference is a length, so it uses units like centimeters or meters. Area uses square units. If \(r\) is measured in meters, πr² is in square meters. The square in the formula is not decoration; it tells us the measurement is two-dimensional.

The technical machinery: cylinder, pyramid, and cone volume

A cylinder can be understood as a stack of identical circular layers. The base has area \(B = πr²\). If the height is \(h\), the total volume is \(Bh = πr²h\). This is the same idea as a rectangular prism: base area times height. The base does not have to be a rectangle. The key is that each horizontal cross-section has the same area as the base.

A pyramid has volume \(V = (1/3)Bh\), where \(B\) is the area of the base and \(h\) is the perpendicular height. The factor \(1/3\) is the part students most need to understand. A pyramid is not simply a prism that gets smaller uniformly from bottom to top; its cross-sections shrink by similarity. At halfway up, a cross-section parallel to the base has side lengths half as large as the base, so its area is one fourth as large. The areas of slices shrink quadratically as height changes. Adding all those shrinking slices gives one third of the matching prism volume.

A cone works the same way as a pyramid, except the base is circular. Its volume is \(V = (1/3)πr²h\). A cone’s horizontal slices are circles whose radii shrink linearly from \(r\) at the base to 0 at the tip. Because area depends on the square of radius, the slice areas shrink quadratically. Again, the accumulated volume is one third of the corresponding cylinder.

Cavalieri’s principle helps students see why pyramids and cones with the same base area and height have the same volume pattern. If cross-sections at corresponding heights have predictable area relationships, volume follows from those slices. This is also why oblique solids can have the same volume as straight ones when the cross-sectional areas match at each height.

What can go wrong, and how to fix it

The most common mistake is treating formulas as unrelated facts. Students memorize πr², 2πr, \(Bh\), and \((1/3)Bh\) but cannot explain which is length, which is area, and which is volume. The fix is to attach each formula to dimensional meaning. Circumference is one-dimensional distance around. Area is two-dimensional coverage. Volume is three-dimensional capacity.

Another mistake is confusing radius and diameter. Since circumference can be written as πd or 2πr, students sometimes mix them and write 2πd. The fix is to draw the circle, label the radius and diameter, and remember that diameter already includes two radii.

A third mistake is forgetting the \(1/3\) for cones and pyramids. Students may write \(Bh\) because they remember base times height. The fix is to compare the solid to its matching prism or cylinder. If the top collapses to a point, the volume is not the full stack; it is one third of that stack.

A fourth mistake is using slant height instead of perpendicular height. Volume formulas require perpendicular height, not the length along the side. This matters especially for cones and pyramids. Slant height belongs to surface area calculations, not volume.

Where this fits into the big map of math

This objective is a major bridge between geometry and calculus. Students are not yet doing formal integration, but they are learning the intuition: cut a shape into pieces, approximate curved objects with simpler ones, compare slices, and let approximations become more accurate. That is the conceptual seed of integral calculus.

It also connects geometry to modeling. Real objects are rarely perfect textbook solids, but they can often be approximated by cylinders, cones, pyramids, or combinations of them. A silo is roughly a cylinder with a cone on top. A paper cup is a truncated cone. A storage tank might be a cylinder with rounded ends. Understanding formula origins helps students adapt, estimate, and reason instead of freezing when a problem is not identical to an example.

In the full map of mathematics, this objective teaches that formulas are compressed structure. They are not arbitrary rules. Circumference expresses similarity of circles. Area expresses two-dimensional accumulation. Volume expresses three-dimensional stacking. The \(1/3\) in pyramids and cones expresses how cross-sections shrink. The use of dissection, Cavalieri, and limits shows students three of the most important mathematical moves: rearrange, slice, and approximate. Those moves will keep returning in geometry, algebra, statistics, physics, engineering, computer graphics, and calculus.

Problem Library

Problems in the App From This Objective

144 problems across 12 archetypes in the app.

connect `C=pi d` and `C=2pi r`.
12 problems Warmup Practice Mixed Review Assessment
Problem 1

Explain the circumference formula for a circle with diameter d.

Problem 2

Explain the circumference formula for a circle with radius r.

Problem 3

Explain the circumference formula for a circle with diameter 10.

Open in simulator
Problem 4

Explain the circumference formula for a circle with radius 6.

Problem 5

Explain the circumference formula for a circle with diameter 7.

Problem 6

Explain the circumference formula for a circle with radius 5.

Problem 7

Explain the circumference formula for a circle with diameter 15.

Problem 8

Explain the circumference formula for a circle with radius 8.

Problem 9

Explain the circumference formula for a circle with diameter 25.

Problem 10

Explain the circumference formula for a circle with radius 12.

Problem 11

Explain the circumference formula for a circle with diameter 0.5.

Problem 12

Explain the circumference formula for a circle with radius 2.5.

connect rearranged sectors to rectangle-like area.
12 problems Warmup Practice Mixed Review Assessment
Problem 13

Explain the circle area formula using sector rearrangement for many equal sectors rearranged alternately.

Problem 14

Explain the circle area formula using sector rearrangement for a circle cut into thin wedges.

Open in simulator
Problem 15

Explain the circle area formula using sector rearrangement for sector pieces becoming narrower.

Problem 16

Explain the circle area formula using sector rearrangement for using circumference in the rearranged shape.

Problem 17

Explain the circle area formula using sector rearrangement for a circle divided into 100 sectors.

Problem 18

Explain the circle area formula using sector rearrangement for infinitely many sectors forming a perfect rectangle.

Problem 19

Explain the circle area formula using sector rearrangement for sectors creating a wavy parallelogram that straightens.

Problem 20

Explain the circle area formula using sector rearrangement for alternating sector tips forming a straight line.

Problem 21

Explain the circle area formula using sector rearrangement for the arc lengths forming the base of the rectangle.

Problem 22

Explain the circle area formula using sector rearrangement for the radius becoming the width of the rearranged figure.

Problem 23

Explain the circle area formula using sector rearrangement for two halves of the circumference forming the top and bottom.

Problem 24

Explain the circle area formula using sector rearrangement for the jagged edges becoming smooth lines.

relate triangle to half a parallelogram or rectangle.
12 problems Warmup Practice Mixed Review Assessment
Problem 25

Explain the triangle area formula using dissection for right triangle.

Problem 26

Explain the triangle area formula using dissection for acute triangle.

Problem 27

Explain the triangle area formula using dissection for obtuse triangle.

Open in simulator
Problem 28

Explain the triangle area formula using dissection for any triangle with base b and height h.

Problem 29

Explain the triangle area formula using dissection for isosceles triangle.

Problem 30

Explain the triangle area formula using dissection for equilateral triangle.

Problem 31

Explain the triangle area formula using dissection for general triangle.

Problem 32

Explain the triangle area formula using dissection for triangle with base B and height H.

Problem 33

Explain the triangle area formula using dissection for triangle within a rectangle.

Problem 34

Explain the triangle area formula using dissection for scalene triangle.

Problem 35

Explain the triangle area formula using dissection for triangle by completing a parallelogram.

Problem 36

Explain the triangle area formula using dissection for a triangle and its corresponding parallelogram.

rearrange to rectangle.
12 problems Warmup Practice Mixed Review Assessment
Problem 37

Explain the parallelogram area formula by dissection for a slanted parallelogram with base b and height h.

Problem 38

Explain the parallelogram area formula by dissection for a parallelogram with base 10 and height 6.

Problem 39

Explain the parallelogram area formula by dissection for an oblique parallelogram.

Open in simulator
Problem 40

Explain the parallelogram area formula by dissection for a parallelogram on grid paper.

Problem 41

Explain the parallelogram area formula by dissection for a parallelogram with base 8 cm and height 5 cm.

Problem 42

Explain the parallelogram area formula by dissection for a general parallelogram ABCD.

Problem 43

Explain the parallelogram area formula by dissection for a parallelogram with a base of 15 units and a height of 10 units.

Problem 44

Explain the parallelogram area formula by dissection for a parallelogram that is not a rectangle.

Problem 45

Explain the parallelogram area formula by dissection for a parallelogram with base 'b' and altitude 'h'.

Problem 46

Explain the parallelogram area formula by dissection for a parallelogram with a base of 20 meters and a height of 7 meters.

Problem 47

Explain the parallelogram area formula by dissection for a parallelogram with a given base and its corresponding altitude.

Problem 48

Explain the parallelogram area formula by dissection for a parallelogram with base length L and perpendicular height H.

use layers of congruent bases.
12 problems Warmup Practice Mixed Review Assessment
Problem 49

Explain prism volume as base area times height for a rectangular prism.

Problem 50

Explain prism volume as base area times height for a triangular prism.

Problem 51

Explain prism volume as base area times height for a prism with base area 20 and height 7.

Problem 52

Explain prism volume as base area times height for an oblique prism.

Problem 53

Explain prism volume as base area times height for a hexagonal prism.

Open in simulator
Problem 54

Explain prism volume as base area times height for a pentagonal prism.

Problem 55

Explain prism volume as base area times height for a prism with base area 15 and height 10.

Problem 56

Explain prism volume as base area times height for a right circular cylinder.

Problem 57

Explain prism volume as base area times height for a prism with a trapezoidal base.

Problem 58

Explain prism volume as base area times height for a prism with base area B and height H.

Problem 59

Explain prism volume as base area times height for a prism whose base is a regular octagon.

Problem 60

Explain prism volume as base area times height for a prism with height 5 and base area 25.

use circular layers.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Explain cylinder volume as base area times height for radius r and height h.

Problem 62

Explain cylinder volume as base area times height for radius 4 and height 10.

Problem 63

Explain cylinder volume as base area times height for diameter 12 and height h.

Problem 64

Explain cylinder volume as base area times height for an oblique cylinder.

Problem 65

Explain cylinder volume as base area times height for radius 5 and height 8.

Problem 66

Explain cylinder volume as base area times height for diameter 6 and height 7.

Open in simulator
Problem 67

Explain cylinder volume as base area times height for radius r and height 2h.

Problem 68

Explain cylinder volume as base area times height for diameter d and height h.

Problem 69

Explain cylinder volume as base area times height for a can with radius 3cm and height 15cm.

Problem 70

Explain cylinder volume as base area times height for a cylinder with base area 25pi and height 6.

Problem 71

Explain cylinder volume as base area times height for a cylinder whose radius is half its height, with height H.

Problem 72

Explain cylinder volume as base area times height for a cylinder with circumference 10pi and height 5.

relate pyramid volume to one-third prism volume.
12 problems Warmup Practice Mixed Review Assessment
Problem 73

Explain the pyramid volume formula informally for a pyramid and prism with the same base and height.

Problem 74

Explain the pyramid volume formula informally for a rectangular pyramid under a matching rectangular prism.

Problem 75

Explain the pyramid volume formula informally for base area 30 and height 9.

Problem 76

Explain the pyramid volume formula informally for triangular pyramid and matching prism.

Open in simulator
Problem 77

Explain the pyramid volume formula informally for a square pyramid and a square prism with identical bases and heights.

Problem 78

Explain the pyramid volume formula informally for a pyramid with base area 45 and height 12.

Problem 79

Explain the pyramid volume formula informally for the general relationship between a pyramid and a prism.

Problem 80

Explain the pyramid volume formula informally for a pentagonal pyramid and a pentagonal prism of equal base area and height.

Problem 81

Explain the pyramid volume formula informally for how many pyramids fit into a prism of the same base and height.

Problem 82

Explain the pyramid volume formula informally for the formula V = (1/3)Bh for a pyramid.

Problem 83

Explain the pyramid volume formula informally for a pyramid with a base area of 60 square units and a height of 15 units.

Problem 84

Explain the pyramid volume formula informally for any pyramid compared to a prism with the same base and height.

relate cone volume to one-third cylinder volume.
12 problems Warmup Practice Mixed Review Assessment
Problem 85

Explain the cone volume formula informally for a cone and cylinder with the same radius and height.

Problem 86

Explain the cone volume formula informally for radius 5 and height 12.

Problem 87

Explain the cone volume formula informally for a cone inside a same-base same-height cylinder.

Problem 88

Explain the cone volume formula informally for diameter 8 and height h.

Problem 89

Explain the cone volume formula informally for radius 3 and height 9.

Problem 90

Explain the cone volume formula informally for diameter 10 and height 6.

Problem 91

Explain the cone volume formula informally for radius r and height h.

Problem 92

Explain the cone volume formula informally for diameter d and height H.

Problem 93

Explain the cone volume formula informally for the ratio of a cone's volume to a cylinder's volume.

Open in simulator
Problem 94

Explain the cone volume formula informally for radius 4 and height 3 times the radius.

Problem 95

Explain the cone volume formula informally for filling a cone with water from a cylinder.

Problem 96

Explain the cone volume formula informally for radius 6 and height 7.5.

connect formula to accumulation/slicing idea.
12 problems Warmup Practice Mixed Review Assessment
Problem 97

Explain the sphere volume formula informally for sphere of radius r.

Problem 98

Explain the sphere volume formula informally for sphere compared with a cylinder of radius r and height 2r.

Problem 99

Explain the sphere volume formula informally for radius 3 sphere.

Problem 100

Explain the sphere volume formula informally for why radius is cubed.

Problem 101

Explain the sphere volume formula informally for the origin of the 4/3 constant.

Problem 102

Explain the sphere volume formula informally for how a sphere's volume relates to a cone.

Problem 103

Explain the sphere volume formula informally for why pi is in the formula for sphere volume.

Problem 104

Explain the sphere volume formula informally for a sphere of arbitrary radius R.

Problem 105

Explain the sphere volume formula informally for visualizing sphere volume with slices.

Problem 106

Explain the sphere volume formula informally for Cavalieri's Principle for sphere volume.

Open in simulator
Problem 107

Explain the sphere volume formula informally for the three-dimensional nature of sphere volume.

Problem 108

Explain the sphere volume formula informally for how volume scales with radius for a sphere.

compare equal-height cross-sections.
12 problems Warmup Practice Mixed Review Assessment
Problem 109

Use Cavalieri's principle to compare volumes for two prisms with equal height and equal cross-sectional area at every level.

Open in simulator
Problem 110

Use Cavalieri's principle to compare volumes for an oblique cylinder and right cylinder with the same base area and height.

Problem 111

Use Cavalieri's principle to compare volumes for two solids with equal heights but different cross-sectional areas at some levels.

Problem 112

Use Cavalieri's principle to compare volumes for two stacks with matching cross-section area at every height.

Problem 113

Use Cavalieri's principle to compare volumes for two cones with equal height and equal base area.

Problem 114

Use Cavalieri's principle to compare volumes for a stack of CDs and a stack of DVDs, both with the same height and diameter.

Problem 115

Use Cavalieri's principle to compare volumes for two prisms with the same base area but different heights.

Problem 116

Use Cavalieri's principle to compare volumes for a sphere and a cube with the same height.

Problem 117

Use Cavalieri's principle to compare volumes for a right pyramid and an oblique pyramid with the same base area and height.

Problem 118

Use Cavalieri's principle to compare volumes for two cylinders with the same height but different base areas.

Problem 119

Use Cavalieri's principle to compare volumes for a shifted stack of identical coins and a straight stack of the same coins, both having the same total height.

Problem 120

Use Cavalieri's principle to compare volumes for a cone and a cylinder with the same height and base area.

match dissection, Cavalieri, or limit idea to formula.
12 problems Warmup Practice Mixed Review Assessment
Problem 121

Choose the informal argument that supports formula A=bh for a parallelogram.

Problem 122

Choose the informal argument that supports formula V=Bh for a prism.

Problem 123

Choose the informal argument that supports formula V=1/3 pi r^2 h for a cone.

Open in simulator
Problem 124

Choose the informal argument that supports formula A=pi r^2 for a circle.

Problem 125

Choose the informal argument that supports formula A = 1/2 bh for a triangle.

Problem 126

Choose the informal argument that supports formula A = 1/2 h (b1 + b2) for a trapezoid.

Problem 127

Choose the informal argument that supports formula A = 1/2 d1 d2 for a kite.

Problem 128

Choose the informal argument that supports formula V = pi r^2 h for a cylinder.

Problem 129

Choose the informal argument that supports formula V = 1/3 Bh for a pyramid.

Problem 130

Choose the informal argument that supports formula V = Bh for an oblique cylinder.

Problem 131

Choose the informal argument that supports formula V = 4/3 pi r^3 for a sphere.

Problem 132

Choose the informal argument that supports formula A = 1/2 a P for a regular polygon.

catch dimension mismatch, wrong base/height, or invalid comparison.
12 problems Warmup Practice Mixed Review Assessment
Problem 133

Correct the formula-explanation error: A student explains triangle area using a slanted side as height.

Open in simulator
Problem 134

Correct the formula-explanation error: A student says cylinder volume is circumference times height.

Problem 135

Correct the formula-explanation error: A student says a cone has half the volume of a matching cylinder.

Problem 136

Correct the formula-explanation error: A student compares two solids by height only using Cavalieri's principle.

Problem 137

Correct the formula-explanation error: A student states the area of a circle is 2 * pi * r.

Problem 138

Correct the formula-explanation error: A student calculates the volume of a rectangular prism as length * width.

Problem 139

Correct the formula-explanation error: A student calculates the area of a parallelogram by multiplying two adjacent sides.

Problem 140

Correct the formula-explanation error: A student says the volume of a pyramid is base area times height.

Problem 141

Correct the formula-explanation error: A student calculates the area of a trapezoid by averaging the non-parallel sides and multiplying by height.

Problem 142

Correct the formula-explanation error: A student explains that the amount of space inside a box is its surface area.

Problem 143

Correct the formula-explanation error: A student states the volume of a sphere is (1/3) * pi * r^3.

Problem 144

Correct the formula-explanation error: A student claims two rectangles have the same area because their perimeters are equal.