HSG.MG.A.1: Modeling Real Objects with Geometric Shapes
In plain English: HSG.MG.A.1 is the Common Core geometry standard that asks students to describe real objects with geometric shapes, their measures and their properties, such as modeling a tree trunk or a human torso as a cylinder. Students choose a shape, take the measurements it needs, compute lengths, areas or volumes, and judge how well the model fits. It is usually taught in high school Geometry.
Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder).
Common Core State Standards for Mathematics · Domain: Modeling with Geometry (MG) · Cluster: Apply geometric concepts in modeling situations Also written as HSG-MG.A.1 or G-MG.1 · Official standard
Students learn to describe real objects with geometric shapes: a tree trunk or a torso as a cylinder, a roof as a triangular prism, a ball as a sphere, a traffic cone as a cone. The point is not only to apply a formula but to choose a shape whose properties match the object, take the measurements that shape needs, and use the shape's measures to answer a real question.
Students also judge their models. They compare two models of the same object, decide whether a model overestimates or underestimates, and explain what the model ignores. The official examples of the standard, a tree trunk and a human torso modeled as cylinders, are worked in full.
Learning Objectives
By the end of this lesson, students will be able to:
Choose a geometric shape or combination of shapes whose properties match a real object
Identify and take the measurements a model needs, such as a circumference, and convert them to the dimensions a formula uses
Use the measures and properties of the shape (lengths, areas, volumes, slants) to answer questions about the object
Compare models of the same object and explain whether each overestimates or underestimates
State the limits of a model in context
Prior Knowledge Required
Students should already be comfortable with:
Area and circumference of a circle 7.G.B.4
Area, surface area and volume of prisms and other solids 7.G.B.6
The Pythagorean Theorem in two and three dimensions 8.G.B.7
Volume formulas for cylinders, cones and spheres HSG.GMD.A.3
Choosing and converting units in a multi-step problem HSN.Q.A.1
Show photos of five objects: a soup can, a tree trunk, a basketball, a tent and a person standing. Ask students to work in pairs for three minutes:
Warm-Up Prompt
"For each object, name one geometric shape that could describe it. What would you need to measure to find how much it holds or how much material covers it? Which object is hardest to describe with one shape, and why?"
Collect answers on the board. Expect cylinder for the can and the trunk, sphere for the basketball, triangular prism for the tent, and a debate about the person. Accept more than one answer and ask what each choice ignores: a trunk is not perfectly round and it gets thinner as it rises. Tell students that a model is a useful simplification, not a perfect copy, and that this lesson is about choosing and using good ones.
Direct Instruction20 minutes
Present the modeling steps students will use for every object in the lesson:
Choose a shape whose properties match the object: a constant circular cross-section suggests a cylinder, a point on top suggests a cone or pyramid, and the same distance from a center in every direction suggests a sphere.
Decide which measures you need and which are easy to take. For a tree, the circumference is easy to measure with a tape; the radius is not.
Compute with the shape's formulas and properties, such as r = C ÷ 2π, the Pythagorean Theorem for a slant, or the lateral area of a cylinder, C × h.
Interpret and check: give units, round sensibly, and ask whether the answer is reasonable and whether the model overestimates or underestimates.
Tree trunk as a cylinder
A forester wraps a tape around a trunk and reads a circumference of 2.2 m. The trunk rises 8 m to the first branch. Estimate the volume of wood in that section.
Equation: r = 2.2 ÷ 2π ≈ 0.35 m; V = πr²h = C²h ÷ 4π ≈ 3.08 m³
Human torso as a cylinder
A torso has a waist circumference of 80 cm and a height of 60 cm from the hips to the shoulders. How much fabric covers it, ignoring overlap, and what is its volume?
Equation: Side area = C × h = 80 × 60 = 4,800 cm²; V = 80² × 60 ÷ 4π ≈ 30,558 cm³ ≈ 30.6 L
Roof as a triangular prism
An A-frame roof is 10 m long. Its cross-section is an isosceles triangle with base 8 m and height 3 m. How many square meters of shingles cover it?
Equation: Slant = √(4² + 3²) = 5 m; area = 2 × 5 × 10 = 100 m²
Comparing models of a tapering trunk
A trunk section is 6 m tall, with diameter 0.8 m at the bottom and 0.5 m at the top. Compare cylinder models that use each diameter and their average.
Equation: Bottom: 0.96π ≈ 3.02 m³; top: 0.375π ≈ 1.18 m³; average d = 0.65 m: 0.634π ≈ 1.99 m³
Pencil as a hexagonal prism
An unsharpened pencil is a regular hexagonal prism with side 4 mm and length 160 mm. How much surface is painted, not counting the ends?
For the tapering trunk, ask which model is most reasonable. The bottom diameter overestimates and the top diameter underestimates, because the true trunk lies between the two cylinders. The average-diameter cylinder is a reasonable middle estimate. Students who want a closer model can treat the trunk as part of a cone (a frustum); its volume, about 2.03 m³, is close to the average model. Use Diagram 1 for the first two examples and Diagram 2 for the roof.
Guided Practice15 minutes
Pairs model three objects. For each, they name the shape, list the measures, compute, and write one sentence on what the model ignores:
A round hay bale 1.5 m in diameter and 1.2 m wide, as a cylinder: V = π(0.75)²(1.2) = 0.675π ≈ 2.12 m³. It ignores the loose outer layer.
A traffic cone 70 cm tall with a base diameter of 25 cm, as a cone (ignore the square base plate): V = ⅓π(12.5)²(70) ≈ 11,454 cm³, about 11.5 L.
A paper party hat with base radius 8 cm and slant height 20 cm, as the lateral surface of a cone: area = πrℓ = 160π ≈ 503 cm² of paper.
Listen for students who use a diameter as a radius, who mix meters and centimeters, and who use the height of the cone where the slant height is needed.
Independent Practice10-15 minutes
Students model four objects on their own:
A drum with diameter 36 cm: the area of its circular head, 324π ≈ 1,018 cm².
An igloo modeled as a hemisphere with inside radius 1.5 m: floor area π(1.5)² ≈ 7.07 m² and air volume ⅔π(1.5)³ ≈ 7.07 m³. Ask why the two numbers match here and why they still describe different things.
A shipping container as a rectangular prism with inside dimensions 12.0 m by 2.35 m by 2.39 m: volume ≈ 67.4 m³.
A pack of 500 sheets of paper, 5 cm thick, as a rectangular prism: each sheet is 5 ÷ 500 = 0.01 cm = 0.1 mm thick.
Closure5-10 minutes
Exit ticket: (1) Name a shape to model a water bottle, a pizza box and a funnel, and one measurement you would take for each. (2) A student models a pine tree, trunk and branches together, as a cone. Name one property of the tree that the cone captures and one it misses. Collect the tickets and use them to open the next class with two strong and two weak models.
Differentiation Strategies
For Struggling Students
Provide a shape card for each solid with its picture, its key property and its formulas in terms of r, d and C
Start with objects in hand (a can, a ball) before objects in photos, so students can measure and see the shape
Use a three-column organizer: object, shape and measures, computation and check
For Advanced Students
Model a tapering trunk as a frustum of a cone and compare the result with the cylinder models
Build a composite model of a person from a sphere and several cylinders and compare its volume with the person's mass in kilograms
Find the error in a model by measuring a real object two ways, such as a can's computed volume and the volume on its label
Assessment Guidance
What to Look For
Look for a named shape with a reason tied to a property of the object, not only a formula. Check that students convert what they measured into what the formula needs (circumference to radius, diameter to radius, slant from height) and keep units consistent. Strong answers state whether the model is too big or too small and why. A student who gives a volume to six decimal places for a tree has not yet thought about the precision a model can support.
02
Classroom Activities
3 Activities
1
Measure and Model Stations
25 minGroups of 3-4
Groups rotate through four stations. At each one they choose a shape, measure the real object, compute a volume, and compare with what the object is known to hold when that information is printed on it.
Stations, with Sample Measurements
Soda can as a cylinder: diameter 6.6 cm and height 12.2 cm give V = π(3.3)²(12.2) ≈ 417 cm³, while the label says 355 mL
Tennis ball as a sphere: diameter 6.7 cm gives V = ⁴⁄₃π(3.35)³ ≈ 157 cm³
Cereal box as a rectangular prism: 19 cm by 7 cm by 28 cm gives V = 3,724 cm³
Paper cone cup as a cone: rim diameter 7 cm and height 9 cm give V = ⅓π(3.5)²(9) ≈ 115 cm³
Discussion Questions
The can model gives about 417 cm³ but the label says 355 mL. Name two reasons for the difference. (The top and bottom are narrowed and domed, and the can is not filled to the top.)
Which object was best described by its shape? Which was worst?
Which measurement was hardest to take accurately, and how does that affect the answer?
Modification for Distance Learning
Students choose four objects at home that match the four shapes, measure them with a ruler or a piece of string, and post a photo with their model and computation to a shared class page.
2
Tree Survey
20 minPairs
Foresters describe a tree by its diameter at chest height, and they get it from the circumference because a tape fits around a trunk and a ruler does not fit through it. Pairs use invented data for four trees, or measure trees on campus, and model each trunk as a cylinder up to its first branch.
Tree Data (Invented)
Tree A: circumference 1.26 m, trunk height 6 m, so d ≈ 0.40 m and V ≈ 0.76 m³
Tree B: circumference 1.88 m, trunk height 7 m, so d ≈ 0.60 m and V ≈ 1.97 m³
Tree C: circumference 2.51 m, trunk height 9 m, so d ≈ 0.80 m and V ≈ 4.51 m³
Tree D: circumference 0.94 m, trunk height 4 m, so d ≈ 0.30 m and V ≈ 0.28 m³
Procedure
Find each diameter with d = C ÷ π and each volume with V = C²h ÷ 4π
Order the trees by volume and compare with the order by circumference
Tree C has twice the circumference of Tree A, rounded. Predict how its cross-section area compares, then check (about 4 times)
Challenge Variation
Give each pair the diameter at the top of the trunk section for Tree C, 0.60 m, and ask them to build a better model. Compare an average-diameter cylinder with the original model and decide which estimate a lumber buyer should trust.
3
The Body as Shapes
20 minGroups of 3
Groups model a person as a sphere for the head and cylinders for the torso, arms and legs, then add the parts. This extends the official torso example to a composite model and shows how one shape per part can describe a complex object.
Sample Measurements (Invented)
Head, as a sphere: circumference 55 cm, so r ≈ 8.75 cm and V ≈ 2,810 cm³
Torso, as a cylinder: circumference 84 cm, height 58 cm, V ≈ 32,567 cm³
Each arm, as a cylinder: circumference 26 cm, length 62 cm, V ≈ 3,335 cm³
Each leg, as a cylinder: circumference 45 cm, length 85 cm, V ≈ 13,697 cm³
Total: about 69,400 cm³, or about 69 L
Discussion Questions
Which body part does its shape describe best? Which one worst?
Arms and legs taper. Does the cylinder model overestimate or underestimate them, if the circumference was measured at the widest point?
The body's density is close to that of water, so a person's volume in liters is close to their mass in kilograms. Is 69 L reasonable for an adult? (This check previews HSG.MG.A.2.)
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Tree Trunk and a Torso Modeled as Cylinders
Left: a trunk section with circumference 2.2 m and height 8 m, drawn at 30 pixels per meter. Right: a torso with circumference 80 cm and height 60 cm, drawn at 4 pixels per centimeter. The scale bars show each scale. In both models the measured circumference gives the radius, r = C ÷ 2π, and the lateral area is simply C × h.
Diagram 2: An A-Frame Roof as a Triangular Prism
The cross-section of the roof is an isosceles triangle with base 8 m and height 3 m, so each sloping side is √(4² + 3²) = 5 m. The two roof panels are 5 m by 10 m rectangles. Everything is drawn to the same scale, 20 pixels per meter.
04
Homework Assignment
~30 min
HSG.MG.A.1 Homework: Modeling Objects with Shapes
Directions: For every problem, name the shape you use and the property of the object that justifies it. Show the formula, the values you substitute and the units. Round to a sensible precision and say whether your model overestimates or underestimates when the problem asks.
Part 1: Cylinders and Spheres (Problems 1-2)
A wooden utility pole stands 12 m above the ground and has a diameter of 30 cm. Model it as a cylinder. (a) How many square meters of surface would a protective coating cover, not counting the top? (b) What volume of wood is above the ground?
A basketball has a circumference of 75 cm. Model it as a sphere and find (a) its radius, (b) the area of leather or rubber on its surface, and (c) the volume of air inside, ignoring the thickness of the ball.
Part 2: Cones and Prisms (Problems 3-4)
A pile of road salt is shaped roughly like a cone with a base diameter of 12 m and a height of 5 m. Estimate the volume of salt in the pile. What property of a pile of loose material makes a cone a reasonable model?
A camping tent is a triangular prism 2.2 m long. Its front is an isosceles triangle with base 1.8 m and height 1.2 m. (a) Find the length of each sloping side. (b) How much fabric is needed for the two sloping sides and the two triangular ends? (c) What is the volume of air inside?
Part 3: Choosing and Comparing Models (Problems 5-6)
A swimming pool is 25 m long and 10 m wide. The bottom slopes evenly from a depth of 1 m at one end to 2 m at the other. Model the water as a prism whose base is the side view of the pool, a trapezoid, and find the volume of water in cubic meters and in liters.
A carrot is 18 cm long and 3 cm across at its top. Model it as a cone and as a cylinder, and find both volumes. Which model is better, and by what factor does the other one differ?
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Choice of Shape
Shape named and justified by a property of the object
Shape named without a reason
No shape or an unsuitable shape
Measures and Units
Correct dimensions (radius, slant) found from the given measures, units consistent
One conversion or unit error
Dimensions misused throughout
Computation
Formulas applied correctly with sensible rounding
Correct method with one arithmetic error
Incorrect formulas
Judging the Model
Over- or underestimate and limits explained in context
Limits mentioned without reasons
No judgment of the model
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
You want to estimate the volume of rubber in a hockey puck. Which shape is the best model?
Answer: A
A puck has a circular cross-section that stays the same from top to bottom, which is the defining property of a cylinder. A rectangular prism has square corners the puck does not have, and a sphere or cone would change width from top to bottom.
Question 2 of 20 · Multiple Choice
A tree trunk has a circumference of 1.57 m. Modeling the trunk as a cylinder, what is its diameter, to the nearest hundredth of a meter?
Answer: C
C = πd, so d = 1.57 ÷ π ≈ 0.50 m. Choice A is the radius. Choice B doubles the circumference, and choice D multiplies it by π instead of dividing.
Question 3 of 20 · Multiple Choice
A log is 5 m long with a diameter of 0.4 m. Modeled as a cylinder, what is its volume, to the nearest hundredth?
Answer: B
r = 0.2 m, so V = π(0.2)²(5) = 0.2π ≈ 0.63 m³. Choice A uses the diameter 0.4 as the radius. Choice C is the lateral area, 2π(0.2)(5), in square meters. Choice D leaves out π.
Question 4 of 20 · Multiple Choice
A roll of paper towels is 28 cm tall, 12 cm across, and has a cardboard tube 4 cm across in the middle. Model the paper as a large cylinder minus the tube. What is the volume of paper, to the nearest cm³?
Answer: D
V = π(6² - 2²)(28) = π(32)(28) = 896π ≈ 2,815 cm³. Choice A forgets to subtract the tube. Choice B is only the tube. Choice C uses the diameters 12 and 4 as radii.
Question 5 of 20 · Multiple Choice
A scoop of ice cream is modeled as a sphere with a diameter of 6 cm. What is its volume, to the nearest tenth?
Answer: B
r = 3 cm, so V = ⁴⁄₃π(3)³ = 36π ≈ 113.1 cm³. Choice A uses the diameter as the radius. Choice C squares the radius instead of cubing it, and choice D leaves out the ⁴⁄₃.
Question 6 of 20 · Multiple Choice
To model a round concrete pillar in a parking garage as a cylinder, which measurement is the most practical to take?
Answer: D
A tape fits around the pillar, and the radius follows from r = C ÷ 2π. A ruler cannot pass through solid concrete (choice A), and the cross-section area and the volume cannot be measured directly on a pillar that holds up a building.
Question 7 of 20 · Multiple Choice
A person's head has a circumference of 56 cm. If the head is modeled as a sphere, what is its radius, to the nearest tenth?
Answer: A
The circumference of the sphere's great circle is 2πr = 56, so r = 56 ÷ 2π ≈ 8.9 cm. Choice B is the diameter. Choice C is half the circumference. Choice D solves πr² = 56, using the area formula instead of the circumference formula.
Question 8 of 20 · Multiple Choice
A torso is modeled as a cylinder with a circumference of 90 cm and a height of 55 cm. About how much fabric covers the side of the cylinder, ignoring overlap?
Answer: C
Unrolled, the side of a cylinder is a rectangle with width equal to the circumference and height h, so its area is C × h = 90 × 55 = 4,950 cm². Choice A halves that. Choice B is the volume, in cubic centimeters. Choice D uses 90 as the diameter in πdh, but 90 cm is already the circumference.
Question 9 of 20 · Multiple Choice
A classroom is 9 m long, 7 m wide and 3 m high. Modeled as a rectangular prism, how much air does it hold?
Answer: B
V = 9 × 7 × 3 = 189 m³. Choice A is the floor area only. Choice C adds the dimensions. Choice D is the area of the four walls, 2(9 × 3) + 2(7 × 3), which is not a volume.
Question 10 of 20 · Multiple Choice
A flower pot widens from a diameter of 20 cm at the bottom to 30 cm at the top. Which cylinder model underestimates the volume of soil it holds when full?
Answer: C
The cylinder built on the bottom diameter fits inside the pot, so it is too small. The cylinder built on the top diameter contains the whole pot and overestimates (choice A). The average model lies between the two, and choice D is false because the bottom-diameter cylinder is too small.
Question 11 of 20 · Multiple Choice
An A-frame cabin is 12 m long. Its front is an isosceles triangle with base 6 m and height 4 m. What is the total area of its two sloping roof panels?
Answer: A
Each sloping side is √(3² + 4²) = 5 m, so each panel is 5 m by 12 m. Total = 2 × 5 × 12 = 120 m². Choice B uses the height 4 m instead of the slant, a common error. Choice C uses the base 6 m, and choice D counts only one panel.
Question 12 of 20 · Multiple Choice
Earth is modeled as a sphere with a radius of 6,371 km. About how long is the equator?
Answer: D
The equator is a great circle, so its length is 2πr = 2π(6,371) ≈ 40,030 km. Choice A is πr, half the circumference. Choice B is the diameter, and choice C is 4πr.
Question 13 of 20 · Multiple Choice
A cylinder model of a tank has its radius doubled while its height stays the same. What happens to its volume?
Answer: B
V = πr²h, and (2r)² = 4r², so the volume is multiplied by 4. Choice A treats volume as proportional to r. Choice C would be correct if every dimension, including the height, doubled.
Question 14 of 20 · Multiple Choice
A beach ball has a diameter of 50 cm. Modeled as a sphere, about how much air does it hold? (1 L = 1,000 cm³)
Answer: C
r = 25 cm, so V = ⁴⁄₃π(25)³ ≈ 65,450 cm³ = 65.4 L. Choice A uses 50 cm as the radius. Choice B comes from the surface area 4πr² ≈ 7,854 cm², which is not a volume. Choice D forgets to divide by 1,000 to convert to liters.
Question 15 of 20 · Short Answer
A drinking glass is 12 cm tall and 7 cm across the inside. Model it as a cylinder and find its volume in milliliters (1 cm³ = 1 mL). Give one reason the glass really holds less.
r = 3.5 cm, so V = π(3.5)²(12) = 147π ≈ 462 mL. The glass holds less because its bottom is thick, it is often narrower at the base, and nobody fills it to the brim.
Question 16 of 20 · Short Answer
A tree trunk has a circumference of 3.0 m and rises 6 m to its first branch. Model the trunk as a cylinder and estimate its volume to the nearest hundredth of a cubic meter.
r = 3.0 ÷ 2π ≈ 0.477 m. V = πr²h = C²h ÷ 4π = (9)(6) ÷ 4π = 13.5 ÷ π ≈ 4.30 m³.
Question 17 of 20 · Short Answer
A doghouse roof is 1.5 m long. Its front is an isosceles triangle with base 1.2 m and height 0.8 m. Find the area of roofing needed for the two sloping panels.
Half the base is 0.6 m, so each slant is √(0.6² + 0.8²) = √1 = 1.0 m. Each panel is 1.0 m by 1.5 m, so the roofing area is 2 × 1.0 × 1.5 = 3.0 m².
Question 18 of 20 · Short Answer
Why is a sphere a reasonable model for an orange? Use it to estimate the area of peel on an orange 8 cm across, and name one way the model differs from a real orange.
An orange is close to the same distance from its center in every direction, the defining property of a sphere. With r = 4 cm, the peel area is 4π(4)² = 64π ≈ 201 cm². A real orange is slightly flattened and its peel is bumpy and has thickness, so the model is an approximation.
Question 19 of 20 · Short Answer
A trunk section is 5 m tall, 0.6 m across at the bottom and 0.4 m across at the top. Estimate its volume with an average-diameter cylinder. Then find how much larger the estimate from a cylinder that uses the bottom diameter is, in cubic meters and as a percent.
Average diameter 0.5 m, r = 0.25 m: V = π(0.25)²(5) = 0.3125π ≈ 0.98 m³. The bottom-diameter cylinder gives π(0.3)²(5) = 0.45π ≈ 1.41 m³, which is about 0.43 m³, or 44%, larger, since 0.45 ÷ 0.3125 = 1.44. It overestimates, because the trunk is narrower than 0.6 m everywhere above the bottom.
Question 20 of 20 · Short Answer
An arm is 60 cm long with an average circumference of 28 cm. Model it as a cylinder and estimate its volume in liters.
V = C²h ÷ 4π = (28²)(60) ÷ 4π = 47,040 ÷ 4π ≈ 3,743 cm³, which is about 3.7 L. The radius is 28 ÷ 2π ≈ 4.46 cm if you prefer to use πr²h directly.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.MG.A.1 mean?
HSG.MG.A.1 means students use geometric shapes, with their measures and properties, to describe real objects. The official examples are modeling a tree trunk or a human torso as a cylinder. Students choose the shape, take measurements, compute lengths, areas or volumes, and judge how well the model fits.
Is HSG.MG.A.1 part of Geometry or a separate modeling course?
HSG.MG.A.1 is usually taught in high school Geometry, often together with the volume standard HSG.GMD.A.3. Common Core also marks it as a modeling standard, so it can appear in any course that uses geometry to answer real questions.
How is HSG.MG.A.1 different from just using volume formulas?
A formula problem gives the shape. HSG.MG.A.1 asks students to decide which shape describes the object, which measurements it needs, and whether the answer is reasonable. Choosing and judging the model is the skill; the formula is only one step.
Why model a tree trunk as a cylinder if trees are not perfect cylinders?
Because a cylinder captures the key property: a roughly circular cross-section along the height. The model gives a useful estimate from one easy measurement, the circumference. Students should say that it overestimates if the trunk tapers and the circumference was measured near the bottom.
How do you find the radius when you can only measure the circumference?
Use C = 2πr, so r = C ÷ 2π. For a cylinder you can skip the radius: V = C²h ÷ 4π and the lateral area is C × h. This is the usual method for trees, poles and people.
What mistakes do students make in modeling problems?
Common errors are using a diameter as a radius, using the height of a cone or roof where the slant is needed, mixing units such as meters and centimeters, and giving far more decimal places than the measurements support. Another is naming a shape without saying which property of the object it matches.
Can an object be modeled with more than one shape?
Yes. A person can be modeled as one cylinder or as a sphere plus several cylinders, and a pencil as a hexagonal prism with a cone at the tip. More shapes usually give a closer model but take more measurements; part of the skill is deciding how much detail the question needs.
How is HSG.MG.A.1 assessed?
Tasks usually describe an object with a few measurements and ask for an area or volume, the choice of shape with a reason, or a comparison of two models. Some ask whether a model overestimates or underestimates. Written justification counts, not only the number.
Which shapes should students know for this standard?
The common ones are rectangular and triangular prisms, cylinders, cones, pyramids and spheres, with circles, rectangles and triangles for flat objects. Students should know each shape's key property and its area and volume formulas.
Is this kind of modeling on the SAT?
Yes. Word problems that describe a real object as a cylinder, cone, sphere or prism appear in the Geometry and Trigonometry domain of the digital SAT Math section, where the formulas are provided on a reference sheet. The judgment part of modeling is tested less there than in classroom tasks.
07
Related Standards
6 standards
These standards connect to HSG.MG.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.6Prerequisite
Solve problems with area, volume and surface area of 2D and 3D objects
