HSG.MG.A.3: Solving Design Problems with Geometric Methods
In plain English: HSG.MG.A.3 is the Common Core geometry standard that asks students to use geometric methods to solve design problems. Students design objects and structures that meet physical constraints, choose dimensions that keep cost or material low, and build typographic grids from ratios. It belongs to the Modeling with Geometry domain and is usually taught in high school Geometry.
Apply geometric methods to solve design problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios).
Common Core State Standards for Mathematics · Domain: Modeling with Geometry (MG) · Cluster: Apply geometric concepts in modeling situations Also written as HSG-MG.A.3 or G-MG.3 · Official standard
Students use geometry to make design decisions. The lesson works through the three kinds of problems the standard names: designing an object or structure that meets a physical constraint, choosing dimensions that minimize cost or material, and building a typographic grid from ratios. In each case students name the quantity they control, write the constraint and the goal in geometric terms (length, area, volume, slope, ratio), compare designs with tables or graphs, and check that the final design satisfies every constraint.
No calculus is needed. Students find best designs by testing values in a table, reading a graph, using symmetry, or using known facts such as "for a fixed perimeter, a square encloses the largest rectangle". They also learn to report a design the way a client would need it: dimensions with units, sensible rounding and a sentence on why it is the best of the options.
Learning Objectives
By the end of this lesson, students will be able to:
Translate a design situation into a geometric model with a variable, a constraint and a goal
Design an object or structure that satisfies a physical constraint, such as a maximum slope, a fixed volume or a space limit
Compare designs with tables and graphs to find the one with the least cost or material
Compute the dimensions of a typographic grid from page size, margins, gutters and ratios
Check a design against every constraint and explain the choice in context
Prior Knowledge Required
Students should already be comfortable with:
Solving problems with scale drawings and ratios 7.G.A.1
Area and perimeter of rectangles and triangles, and surface area of prisms 7.G.B.6
The Pythagorean theorem in real-world problems 8.G.B.7
Volume formulas for cylinders and prisms HSG.GMD.A.3
Evaluating an expression for several input values and reading a graph
Students work alone for three minutes, then compare with a partner:
Warm-Up Prompt
"You have 36 meters of garden edging. List at least four rectangles you could outline with all of it, and find each area. Which rectangle gives the most garden? Do you think any rectangle could do better?"
Record the class list, such as 2 by 16 (32 m²), 6 by 12 (72 m²), 8 by 10 (80 m²) and 9 by 9 (81 m²). Ask what the edging length does in this problem (it is a constraint) and what the area does (it is the goal). Tell students that every design problem today has these two parts, and that the job is to find the design that meets the constraint and does best on the goal.
Direct Instruction20-25 minutes
Introduce a four-step design routine and use it for every example:
Model: sketch the object and label the dimensions you control.
Constrain: write each requirement as an equation or inequality (a fixed area, a maximum slope, a page width).
Compare: express the goal (cost, material, space) in one variable and test designs with a table or a graph.
Check and report: confirm every constraint, round to buildable numbers, and state the design with units.
Physical constraint: a ramp
An accessible ramp must rise 30 in to a door, with a slope no steeper than 1:12 (the maximum running slope in the 2010 ADA Standards).
Equation: Run ≥ 12 · 30 = 360 in = 30 ft; ramp length √(360² + 30²) ≈ 361 in
Minimize cost: a pen along a barn
A 288 ft² rectangular pen uses the barn wall as one long side. Fence costs $15 per foot. With width w, fence F = 2w + 288/w.
Equation: w = 8: 52 ft; w = 12: 48 ft; w = 16: 50 ft. Best: 12 ft by 24 ft, 48 ft, $720
Typographic grid: columns
A letter page is 8.5 in wide with 1 in side margins. The designer wants 3 equal columns with 0.25 in gutters.
A 6 in by 9 in book page (ratio 2:3) gets a text block with the same 2:3 ratio, 4 in wide. The margins split the leftover space 1:2 (inner:outer, top:bottom).
Equation: Text block 4 in by 6 in; inner 2/3 in, outer 4/3 in, top 1 in, bottom 2 in
Minimize material: a can
A cylindrical can must hold 1,000 cm³. Material is the surface area S = 2πr² + 2000/r.
Equation: r = 4: 600.5 cm²; r = 5: 557.1 cm²; r = 6: 559.5 cm²; r = 7: 593.6 cm². Of these, r = 5 cm uses the least
Use Diagram 1 with the column example and Diagram 2 with the pen example. For the pen, point out that the table alone cannot prove w = 12 is the best, but the graph shows the fence length falling and then rising, with the lowest point at w = 12. For the book page, explain that the text block is similar to the page, so the ratio of the page is repeated inside it: a ratio-based grid is an application of similarity. For the can, discuss why a real can may differ from the best design: thicker tops and bottoms, seams and shelf height are constraints the simple model leaves out.
Guided Practice15 minutes
Pairs solve one problem with the design routine, writing each step on a whiteboard: A community garden along a river needs a rectangular plot fenced on three sides, with the river as the fourth side. There are 120 m of fencing. What dimensions give the largest plot?
Guided practice: plot width w (the two sides perpendicular to the river), length 120 - 2w, and area
w (m)
length (m)
area (m²)
20
80
1600
25
70
1750
30
60
1800
35
50
1750
40
40
1600
Pairs should conclude that the 30 m by 60 m plot gives 1,800 m², and notice the symmetry in the table around w = 30. Ask why the best plot is not a square (the river side needs no fence, so the length can be longer). Watch for students who use 2w + 2l = 120, which fences the river side too.
Independent Practice15 minutes
A poster is 24 in by 36 in with 2 in margins on every side. Design a 4-column grid with 1 in gutters. How wide is each column? (4.25 in.)
A ramp must rise 21 in with a slope no steeper than 1:12. What is the shortest horizontal run? (252 in = 21 ft.)
A 128 ft² dog run is fenced on three sides against a garage wall. Test several widths and find the design with the least fence. (8 ft by 16 ft, 32 ft of fence.)
For one of the problems above, write two sentences a client could read: the design, and why it is the best choice.
Closure5-10 minutes
Exit ticket: (1) In the garden problem, what was the constraint and what was the goal? (2) Name one constraint a real builder would add to the pen problem that the model left out. (3) A page has a 7 in text block and needs 2 columns with a 0.5 in gutter. How wide is each column? (3.25 in.)
Differentiation Strategies
For Struggling Students
Provide partly filled tables so that students compute the goal for given dimensions before choosing their own
Use grid paper to draw every candidate design to scale, so that "long and thin" and "close to square" are visible
Give a constraint-and-goal organizer with the four steps of the design routine as headings
For Advanced Students
Ask students to prove the pen result with algebra: show that 2w + 288/w - 48 = 2(w - 12)²/w, which is never negative
Have students redo the can problem when the top and bottom cost twice as much per cm² as the side, and compare the best radius
Ask students to design a 12-column web layout for a 1,200-pixel screen with 24-pixel gutters and list every card width that spans a whole number of columns
Assessment Guidance
What to Look For
Look for a model before any arithmetic: a labeled sketch, a clear variable, the constraint and the goal. Check that students compare several designs rather than guessing one, and that the final design meets every constraint, including ones that bind, such as a maximum width. In typographic problems, check that students count gutters correctly (one fewer than the number of columns when there are no outer gutters). Answers should be reported with units and rounded to values someone could build.
02
Classroom Activities
3 Activities
1
Typographic Grid Workshop
20 minPairs
Pairs design the grid for a one-page school event flyer on letter paper, compute every column and module size from the margins and gutters, and then lay out a headline, a photo and two text blocks on the grid.
Design Brief
Page 8.5 in by 11 in, margins 0.75 in on all sides, 4 columns with 0.2 in gutters
The photo must span 2 columns and keep a 4:3 ratio (width:height)
The headline spans all 4 columns; each text block spans 1 or 2 columns
Procedure
Compute the text block (7 in by 9.5 in) and the column width ((7 - 0.6) ÷ 4 = 1.6 in)
Compute the photo size: 2(1.6) + 0.2 = 3.4 in wide, and 3.4 · 3/4 = 2.55 in tall
Draw the grid to scale on letter paper with a ruler, then sketch the layout on it
Trade flyers with another pair and check each other's measurements with a ruler
Discussion Questions
Why is the column width not simply 7 ÷ 4?
If you change to 3 columns, what happens to the photo width when it spans 2 columns?
How does using the same ratio for the photo and for other boxes make a page look consistent?
Modification for Distance Learning
Pairs build the grid in a slide or document editor using its ruler and guides, and share a screenshot with the computed sizes written beside the grid.
2
Build the Best Box
25 minGroups of 3-4
Groups make open-top boxes from letter-size cardstock (21.6 cm by 27.9 cm) by cutting equal squares of side x from the corners, predict each volume, and test the prediction by filling the boxes with rice. The goal is the largest volume from one sheet.
Procedure
Each group writes the volume model V = x(21.6 - 2x)(27.9 - 2x) and predicts V for x = 2, 3, 4 and 5 cm
Each group builds one box per value of x (4 boxes), fills each with rice and pours the rice into a measuring cup (1 mL = 1 cm³)
Groups record predicted and measured volumes in a table and circle the best design
Predicted volumes: about 841, 1025, 1083 and 1038 cm³, so x = 4 cm is the best of the four
Discussion Questions
Why do the measured volumes come out a little lower than the predictions?
Why does the volume rise and then fall as x increases?
Which value of x between 3 and 5 would you test next, and why?
Challenge Variation
The box must also fit in a drawer that is only 5 cm deep. Which design is best now? (x = 4 cm still fits, since the height of the box equals x.) Then change the drawer to 3.5 cm deep and decide again.
3
Accessible Ramp Design Challenge
20 minGroups of 3-4
Groups design a ramp for a porch that is 28 in above the ground, in a front yard only 20 ft deep. A straight ramp does not fit, so groups must design a ramp with a turn and a landing that meets every rule on the design card.
Design Card
Slope no steeper than 1:12 on every sloped part
A level landing at least 5 ft by 5 ft wherever the ramp turns
The whole ramp, landing included, must fit within 20 ft of yard depth
Procedure
Compute the total horizontal run needed: 12 · 28 = 336 in = 28 ft, which is longer than the yard
Split the rise into two sloped parts of 14 in each, with a run of 14 ft each, joined by a 5 ft landing: 14 + 5 = 19 ft of depth, which fits
Draw the design to scale on grid paper (1 square = 1 ft) and label every length, rise and slope
Compute the length of each sloped surface with the Pythagorean theorem (√(168² + 14²) ≈ 168.6 in each)
Discussion Questions
Why can the two sloped parts not share the landing's space?
Would a gentler slope, such as 1:16, fit in the same yard? What would it change?
What other constraints would a real builder check (handrails, drainage, the path to the sidewalk)?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Three-Column Grid on a Letter Page
An 8.5 in by 11 in page with 1 in margins and three 2 in columns separated by 0.25 in gutters, drawn to scale. The column width comes from subtracting the margins and gutters from the page width and dividing the rest equally.
Diagram 2: Least Fence for a 288 ft² Pen
Fence length F = 2w + 288/w for a pen that uses a barn wall as one long side, drawn to scale. The fence needed falls and then rises as the width grows, and it is least at w = 12 ft, where the pen is 12 ft by 24 ft and needs 48 ft of fence.
04
Homework Assignment
~30 min
HSG.MG.A.3 Homework: Geometric Design Problems
Directions: For every problem, sketch the design, name the constraint and the goal, and show the calculations or table you used to compare designs. Give final dimensions with units and one sentence explaining your choice.
Part 1: Physical Constraints and Cost (Problems 1-2)
A loading door is 18 in above the ground, and there are 20 ft of level space in front of it. A ramp may be no steeper than 1:12. (a) What is the shortest horizontal run the ramp can have? (b) Does a straight ramp fit? (c) How long is the sloped surface, to the nearest tenth of an inch?
A rectangular storage yard must have an area of 2,400 ft². The front side (one of the two sides of length x) uses decorative fence at $20 per foot, and the other three sides use chain-link at $10 per foot. (a) Write the cost in terms of x. (b) Find the cost for x = 30, 40, 48 and 60 ft. (c) Which of these designs costs least, and what are its dimensions?
Part 2: Typographic Grids and Ratios (Problems 3-4)
A magazine page is 9 in wide with 0.75 in side margins. The designer wants 4 equal columns with 0.3 in gutters. (a) How wide is each column? (b) A photo spans 2 columns and one gutter. How wide is it? (c) The photo has a 3:2 ratio (width:height). How tall is it?
A book page is 8 in by 12 in. The designer wants a text block similar to the page, 5.6 in wide. (a) How tall is the text block? (b) The leftover width is split between the inner and outer margins in the ratio 1:2, and the leftover height between the top and bottom margins in the ratio 1:2. Find all four margins. (c) Check that the margins and text block add up to the page size.
Part 3: Designing for Least Material (Problems 5-6)
An open-top planter box has a square base with side x cm and must hold 32,000 cm³. (a) Write the height and the amount of material (base plus four sides) in terms of x. (b) Find the material for x = 30, 40 and 50 cm. (c) Which design uses least material? If the wood costs $0.50 per 100 cm², what does that planter cost?
A café wants a rectangular patio against its front wall with an area of at least 450 ft². City rules limit the depth (the distance out from the wall) to 12 ft. Railing is needed on the three open sides and costs $40 per foot. (a) Write the railing length in terms of the depth d when the area is exactly 450 ft². (b) Find the railing for d = 9, 10 and 12 ft. (c) Which design do you recommend, and what does the railing cost? Explain how the depth rule affects your choice.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Model
Labeled sketch, variable, constraint and goal written
Sketch or equations incomplete
No model
Comparing Designs
Several designs tested correctly in a table or graph
Some designs tested, one calculation error
Single guess with no comparison
Constraints Checked
Final design meets every constraint, including limits that bind
One constraint not checked
Design breaks a constraint
Report
Dimensions with units and a clear reason for the choice
Dimensions given without a reason
No final design stated
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Choose an answer to see whether it is right and why. For the short-answer questions, sketch the design and work it out before you open the answer. Reset quiz clears all answers.
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
A ramp must rise 24 in with a slope no steeper than 1:12. What is the shortest horizontal run it can have?
Answer: C
A 1:12 slope needs 12 in of run for every inch of rise: 12 · 24 = 288 in, and 288 ÷ 12 = 24 ft. Choice A divides the rise by 12 instead of multiplying. Choice D gives 288 but labels it in feet instead of inches.
Question 2 of 20 · Multiple Choice
Of all rectangles with a perimeter of 60 m, which area is the largest?
Answer: A
For a fixed perimeter, the square encloses the largest rectangle: sides 60 ÷ 4 = 15 m, area 225 m². Choices B and D come from 10 by 20 and 12 by 18 rectangles, which have the same perimeter but less area. Choice C uses 60 ÷ 2 = 30 as the side, which would need 120 m of perimeter.
Question 3 of 20 · Multiple Choice
A text block is 7.2 in wide and holds 3 equal columns with 0.3 in gutters between them. How wide is each column?
Answer: D
Three columns have two gutters: (7.2 - 2 · 0.3) ÷ 3 = 6.6 ÷ 3 = 2.2 in. Choice A ignores the gutters. Choice B subtracts three gutters instead of two.
Question 4 of 20 · Multiple Choice
A page is 8 in wide and 10 in tall. The designer wants a text block similar to the page that is 6 in wide. How tall should it be?
Answer: B
Similar rectangles keep the ratio of height to width: 10/8 = h/6, so h = 7.5 in. Choice A subtracts 2 from the height, the amount taken from the width, but similarity multiplies lengths. Choice C uses the ratio upside down.
Question 5 of 20 · Multiple Choice
A 200 ft² rectangular pen uses a barn wall as one long side, so fence is needed on the other three sides. Which design uses the least fence?
Answer: A
With width w, the fence is 2w + 200/w: 40 ft at w = 10, compared with 50 ft at w = 20 or w = 5 and about 42.4 ft for the square. Choice C applies the "square is best" rule, which holds when all four sides are fenced, not three.
Question 6 of 20 · Multiple Choice
A cylindrical can must hold 500 cm³. Its surface area is S = 2πr² + 1000/r. Which radius uses the least material?
Answer: C
S(2) ≈ 525.1, S(3) ≈ 389.9, S(4.3) ≈ 348.7 and S(6) ≈ 392.9 cm², so r = 4.3 cm uses the least. Choice A makes a tall, thin can with a lot of side area. Choice D makes a wide, flat can with large top and bottom.
Question 7 of 20 · Multiple Choice
A triangular garden bed fits in the corner where two walls meet at a right angle. One edge along a wall is 6 m and the bed must have an area of 24 m². How much edging is needed for the third side, away from the walls?
Answer: D
Area = (1/2)(6)(b) = 24, so the other leg is b = 8 m, and the third side is √(6² + 8²) = 10 m. Choice A adds the legs. Choice B is the second leg, not the edge away from the walls.
Question 8 of 20 · Multiple Choice
A safety guideline for ladders says the base should be 1 ft from the wall for every 4 ft of height. A ladder must reach a point 16 ft up a wall. About how long must the ladder be?
Answer: C
The base is 16 ÷ 4 = 4 ft from the wall, so the ladder is √(16² + 4²) = √272 ≈ 16.5 ft. Choice A ignores the distance at the base. Choice B adds 16 and 4, but the ladder is the hypotenuse of a right triangle, not the sum of its legs.
Question 9 of 20 · Multiple Choice
You have 24 m of flexible garden edging. Which shape encloses the most area?
Answer: D
The areas are 36 m², 32 m², about 27.7 m² and 24²/(4π) ≈ 45.8 m². For a fixed perimeter, a circle encloses more area than any polygon. Choice A is the best rectangle, but not the best shape.
Question 10 of 20 · Multiple Choice
A text block 6.4 in wide and 8.6 in tall is divided into a modular grid of 3 columns and 4 rows with 0.2 in gutters between columns and between rows. What size is each module?
Answer: A
Width: (6.4 - 2 · 0.2) ÷ 3 = 2 in. Height: (8.6 - 3 · 0.2) ÷ 4 = 2 in. Choice B ignores the gutters. Choice C subtracts three gutters from the width, but 3 columns have only 2 gutters.
Question 11 of 20 · Multiple Choice
An open box is made from a 20 cm by 20 cm sheet by cutting squares of side x from the corners. Its volume is V = x(20 - 2x)². Which cut gives the largest volume?
Answer: B
V(2) = 512, V(3) = 588, V(4) = 576 and V(5) = 500 cm³, so x = 3 cm is best of these. Choice D is a common guess because deeper boxes seem larger, but the base shrinks faster than the height grows.
Question 12 of 20 · Multiple Choice
A client wants a rectangular garden with a perimeter of 30 m and an area of 100 m². What should the designer say?
Answer: C
The largest rectangle with perimeter 30 m is the 7.5 m square, with area 56.25 m², so no rectangle can reach 100 m². Choice A has area 100 m² but perimeter 40 m. Choice D has the right perimeter but only 56.25 m² of area.
Question 13 of 20 · Multiple Choice
A cable runs from a pole to a house. The house is 40 m along a straight road from the pole and 30 m back from the road across a field. Cable along the road costs $8 per meter, and across the field $10 per meter. Which plan costs least?
Answer: D
The straight route is √(40² + 30²) = 50 m of field cable, $500. The other plans cost $620, about $520.56 and about $504.26. Choice A seems cheaper per meter but uses 70 m of cable. Here the saving on the road does not make up for the extra length.
Question 14 of 20 · Multiple Choice
An 8 in text block holds 2 columns and one 0.5 in gutter. The wide column must be twice as wide as the narrow one. How wide is each column?
Answer: B
n + 2n + 0.5 = 8, so 3n = 7.5, n = 2.5 in and the wide column is 5 in. Check: 2.5 + 5 + 0.5 = 8. Choice A ignores the gutter. Choice C does not add up to 8 in with the gutter. Choice D splits the space equally and ignores the 1:2 ratio.
Question 15 of 20 · Short Answer
A rectangular garden is fenced on three sides, with a wall as the fourth side. There are 80 m of fencing. Find the dimensions that give the largest area and state that area.
With width w (the two sides perpendicular to the wall), the length is 80 - 2w and A = w(80 - 2w). A table shows A(15) = 750, A(20) = 800, A(25) = 750, and the values are symmetric around w = 20. The best garden is 20 m by 40 m, with area 800 m².
Question 16 of 20 · Short Answer
A porch is 27 in above the ground. There are 25 ft of space in front of it. Can a straight ramp with a slope no steeper than 1:12 fit? If not, describe a design that could work.
The run must be at least 12 · 27 = 324 in = 27 ft, which is more than 25 ft, so a straight ramp does not fit. A design that turns, with two sloped parts and a level landing between them, keeps the slope at 1:12 while using less depth. For example, two parts rising 13.5 in each need 13.5 ft of run each.
Question 17 of 20 · Short Answer
A presentation slide is 1920 px wide. The designer uses 12 columns with 40 px outer margins and 20 px gutters between columns. (a) How wide is each column? (b) An image spans 4 columns and the gutters between them. How wide is it, and how tall is it if its ratio is 3:2 (width:height)?
A closed box with a square base must hold 1,000 cm³. Its surface area is S = 2x² + 4000/x, where x is the side of the base. Compare x = 5, 10 and 20 cm and choose a design.
S(5) = 50 + 800 = 850 cm², S(10) = 200 + 400 = 600 cm², S(20) = 800 + 200 = 1000 cm². The 10 cm by 10 cm by 10 cm cube uses the least material, 600 cm².
Question 19 of 20 · Short Answer
A bookshelf is 7 ft tall and 2 ft deep. It is built lying on the floor and must be tipped upright in a room with an 8 ft ceiling. Will it clear the ceiling while it is tipped? Explain with geometry.
While it tips, the highest point is the far corner, at a distance equal to the diagonal of its side: √(7² + 2²) = √53 ≈ 7.28 ft. 7.28 ft is less than 8 ft, so it clears the ceiling, with about 8.6 in to spare.
Question 20 of 20 · Short Answer
A rectangular frame for a poster must be made from exactly 2.4 m of wood trim, and its width and height must be in the ratio 2:3. Find the dimensions.
Width 2k and height 3k, so the perimeter is 2(2k + 3k) = 10k = 2.4 and k = 0.24. The frame is 0.48 m wide and 0.72 m tall.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.MG.A.3 mean?
HSG.MG.A.3 means students use geometry to solve design problems. The standard names three kinds: designing an object or structure to meet physical constraints, designing to minimize cost, and working with typographic grid systems based on ratios. Students choose dimensions, check them against the constraints and justify the design.
Is HSG.MG.A.3 taught in Geometry?
It is usually taught in high school Geometry, often spread across units instead of taught once: area and perimeter design problems, volume and surface area problems, and similarity and ratio problems. It belongs to the Modeling with Geometry domain with HSG.MG.A.1 and HSG.MG.A.2.
What is a typographic grid system?
A typographic grid is the set of margins, columns, rows and gutters that designers use to place text and images on a page or screen. The sizes come from ratios and subtraction: the page width minus the margins and gutters, divided equally among the columns. Some grids also repeat the page's own ratio in the text block, which makes the text block similar to the page.
How can students minimize cost without calculus?
By comparing designs. Students write the cost or material in terms of one dimension, test several values in a table, and use a graph to see where the value is lowest. Symmetry and known facts help: for a fixed perimeter, the square is the rectangle with the largest area. Quadratic models can also be solved exactly with the vertex from Algebra I.
What counts as a physical constraint in a design problem?
Any requirement the design must meet: a maximum slope for a ramp, a fixed volume for a container, a space the object must fit in, a clearance under a ceiling, or a fixed amount of material. Students write each one as an equation or inequality and check the final design against all of them.
Why does a square give the largest area for a fixed perimeter?
Because the area of a rectangle with perimeter P and width w is w(P/2 - w), a quadratic whose largest value is at w = P/4, the midpoint of its zeros. Then both sides equal P/4, so the rectangle is a square. When one side needs no fence, as with a wall or a river, the best shape is no longer a square.
Is the 1:12 ramp slope a real rule?
Yes. The 2010 ADA Standards for Accessible Design in the United States set a maximum running slope of 1:12 for ramps, which means at least 12 in of horizontal run for every inch of rise. The standards also set other rules, such as landing sizes and a maximum rise for each ramp run, which make good extra constraints for advanced design problems.
How is HSG.MG.A.3 different from HSG.MG.A.1 and HSG.MG.A.2?
HSG.MG.A.1 uses shapes to describe real objects, such as modeling a tree trunk as a cylinder, and HSG.MG.A.2 applies density based on area or volume. HSG.MG.A.3 goes a step further: students make decisions, choosing a design that meets constraints or minimizes cost, instead of only describing or measuring.
What should a complete answer to a design problem include?
It should include a labeled sketch, the variable, the constraint and the goal, the comparison of designs, and a final design with units. It should also say whether the answer is practical, for example rounding to dimensions that can be built and mentioning constraints the model left out.
How does HSG.MG.A.3 connect to later courses?
The same reasoning returns in Algebra, where students write constraints as equations and inequalities (HSA.CED.A.3) and find the maximum or minimum of a quadratic, and in calculus, where optimization problems use derivatives to find the best design exactly. Graphic design, engineering and architecture courses use typographic grids and constraint-based design directly.
07
Related Standards
5 standards
These standards connect to HSG.MG.A.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.G.A.1Prerequisite
Solve problems with scale drawings, computing actual lengths and areas
