Secondary 2 Missing Measurements and Unit Conversion
Every measurement formula you have met so far has been run forwards: put the dimensions in, get a perimeter, an area or a volume out. This topic runs them backwards. You are handed the result and one of the dimensions, and asked for the one that is missing — in plane figures, in solids, and in decomposable and truncated ones. The second half then deals with the conversions those questions keep needing, where an area does not convert the way a length does and time does not convert like anything else. It reads on the page, and the PDF is free to print.
Practice worksheet — free PDF
No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 9 harder problems come with the Secondary 2 Math bundle.
All 10 questions
Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.
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Q1Finding Missing Measurements in Plane Figures From the Perimeter
Find the missing length in each figure. Show the equation you solve.
- A rectangular banner has a perimeter of and is wide. How long is it?
- An isosceles triangle has a perimeter of . Its two equal sides each measure . How long is the base?
- A rectangular hockey-card display has a perimeter of , and its length is twice its width. Find both dimensions.
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Q2Finding Missing Measurements in Plane Figures from their Area
Each figure's area is given. Find the missing measurement.
- A parallelogram has an area of and a base of . Find its height.
- A triangle has an area of and a height of . Find the base that goes with that height.
- A trapezoid has an area of and parallel sides measuring and . Find its height.
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Q3Missing Measurements According to Area: Decomposable and Truncated Solids
A storage bin is an open-topped rectangular prism with a base measuring by . The inside of the base and all four walls are coated, and the coated surface measures . Find the height of the bin.
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Q4Missing Measurements in Plane Figures
For each situation, name the measurement formula you would start from, say which letter you are solving for, and state the first step you would take — or explain why the information you are given is not enough. Do not carry out the arithmetic.
- You know a triangle's area and its base, and you want its height.
- You know a rectangle's perimeter and that it is three times as long as it is wide, and you want its width.
- You know a trapezoid's area, its height and one of its parallel sides, and you want the other parallel side.
- You know only a rectangle's perimeter, and you want its area.
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Q5Missing Measurements of Solids from the Area
Find the missing measurement of each solid.
- A cube has a total surface area of . Find its edge.
- The label on a cylindrical can covers the curved side only and has an area of . The can's radius is . Find its height.
- A pyramid with a square base of edge has a lateral area (the four triangular faces) of . Find the slant height of one triangular face.
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Q6Missing Measures from a Volume: Decomposable and Truncated Solids
A water trough is made by joining two rectangular prisms end to end. The first measures long, wide and deep. The second is long, wide and cm deep. Together they hold . Find .
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Q7Units for Measuring Area and Their Conversion
Convert each measurement.
- into
- into
- into
- into
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Q8Units for Measuring Mass and Their Conversion
Convert or compute.
- into
- into
- into
- A camping bag holds six pouches of each plus a stove of . Give the total mass in kilograms.
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Q9Units of Time and Their Conversion
Convert or compute.
- into minutes
- into hours and minutes
- into minutes, then into seconds
- A film lasting starts at . At what time does it end?
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Q10Synthesis — drawing on several sheets in this topic
A rectangular patio is to have an area of , and the space available makes it exactly wide.
- Find the length of the patio in metres.
- Give that length in centimetres.
- Give the patio's area in , and check that it agrees with your two side lengths in centimetres.
The 9 challenge problems for this topic are a separate, paid sheet and are not reproduced here.
Which stream is this for? There is no stream to choose at this level. Secondary 2 is the second year of Cycle One and every student follows the same program, so this sheet is simply Secondary 2 mathematics as the Québec Education Program's Progression of Learning sets it out — the streams begin in Secondary 4.
How to do every concept on this sheet
This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.
One method, run on six kinds of figure
Q1, Q2, Q3, Q5 and Q6 are the same four steps applied to a rectangle, a triangle, a trapezoid, a cube, a cylinder, a pyramid, an open bin and a pair of joined prisms. Learning them as eight separate recipes is what makes this topic feel long. It is one recipe.
Working a measurement formula backwards
The same four steps, whichever figure it is.
- 1Write the formula out in full first
Before any number goes anywhere. The area of a trapezoid, the total area of a cube, the volume of a prism — written as a formula with letters. Half the errors in this topic are a half-remembered formula, and they are invisible once numbers have replaced the letters.
- 2Substitute everything you know, and name the letter you don't
You now have an equation with one unknown. Say which letter it is out loud: "I am solving for h." Q4 asks for exactly this and nothing more, because deciding what to solve for is a separate skill from solving.
- 3Simplify the known side before you undo anything
Turn 2(40 + 30)h into 140h, or 6x² into a single term. Undoing an expression you have not tidied is how a factor gets dropped — and it is exactly the mistake Q5's cylinder question is built around.
- 4Undo the operations, then check in the original wording
Divide where the unknown was multiplied, take a square root where it was squared. Then put your answer back into the question, not into your own equation — which is where the mistake would be if there is one.
Q4 asks for the plan, not the answer
Q4 gives four situations and forbids arithmetic. Name the formula, name the letter you are solving for, state the first step — or explain why what you have been given is not enough. That last branch is the reason the question exists.
A perimeter alone does not determine a rectangle. Knowing only that the perimeter is 54 dm leaves infinitely many rectangles; 20 by 7 and 15 by 12 both fit. It becomes solvable the moment a second fact links the two dimensions — "the length is twice the width" in Q1, "three times as long as it is wide" in Q4. One equation, one unknown: the second fact is what reduces two unknowns to one.
Say which height. A triangle has three, one for each side, and the area formula pairs a base with its own height. Q2 asks for "the base that goes with that height" in exactly those words for that reason.
Lateral area or total area — the question the wording answers (Q5)
Q5 and its harder partner turn on one distinction. A label wrapping a can covers the curved surface only. Sheet metal for a closed can covers the curved surface and both discs. The two lead to different equations from the same numbers:
lateral only: 2πrh = A total: 2πrh + 2πr² = AGiven a cylinder's total area, a student who writes 10πh = 100π has quietly used the lateral formula on a total area. The fix is to subtract the two discs from the given total first, and only then divide. Read the object: a rain barrel open at the top has one disc, a can has two, a label has none.
The same trap on an open bin. Q3's storage bin is coated on the base and the four walls — five surfaces, not six. The equation is 40 × 30 + 2(40h) + 2(30h) = 4000, and starting from the closed-prism formula puts an extra 1200 cm² into it before you begin.
Decomposable and truncated solids: subtract before you solve (Q6)
When part of the solid is already fully known, take it out of the total first and you are left with the ordinary one-unknown problem you already know how to do. Q6's trough is two prisms holding 168 000 cm³ between them: the first prism's volume is computable outright, so subtract it and the second prism's depth falls out in a single division.
A solid with a channel cut through it works the same way in reverse — start from the whole block, subtract the removed piece, and set the result equal to what is left. The only extra care it needs is the units: a volume quoted in dm³ and dimensions given in cm have to be brought to the same unit before the equation means anything, and that conversion is not the one most people expect.
Area units square the factor, volume units cube it (Q7)
This is the idea the whole conversion half rests on, and it is worth being able to justify rather than just remember. One metre is 100 cm. A square of one metre on each side is therefore 100 cm by 100 cm:
1 m² = 100 cm × 100 cm = 10 000 cm² 1 m³ = 1 000 000 cm³So the factor between two length units gets squared for areas and cubed for volumes. Converting an area with the plain length factor instead gives 2.4 m² = 240 cm², which is off by a factor of a hundred. Drawing that one-metre square once is the fastest way to see why, and it is worth doing on paper rather than taking on trust.
Check the direction before you compute. Going to a smaller unit gives a bigger number, always. 4.5 m² into cm² must grow; 320 000 cm² into m² must shrink. Deciding that in advance catches a multiplication done the wrong way round without any arithmetic at all.
Volume and capacity meet here too. 1 dm³ is one litre and 1000 cm³. A volume given in dm³ next to lengths in cm is a conversion, not a coincidence.
Mass, and the rounding that goes downwards (Q8)
Mass converts on plain powers of ten — milligram, gram, kilogram, tonne — and Q8 mixes a conversion into a computation: six pouches in grams plus a stove in kilograms, answered in kilograms. Convert everything to one unit before adding a single pair of numbers.
Not every answer rounds to the nearest whole number. A crate rated for 250 kg carrying 14 kg boxes gives 17.8…, and the answer is 17. The eighteenth box would put the crate over its limit, so the context — not the decimal — decides the direction. The mirror situation rounds the other way: asked how many crates are needed to ship a load, a decimal always rounds up, because a part-full crate is still a crate. Say in one sentence which of the two situations you are in.
Time is not decimal (Q9)
An hour is 60 minutes, not 100, and that is the whole difficulty. Typing 2.47 + 1.55 into a calculator to add 2 h 47 min and 1 h 55 min treats minutes as hundredths of an hour and lands 40 minutes short — a single mistake that accounts for most wrong time totals.
Add the units separately, then carry. 47 + 55 = 102 minutes, which is 1 h 42 min, so the total is 2 + 1 + 1 = 4 h and 42 min. Carrying happens at 60, not at 100.
Converting to a decimal is legitimate — with the right factor. 1.75 h is 1.75 × 60 = 105 minutes, and 105 × 60 = 6300 seconds. The error is never in converting to decimals; it is in assuming 2 h 47 min already is the decimal 2.47.
Finish times need the same carry. A film of 2 h 15 min starting at 18:50 ends at 21:05, not 20:65.
Q10: two units in one problem
The synthesis question sets an area in square metres against a width given in centimetres, and asks for the answer both ways. Convert to a single unit before you divide — mixing metres and centimetres inside one formula produces a number that is wrong by a factor of a hundred and looks entirely reasonable.
Then do what part c) asks even when a question does not ask it: recompute the area from the two side lengths in the other unit and check that the two agree. That agreement is the whole point of the squaring rule, and it is a check you can run on any conversion you are unsure of.
Getting the most out of it
Write the formula with letters before any number appears
Every question in the first half is a formula run backwards, so the formula has to be on the page before it can be run in either direction. Write it, substitute, then solve — in that order, every time.
Decide which surfaces are actually counted
Open at the top, coated on the inside, painted except where it rests on the ground, labelled on the curved side only: each of those phrases changes the equation. Underline the phrase in the question and list the surfaces before writing anything.
Put your answer back into the question, not into your equation
Substituting into your own equation only confirms your algebra. Recomputing the perimeter, area or volume from the original wording confirms the whole solution, and it is the only check that catches a formula misremembered in the first line.
Want the solutions, or something more challenging?
The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the Secondary 2 Math Solutions Bundle, which is what keeps the rest of the series free.
What else exists for Missing Measurements and Unit Conversion
Three PDFs · 10 pages · all three are in the bundle below.
- Answer key — 2 pages. All 10 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
- Challenge problems — 5 pages, 9 problems. A separate sheet at exam-plus difficulty covering the same 9 concepts. Harder than anything on the free sheet.
- Challenge answer key — 3 pages. Every challenge problem worked to the same standard, with the checks shown.
- PDF, letter size, print-ready.
The one thing that's for sale
Every Secondary 2 Math topic — the complete Solutions Bundle
One download, one payment, the whole program. Every answer key and every challenge set for all 14 Secondary 2 Math worksheet sets — including this one.
- Worked solutions, not answer lists — every step written out
- Covers the whole year's program at this level
- Less than the price of one hour of tutoring — for the entire year's solutions
Taking Secondary 1 Math as well? The Secondary 1 Math bundle covers all 15 of its sets — 45 PDFs, 182 pages — on the same terms.
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Taking Secondary 4 Math as well? The Secondary 4 Math bundle covers all 17 of its sets — 51 PDFs, 154 pages — on the same terms.
Taking Secondary 5 Math as well? The Secondary 5 Math bundle covers all 21 of its sets — 63 PDFs, 228 pages — on the same terms.
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Taking AP Calculus AB as well? The AP Calculus AB bundle covers all 8 of its sets — 24 PDFs, 126 pages — on the same terms.
Common questions
Is this worksheet really free?
Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete Secondary 2 Solutions Bundle, which covers every set at this level.
Why is 1 m² equal to 10 000 cm² and not 100 cm²?
Because a square metre is a square measuring 100 cm on each side, so it holds 100 × 100 = 10 000 square centimetres. The factor between two length units gets squared for areas and cubed for volumes, which is why 1 m³ is a million cubic centimetres. Drawing the one-metre square once is usually enough to stop the mistake for good.
How do I find a missing measurement from an area?
Write the area formula with letters, substitute everything you know, simplify the side that is fully known, then undo the operations acting on the unknown. Divide where it was multiplied, take a square root where it was squared, and finish by recomputing the area from your answer to check it against the original wording.
Why can't I add 2 h 47 min and 1 h 55 min on a calculator as 2.47 + 1.55?
Because an hour has 60 minutes, not 100, so 47 minutes is not 0.47 of an hour. Add the hours and the minutes separately and carry at 60: 47 + 55 = 102 min, which is 1 h 42 min, giving 4 h 42 min in total. Converting to decimals is fine as long as you convert with 60 rather than assuming the minutes are already decimals.
When do I round down instead of to the nearest whole number?
When the context forbids the extra one. A crate with a mass limit takes only whole boxes that fit inside the limit, so 17.8 boxes means 17. The mirror question — how many crates are needed to carry a load — rounds up instead, because a part-full crate is still a crate. Decide from the situation and write the reason down in a clause.
Which year is this sheet for?
Secondary 2, the second year of Cycle One, and it stays inside that year. Missing measurements here come from one equation in one unknown; recovering them by solving a system of equations is a Secondary 3 method, and nothing on this sheet needs it.
Can teachers use this in class?
Yes. Print and photocopy it for your own classes freely — I just ask that the tutorinmontreal.ca footer stays on the page.
I'm stuck on one question. Can you help?
Yes — through one-on-one tutoring, in Montreal or online. Get in touch to arrange a session, or see the current rates.
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