Finding and Fixing Errors in Scale Drawing Work | Grade 7 Math | LevelKite

Error analysis means reading a finished solution, finding the step that goes wrong, explaining why it goes wrong, and fixing it. It is detective work. You do more than say the answer is off: you name the line and the condition it breaks.

The solutions in this lesson are about scale drawings. A scale drawing represents a real object, usually at a different size, with every length changed by the same multiplier. You already know how to find actual areas from a scale drawing and how to redraw one at a new size. Now you will judge work that uses those skills.

Every problem gives all of its measurements in words and numbers, so you can test each line by calculating.

Let $d$ be a length on the drawing and let $a$ be the actual length of the corresponding side. The scale factor $k$ is the number you multiply $d$ by to get $a$ , so $a = k \cdot d$ . Lengths are positive, so $k$ is positive.

The value of $k$ depends on the units:

With $d$ in centimeters and $a$ in meters, the scale $1\text{ cm} : 4\text{ m}$ gives $k = 4$ .
With both in centimeters, $4\text{ m} = 400\text{ cm}$ , so the same scale reads $1\text{ cm} : 400\text{ cm}$ and gives $k = 400$ .

Both are correct, but the units must stay consistent: once you choose the units for $k$ , every length must use them. In this lesson, $k$ is always written for drawing lengths in centimeters and actual lengths in meters. With the units fixed, a scale drawing has exactly one $k$ . Every pair of corresponding sides gives the same value.

Area needs more care, because a flat shape has two dimensions. Take the scale $1\text{ cm} : 4\text{ m}$ , so $k = 4$ . A square on the drawing that is $1$ cm on each side has an area of $1\text{ cm}^2$ . It stands for a square that is $4$ m on each side, and that square covers $4 \times 4 = 16\text{ m}^2$ , not $4\text{ m}^2$ . So the actual area in square meters is $16$ times the drawing's area in square centimeters. The factor $16$ is $k \cdot k$ , written $k^2$ .

Symbols show the same thing. A rectangle on the drawing has length $L$ and width $W$ , both in centimeters. Each side is a length, so the corresponding actual sides are $k \cdot L$ and $k \cdot W$ , both in meters. Multiply them to get the actual area. You can reorder and regroup the factors of a product without changing its value, so the two factors of $k$ come together. The actual area is $k^2$ times the drawing's area, $L \times W$ :

(k \cdot L) \times (k \cdot W) = (k \cdot k) \times (L \times W) = k^2 \cdot (L \times W)

Key concept

Test each line against what the scale means

A scale compares a drawing length $d$ to the actual length $a$ of the corresponding side, in consistent units. Let $k$ be the positive scale factor in those units.

Lengths: from the drawing to the real object, multiply by $k$ .
Areas: from the drawing to the real object, multiply by $k^2$ . The area units are the squares of the length units.

From the real object back to the drawing, divide instead: by $k$ for a length and by $k^2$ for an area. Dividing is allowed because $k$ is positive, so $k$ and $k^2$ are not $0$ .

A line of work is an error when it breaks one of these conditions.

a = k \cdot d \qquad \text{actual area} = k^2 \cdot \text{drawing area}

To analyze a solution, work through it in order:

List what is given, with units: the scale and its factor $k$ if they are stated, and the side that each measurement belongs to.
Read the solution one line at a time. For each line, ask the questions in the table below.
Stop at the first line that fails a check. Say which check it fails and why.
Redo the work from that line. Then confirm your new answer a second way.

Correct arithmetic is not enough to make a line correct. The line must also use the right factor, on corresponding sides, in consistent units.

A correct line of scale-drawing work passes all four checks

Check	Question to ask	A sign that the line fails
Corresponding sides	Is each drawing length paired with the actual length of the side it stands for?	A drawing's length is paired with an actual width.
Consistent units	Is every length in the units that $k$ was written for?	A length in millimeters is multiplied by a $k$ written for centimeters.
Lengths	Drawing to actual: is the length multiplied by $k$ ? Actual to drawing: is it divided by $k$ ?	An actual length is multiplied by $k$ when it should be divided.
Areas	Drawing to actual: is the area multiplied by $k^2$ ? Actual to drawing: is it divided by $k^2$ ?	A drawing area is multiplied by $k$ instead of $k^2$ .

Worked example

A scale built from the wrong pair of sides

A rectangular room is $8$ m long and $6$ m wide. On a scale drawing, the room's length is $4$ cm. Kai wants the room's width on the drawing. Line 1 of his work: $4\text{ cm} : 6\text{ m}$ , so the scale is $1\text{ cm} : 1.5\text{ m}$ . Line 2: width on the drawing $= 6 \div 1.5 = 4$ cm. Find Kai's error and correct it.

Name what each number measures. The 4 cm is the room's length on the drawing, so its corresponding side is the room's actual length, 8 m. The 6 m is the room's actual width, which is a different side.
Run the corresponding-sides check on Line 1. Kai pairs the drawing's length with the actual width. A scale compares corresponding sides only, so Line 1 is the error. His drawing would be 4 cm by 4 cm. Dividing each actual length in meters by its drawing length in centimeters then gives two different scale factors, but one scale drawing has only one.
$8 \div 4 = 2 \quad\text{but}\quad 6 \div 4 = 1.5$
Build the scale from the corresponding pair: drawing length 4 cm and actual length 8 m. Divide both parts of the ratio by 4. Dividing both parts by the same nonzero number keeps the ratio the same. With drawing lengths in centimeters and actual lengths in meters, the scale factor is k = 2.
$4\text{ cm} : 8\text{ m} = 1\text{ cm} : 2\text{ m}$
Find the width on the drawing. You are going from an actual length to a drawing length, so divide the actual width, 6 m, by k. Dividing is valid because k = 2 is not 0. The width on the drawing is 3 cm.
$d = a \div k = 6 \div 2 = 3$
Confirm with both pairs of corresponding sides. Each drawing length in centimeters, multiplied by k, must give the matching actual length in meters. You get 8 m and 6 m, which match the room.
$4 \times 2 = 8 \quad\text{and}\quad 3 \times 2 = 6$

Answer

Kai's error is in Line 1: he compared the drawing's length with the room's actual width, and those are not corresponding sides. Line 2 then used that wrong scale. The correct scale is $1\text{ cm} : 2\text{ m}$ , so the room is $3$ cm wide on the drawing.

Worked example

An area multiplied by the wrong factor

A scale drawing of a rectangular patio uses the scale $1\text{ cm} : 3\text{ m}$ . On the drawing, the patio is $5$ cm long and $4$ cm wide. Dana finds the actual area. Line 1 of her work: drawing area $= 5 \times 4 = 20\text{ cm}^2$ . Line 2: actual area $= 20 \times 3 = 60\text{ m}^2$ . Find Dana's error and correct it.

Write what the scale says. Each 1 cm on the drawing stands for 3 m. With drawing lengths in centimeters and actual lengths in meters, the scale factor is k = 3, and its square is 9.
$k = 3 \qquad k^2 = 3 \times 3 = 9$
Check Line 1. Both numbers are drawing lengths in centimeters, so their product is the drawing's area in square centimeters. Line 1 passes.
$5 \times 4 = 20$
Check Line 2. It turns a drawing area into an actual area, so the areas check calls for the square of k. Dana multiplies by k itself. Line 2 is the error.
Show why it fails by using corresponding sides. Each side is a length, so each one is multiplied by k. The actual patio is 15 m long and 12 m wide, and its area is 180 square meters, not 60.
$(5 \times 3) \times (4 \times 3) = 15 \times 12 = 180$
Correct Line 2 by multiplying the drawing area by the square of k. The result agrees with the area found from the sides, because the factor 3 is used once for the length and once for the width.
$20 \times 9 = 180$

Answer

Dana's error is in Line 2: she multiplied the drawing area by $k = 3$ instead of $k^2 = 9$ . The actual area of the patio is $180\text{ m}^2$ .

Worked example

A redraw that shrinks when it should grow

Plan A is a scale drawing of a rectangular field at the scale $1\text{ cm} : 10\text{ m}$ . On Plan A the field is $6$ cm long and $4$ cm wide. Leo redraws the field as Plan B at the scale $1\text{ cm} : 5\text{ m}$ . Line 1 of his work: the number of meters in the scale goes from $10$ to $5$ , and $10 \div 5 = 2$ . Line 2: divide each length by $2$ , so Plan B is $3$ cm long and $2$ cm wide. Find Leo's error and correct it.

Write what each scale says, with drawing lengths in centimeters and actual lengths in meters. Plan A's scale factor is 10 and Plan B's is 5. Both plans represent the same field, so corresponding sides on the two plans stand for the same actual length.
$\text{Plan A: } k = 10 \qquad \text{Plan B: } k = 5$
Test Leo's result with the lengths check. On Plan B, a drawing length multiplied by 5 gives the actual length, so his 3 cm stands for 15 m. On Plan A, the corresponding side is 6 cm, which stands for 60 m. One side cannot be both 15 m and 60 m long, so Line 2 is the error.
$3 \times 5 = 15 \quad\text{but}\quad 6 \times 10 = 60$
Explain the error. The numbers 10 and 5 count actual meters for each 1 cm of drawing. Line 1 is true: that number is halved. But when each centimeter stands for fewer meters, the same side needs more centimeters. Leo made the drawing smaller when it must get larger.
Correct the work. First go from Plan A to the real field: multiply each Plan A length by Plan A's scale factor, 10. The field is 60 m long and 40 m wide.
$6 \times 10 = 60 \quad\text{and}\quad 4 \times 10 = 40$
Now go from the real field to Plan B: divide each actual length by Plan B's scale factor, 5. Dividing is valid because 5 is not 0. Plan B is 12 cm long and 8 cm wide. Each length is 2 times the Plan A length, so Leo's 2 should multiply, not divide.
$60 \div 5 = 12 \quad\text{and}\quad 40 \div 5 = 8$
Confirm with the areas check. Plan A's drawing area is 6 times 4 = 24 square centimeters. Multiply by the square of Plan A's factor, 100, to get the actual area: 2400 square meters. Going back to a drawing, divide by the square of Plan B's factor, 25, which is not 0. That gives 96 square centimeters, and a 12 cm by 8 cm rectangle has exactly that area.
$24 \times 100 = 2400 \qquad 2400 \div 25 = 96 \qquad 12 \times 8 = 96$

Answer

Leo's error is in Line 2: he divided the Plan A lengths by $2$ when he should have multiplied by $2$ , because when each centimeter stands for fewer meters, the same field needs longer drawing lengths. On Plan B the field is $12$ cm long and $8$ cm wide.

Takeaway: To find an error in scale-drawing work, test each line against what the scale means. A scale compares a drawing length to the actual length of the corresponding side, in consistent units. From the drawing to the real object, lengths multiply by the scale factor $k$ and areas multiply by $k^2$ . In the other direction, you divide. Name the first line that breaks one of these conditions, say which condition it breaks, fix it, and confirm your new answer a second way.