AutoCAD Quiz: Scale

10 questions · exam conditions

0:00

ScaleQuestion 1 of 10

An imported part has a measured control distance of $47.6$ units, but that distance should be $50$ units. Another spacing within the same selected geometry currently measures $9.52$ units. The user applies SCALE with Reference, entering $47.6$ as the reference length and $50$ as the new length.

What should the other spacing measure after the operation?

It should measure

9.066

units.

It should measure

9.52

units.

It should measure

10

units.

It should measure

11.92

units.

← Back to quizzes

AutoCAD Quiz

AutoCAD Quiz: Scale

Practice Scale in AutoCAD with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Scale, giving you a quick way to practice the rules, question types, and explanations that matter most for AutoCAD.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What should the other spacing measure after the operation?

It should measure $9.066$ units.
It should measure $9.52$ units.
It should measure $10$ units. (correct answer)
It should measure $11.92$ units.

Explanation: When you use AutoCAD's SCALE command with the Reference option, you're telling the software to resize selected geometry by a ratio derived from two lengths: the current (reference) measurement and the desired target measurement. Every dimension within the selected geometry scales by that same uniform factor — that's the key insight here. The scale factor is calculated as:

\text{Scale Factor} = \frac{\text{New Length}}{\text{Reference Length}} = \frac{50}{47.6} \approx 1.0504

Applying this to the second spacing:

9.52 \times 1.0504 \approx 10.0 \text{ units}

So the correct answer is C — the spacing becomes exactly

10

units, which makes sense: both measurements were proportionally "undersized" by the same ratio, and scaling corrects them simultaneously. A (

9.066

units) represents a shrinkage, as if you divided instead of multiplied — flipping the scale factor upside down. B (

9.52

units) reflects the misconception that only the referenced dimension changes while everything else stays fixed; in reality, SCALE transforms all selected geometry uniformly. D (

11.92

units) comes from mistakenly doubling the scale factor or applying an additive offset rather than a multiplicative ratio — a common arithmetic trap. A helpful strategy: whenever you see SCALE with Reference on the exam, immediately compute

\frac{\text{new}}{\text{reference}}

to find the scale factor, then multiply every other dimension by that same value. Clean ratios like

\frac{50}{47.6}

that produce round results on secondary dimensions are a strong signal you've set up the problem correctly.

Question 2

A closed rectangular object has an area of $48$ square units. Its proportions must remain unchanged, but its area must become $108$ square units.

What scale factor should be applied to the rectangle?

Apply a scale factor of $0.6667$ .
Apply a scale factor of $1.25$ .
Apply a scale factor of $1.5$ . (correct answer)
Apply a scale factor of $2.25$ .

Explanation: When scaling an object in AutoCAD, it's critical to understand that area scales by the square of the linear scale factor. So if you apply a scale factor of

k

, the new area equals the original area multiplied by

k^2

. This means you can't simply divide the new area by the old area to find your scale factor — that gives you the area ratio, not the linear scale factor. Here's how to work it correctly. First, find the area ratio:

\frac{108}{48} = 2.25

. This tells you the area must increase by a factor of

2.25

. Since area scales as

k^2

, you need to take the square root to find the linear scale factor:

k = \sqrt{2.25} = 1.5

. That confirms C is correct — applying a scale factor of

1.5

in AutoCAD will produce a rectangle with

48 \times 1.5^2 = 48 \times 2.25 = 108

square units. Looking at the distractors: A (

0.6667

) would actually shrink the object, reducing the area below 48 — the opposite of what's needed. B (

1.25

) is a plausible-looking number but doesn't produce the correct result:

48 \times 1.25^2 = 75

, not 108. D (

2.25

) is the area ratio itself — this is the most tempting trap, because students who skip the square root step land here. A reliable memory tip: area problems require a square root, volume problems require a cube root. Whenever a question gives you two areas and asks for a scale factor, always compute the ratio first, then take the square root.

Question 3

A bracket edge is $120$ units long. During SCALE, the user selects Reference, identifies two points that are $60$ units apart, and specifies a new reference length of $90$ units. The two reference points are not the endpoints of the bracket edge.

What will be the resulting length of the bracket edge?

The edge becomes $80$ units long.
The edge becomes $90$ units long.
The edge becomes $150$ units long.
The edge becomes $180$ units long. (correct answer)

Explanation: When using the Reference option in AutoCAD's SCALE command, you're defining a scaling ratio — not anchoring specific points or setting an absolute length for a particular edge. The ratio is:

\text{scale factor} = \dfrac{\text{new reference length}}{\text{old reference length}}

. This factor then applies uniformly to the entire object, regardless of which two points you used to measure the reference distance. Here, the reference points are

60

units apart, and you specify a new length of

90

units for that same distance. That gives a scale factor of

\dfrac{90}{60} = 1.5

. AutoCAD then multiplies every dimension in the object by

1.5

, so the bracket edge becomes

120 \times 1.5 = 180

units — confirming D is correct. Choice A (

80

units) likely comes from subtracting:

120 - (90 - 60) = 90

… actually it may come from dividing

120 \div 1.5 = 80

, confusing which direction the scale goes — dividing instead of multiplying. Choice B (

90

units) is the most tempting trap: students assume the Reference option directly sets the edge to the new reference length of

90

, but that's not how it works — it only defines the scale ratio. Choice C (

150

units) might come from adding the

30

-unit increase from the reference (

90 - 60 = 30

) directly onto

120

, treating it as an absolute offset rather than a proportional scale. The key study tip: whenever you see Reference in a SCALE question, immediately compute the ratio

\dfrac{\text{new}}{\text{old}}

and multiply it by the actual object dimension — never assume the new reference value directly becomes any edge's length.

Question 4

A rectangular object is $120$ units wide and $60$ units high. It must become $150$ units wide while remaining exactly $60$ units high. The user intends to use one standard SCALE operation.

Which statement correctly describes the expected result?

A reference change from $120$ to $150$ produces $150$ by $60$ because only the referenced direction changes.
A factor of $1.25$ produces $150$ by $75$ , so one SCALE operation cannot meet both dimensions. (correct answer)
A factor of $1$ produces $150$ by $60$ because the unchanged height controls the operation.
A factor of $0.8$ produces $150$ by $60$ because the inverse width ratio preserves the height.

Explanation: Whenever you see a question about AutoCAD's SCALE command, remember one fundamental rule: SCALE applies a single uniform factor to all dimensions simultaneously. There is no way to stretch width independently of height using a standard SCALE operation — that requires STRETCH or non-uniform scaling tools. Here's the math that makes B correct. To bring the width from

120

150

, you need a scale factor of

\frac{150}{120} = 1.25

. Because SCALE is uniform, that same

1.25

multiplier applies to the height as well:

60 \times 1.25 = 75

. The result is

150 \times 75

, not

150 \times 60

. Since the height requirement of

60

units is violated, one standard SCALE operation simply cannot satisfy both constraints at once — exactly what B states. Answer A describes the Reference option within SCALE, which lets you type an existing length and a new length instead of a direct factor. However, Reference still computes a uniform factor (

\frac{150}{120} = 1.25

) and applies it to every axis. The height would still become

75

, not

60

. A incorrectly implies Reference scales only the referenced direction. Answer C is nonsensical — a factor of

1

changes nothing at all, so the width stays at

120

, not

150

. Answer D inverts the ratio (

\frac{120}{150} = 0.8

), which would actually shrink the width to

96

, not enlarge it to

150

. Study tip: Anytime a question asks whether SCALE can resize an object to specific dimensions in one step, check whether the width ratio and height ratio are equal. If they differ, SCALE alone won't work — and that's almost always the trap being set.

Question 5

Two circles have center x-coordinates of $20$ and $50$ and diameters of $8$ and $12$ units, respectively. Both circles are selected and scaled together about a base point at $x=0$ using a factor of $0.5$ .

Which result correctly describes the circles after scaling?

Centers at $10$ and $25$ ; diameters of $4$ and $6$ . (correct answer)
Centers at $20$ and $50$ ; diameters of $4$ and $6$ .
Centers at $10$ and $25$ ; diameters of $8$ and $12$ .
Centers at $20$ and $35$ ; diameters of $4$ and $6$ .

Explanation: Whenever you see a scaling question in AutoCAD, remember that the SCALE command transforms geometry relative to the base point — meaning every point in the drawing moves closer to or farther from that base point by the scale factor. This affects both the position of objects and their size. Here's how it works mathematically: when you scale by

0.5

about

x = 0

, every x-coordinate is multiplied by

0.5

. So the circle centers move from

20

and

50

20 \times 0.5 = 10

and

50 \times 0.5 = 25

. The diameters also scale:

8 \times 0.5 = 4

and

12 \times 0.5 = 6

. That gives you centers at

10

and

25

with diameters of

4

and

6

— confirming answer A is correct. Answer B keeps the centers at their original positions, which would only be true if the base point were at each circle's own center — it isn't. The centers absolutely move when the base point is external. Answer C correctly recalculates the center positions but leaves the diameters unchanged, confusing a move operation with a scale operation — scaling always resizes the geometry too. Answer D uses incorrect center values (

20

and

35

) that don't follow from any standard scaling logic, likely representing a random arithmetic error. A reliable tip: always ask yourself two questions when scaling — "Where do the centers go?" (multiply coordinates by the factor relative to the base point) and "How do the sizes change?" (multiply all dimensions by the same factor). Both change together.

Question 6

A door symbol is $34$ units wide and $80$ units high. The hinge point is selected as the SCALE base point. Using Reference, the user enters the current width of $34$ units and a new width of $36$ units.

Which result should the user expect?

Width $36$ , height about $84.71$ , with the center point fixed.
Width $36$ , height $80$ , with the hinge point fixed.
Width $36$ , height $82$ , with the hinge point fixed.
Width $36$ , height about $84.71$ , with the hinge point fixed. (correct answer)

Explanation: When using AutoCAD's SCALE command with the Reference option, the key concept to understand is that the scale factor applies uniformly in all directions — and the base point remains fixed while everything else stretches proportionally around it. Here's how to think through it: the Reference option calculates a scale factor by dividing the new value by the reference value. In this case, the factor is

\frac{36}{34} \approx 1.0588

. AutoCAD then applies this same factor to every dimension of the object. So the new height becomes

80 \times \frac{36}{34} \approx 84.71

units. The selected base point — the hinge point — stays locked in place while the rest of the symbol scales outward from it. That confirms D as the correct answer: width

36

, height approximately

84.71

, with the hinge point fixed. A gets the geometry right (width

36

, height ≈

84.71

) but incorrectly states that the center is fixed. The base point you select — not the center — is what remains stationary. B assumes only the width changes while the height stays at

80

. This would describe a non-uniform stretch (like the STRETCH command), not a proportional scale. SCALE always affects all axes equally. C invents a height of

82

, which has no mathematical basis. It may seem like a reasonable "add 2 to match the width increase," but that's arithmetic, not proportional scaling. A useful tip: on AutoCAD exam questions involving SCALE with Reference, always compute

\frac{\text{new}}{\text{old}}

and apply it universally — and remember that your chosen base point never moves.

Question 7

A line extends from $x=20$ to $x=50$ . The user starts SCALE, selects the line, specifies $x=20$ as the base point, chooses the Copy option, and enters a factor of $2$ .

Which geometry exists when the command is completed?

Only a scaled line extending from $x=20$ to $x=80$ .
The original line from $x=20$ to $x=50$ and a copy from $x=20$ to $x=80$ . (correct answer)
The original line from $x=20$ to $x=50$ and a copy from $x=40$ to $x=100$ .
Two coincident lines, both extending from $x=20$ to $x=50$ .

Explanation: When working with AutoCAD's SCALE command, the key concept to understand is the difference between a standard scale operation and one that uses the Copy option. Normally, SCALE transforms the original object in place — it's destructive. The Copy option changes this behavior entirely, preserving the original geometry while producing a scaled duplicate. Here's how the math works: your original line spans from

x=20

x=50

, giving it a length of

30

units. The base point is set at

x=20

. When you scale by a factor of

2

from that base point, every point on the object moves to twice its distance from

x=20

. The left endpoint stays at

x=20

(distance of

0 \times 2 = 0

), and the right endpoint moves to

x=20 + (30 \times 2) = x=80

. Because you used Copy, the original line from

x=20

x=50

is retained, and the new scaled line from

x=20

x=80

is added — making B the correct answer. Choice A describes the result of a standard SCALE without Copy — the original would be gone, replaced by the scaled version. Choice C reflects a misunderstanding of how the base point works; if the base point were at

x=0

, the endpoints would double to

x=40

and

x=100

, but anchoring at

x=20

keeps that endpoint fixed. Choice D would only make sense if a scale factor of

1

were used, which would produce an identical overlapping copy. A helpful rule of thumb: whenever you see the Copy option inside a modify command (SCALE, ROTATE, MIRROR), think "original stays, transformed version is added."

Question 8

A selected component is currently $24$ drawing units long and must become exactly $37.5$ drawing units long. One endpoint will be used as the base point.

Which scale factor should be entered in the SCALE command?

Enter a scale factor of $1.5625$ . (correct answer)
Enter a scale factor of $0.64$ .
Enter a scale factor of $13.5$ .
Enter a scale factor of $0.5625$ .

Explanation: When using AutoCAD's SCALE command, you're applying a multiplier to the object's current size — so the core calculation is always:

\text{Scale Factor} = \dfrac{\text{New Length}}{\text{Old Length}}

. Keeping this formula in mind will cut through any confusion the question tries to create. Here, you need to go from

24

units to

37.5

units, so:

\dfrac{37.5}{24} = 1.5625

. Since the result is greater than

1

, the object gets larger — which makes sense, because

37.5 > 24

. Answer A is correct. As for the wrong answers: B (

0.64

) is the reciprocal trap — it's roughly

\frac{24}{37.5}

, which is what you'd use if you were shrinking from

37.5

down to

24

, not growing in the other direction. Students who flip the numerator and denominator land here. C (

13.5

) comes from subtracting instead of dividing:

37.5 - 24 = 13.5

. The SCALE command uses multiplication, not addition, so the difference between the two lengths is completely irrelevant. D (

0.5625

) is a believable distractor because it looks like a proper decimal scale factor, but it would shrink the object to about half its size rather than enlarging it. A quick study tip: whenever a SCALE question gives you both a current and target dimension, always divide new ÷ old. If the result is above

1

, you're enlarging; below

1

, you're shrinking. That sanity check alone will help you eliminate wrong answers instantly.

Question 9

Geometry was drafted so that one drawing unit represented one inch. It is copied without automatic unit conversion into a drawing where one unit must represent one millimeter. A selected edge currently has a numeric length of $8$ units.

Which scaling result correctly converts the selected geometry to millimeters?

Use factor $0.03937$ ; the edge becomes about $0.315$ units.
Use factor $8$ ; the edge becomes $64$ units.
Use factor $25.4$ ; the edge becomes $203.2$ units. (correct answer)
Use factor $203.2$ ; the edge becomes $1625.6$ units.

Explanation: When geometry moves between drawings with different unit assumptions in AutoCAD, no automatic rescaling happens — you must manually apply a scale factor to convert the physical meaning of the numbers. The key relationship to memorize is that

1 \text{ inch} = 25.4 \text{ mm}

, which drives every inch-to-millimeter conversion. Here, the original drawing treats 1 unit as 1 inch, but the destination drawing treats 1 unit as 1 mm. To preserve the real-world size, every unit value must be multiplied by

25.4

. An edge reading

8

units (meaning 8 inches) must become

203.2

units (meaning 203.2 mm), because

8 \times 25.4 = 203.2

. That makes C correct. Choice A applies the factor

0.03937

, which is actually

\frac{1}{25.4}

— the conversion going the opposite direction, from millimeters to inches. Using it shrinks the geometry drastically instead of enlarging it. Choice B uses a factor of

8

, which is simply the current edge length — there is no logical reason to scale by the length of one edge; this produces a meaningless result. Choice D multiplies by

203.2

, which is itself the target length of that one edge, not a unit-conversion factor. Applying it cascades a single edge's converted value as if it were a ratio, producing a wildly oversized result. A reliable study tip: always anchor unit-conversion scaling to the ratio

\frac{\text{destination unit size}}{\text{source unit size}}

. Inches to millimeters → multiply by

25.4

. Millimeters to inches → multiply by

0.03937

. Getting the direction backwards is the most common trap on this type of question.

Question 10

A line has endpoints at $x=10$ and $x=70$ . The line is scaled with the SCALE command using the endpoint at $x=10$ as the base point and a scale factor of $1.5$ .

What is the resulting x-coordinate of the other endpoint?

The endpoint moves to $x=100$ . (correct answer)
The endpoint moves to $x=105$ .
The endpoint moves to $x=90$ .
The endpoint remains at $x=70$ .

Explanation: When working with AutoCAD's SCALE command, the critical concept to understand is that scaling stretches or shrinks an object relative to its base point. The base point itself never moves — only the distances from the base point change. Here's how to think through it: the original endpoint sits at

x = 70

, and the base point is at

x = 10

. First, find the distance from the base point to the endpoint:

70 - 10 = 60

units. Applying a scale factor of

1.5

multiplies that distance:

60 \times 1.5 = 90

units. Finally, add that new distance back to the base point:

10 + 90 = 100

. The endpoint moves to

x = 100

, confirming that A is correct. As for the wrong answers: B (

x = 105

) likely comes from mistakenly multiplying the entire original coordinate by

1.5

(

70 \times 1.5 = 105

), which ignores the role of the base point entirely. C (

x = 90

) is a partial calculation — it's the scaled distance (

90

units) without adding the base point's position back in, making it an easy arithmetic trap. D claims the endpoint doesn't move at all, which would only be true if the scale factor were

1.0

. A reliable strategy: always break SCALE problems into three steps — find the distance from base point to the object, multiply by the scale factor, then add the result back to the base point coordinate. Never scale the raw coordinate directly.