Autodesk Fusion 360 Quiz: Sketch Dimensions

10 questions · exam conditions

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Sketch DimensionsQuestion 1 of 10

Two adjacent horizontal line segments share an endpoint. Their outer-left endpoint is coincident with the sketch origin, and an Equal constraint makes the segments the same length. A driving dimension places the outer-right endpoint $100\,\mathrm{mm}$ horizontally from the origin.

A designer now applies a dimension of $50\,\mathrm{mm}$ to the left segment. How should Fusion 360 treat this new dimension?

As a driving dimension, because each individual segment still requires its own independently editable length value.

As a driven dimension, because the total length and Equal constraint already determine each segment's length.

As a driving dimension, because the Equal constraint controls orientation but does not control the segment lengths.

As an invalid dimension, because dimensions cannot be applied to segments that participate in an Equal constraint.

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Autodesk Fusion 360 Quiz

Autodesk Fusion 360 Quiz: Sketch Dimensions

Practice Sketch Dimensions in Autodesk Fusion 360 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 Sketch Dimensions, giving you a quick way to practice the rules, question types, and explanations that matter most for Autodesk Fusion 360.

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

A designer now applies a dimension of $50\,\mathrm{mm}$ to the left segment. How should Fusion 360 treat this new dimension?

As a driving dimension, because each individual segment still requires its own independently editable length value.
As a driven dimension, because the total length and Equal constraint already determine each segment's length. (correct answer)
As a driving dimension, because the Equal constraint controls orientation but does not control the segment lengths.
As an invalid dimension, because dimensions cannot be applied to segments that participate in an Equal constraint.

Explanation: Whenever you see a question about sketch constraints and dimensions in Fusion 360, ask yourself: how many degrees of freedom remain, and does adding this dimension resolve a new one or duplicate an existing one? Here, the setup already fully determines every measurement in the sketch. The outer-left endpoint is fixed at the origin (

0\,\mathrm{mm}

), the outer-right endpoint is fixed at

100\,\mathrm{mm}

by the driving dimension, and the Equal constraint forces both segments to be identical in length. That means each segment is already mathematically locked at

50\,\mathrm{mm}

—

100\,\mathrm{mm} \div 2

. When you attempt to add a

50\,\mathrm{mm}

dimension to the left segment, Fusion 360 recognizes it carries no new information and marks it as a driven (reference) dimension — it displays the value but cannot be edited to change the geometry. Answer B is correct. A is wrong because "independently editable" doesn't matter here — the Equal constraint ties the two lengths together, so specifying one implicitly specifies the other. No independent degree of freedom exists for that segment. C is wrong because Equal constraints absolutely do control length, not just orientation; that's the entire purpose of the constraint — to enforce identical magnitudes between selected entities. D is wrong because Fusion 360 never invalidates a dimension solely because the segment participates in an Equal constraint; driven dimensions on constrained geometry are a normal, supported workflow. Study tip: When a sketch is fully constrained, any additional dimension becomes driven automatically — Fusion 360 won't error out, it simply switches the dimension to a reference mode shown in gray.

Question 2

The width of a fully constrained base rectangle is controlled by a driving dimension of $60\,\mathrm{mm}$ . A second rectangle has an Equal constraint relating its width to the base rectangle. A driven dimension on the second rectangle also displays $60\,\mathrm{mm}$ .

The base rectangle's driving width is changed to $75\,\mathrm{mm}$ . What is the expected result?

The second rectangle remains $60\,\mathrm{mm}$ wide because driven dimensions preserve the measured geometry when parameters change.
The Equal constraint is suppressed because the two displayed dimension values no longer agree after the edit.
The second rectangle becomes $75\,\mathrm{mm}$ wide, and its driven dimension updates to report that value. (correct answer)
The sketch becomes over-constrained because the driven dimension continues enforcing its original value of $60\,\mathrm{mm}$ .

Explanation: When working with sketch constraints in Fusion 360, it helps to clearly distinguish between two types of dimensions: driving dimensions, which control geometry, and driven dimensions, which merely report the current state of geometry that is already controlled by something else. In this scenario, the Equal constraint is the active geometric relationship — it forces the second rectangle's width to always match the base rectangle's width. The driven dimension on the second rectangle is simply a readout; it has no authority to lock or preserve any value. So when you change the base rectangle's driving dimension from

60\,\mathrm{mm}

75\,\mathrm{mm}

, the Equal constraint propagates that change instantly, making the second rectangle

75\,\mathrm{mm}

wide. The driven dimension then updates automatically to display

75\,\mathrm{mm}

, confirming the new geometry. That makes C correct. A is wrong because it reverses the role of a driven dimension — driven dimensions never "preserve" geometry; they passively reflect it. B is wrong because Fusion 360 doesn't suppress constraints based on mismatched displayed values; constraints and dimensions operate independently, and the Equal constraint remains active. D is wrong because a driven dimension cannot cause over-constraining. Over-constraining occurs when two driving inputs compete to control the same degree of freedom — a driven dimension contributes no constraint at all. A useful rule of thumb: driven dimensions are read-only reporters. On exam questions, any answer suggesting a driven dimension can lock, preserve, or conflict with geometry is a trap — only driving dimensions and constraints actively control sketch behavior.

Question 3

Two circles have their centers coincident with the horizontal sketch axis. Their centers are constrained symmetrically about the vertical sketch axis, and the circles have an Equal constraint. A driving dimension places the right center $30\,\mathrm{mm}$ from the vertical axis, and another driving dimension sets one circle's diameter to $12\,\mathrm{mm}$ .

How should a dimension between the two circle centers be applied, and what value should it report?

Apply it as driven; it should report $60\,\mathrm{mm}$ because symmetry fixes equal offsets on opposite sides. (correct answer)
Apply it as driving; it should report $30\,\mathrm{mm}$ because only the right-center offset controls the spacing.
Apply it as driven; it should report $48\,\mathrm{mm}$ after subtracting one circle diameter from each offset.
Apply it as driving; it should report $72\,\mathrm{mm}$ after adding both radii to the center offsets.

Explanation: Whenever you encounter a sketch with symmetry constraints and existing driving dimensions, your first task is to count the degrees of freedom. If every geometric property is already fully determined by existing constraints and dimensions, any additional dimension you add must be driven — it displays a value calculated from the geometry rather than controlling it. Here, the setup is fully defined: the right center sits

30\,\mathrm{mm}

from the vertical axis, and the symmetric constraint forces the left center to mirror it at

-30\,\mathrm{mm}

. The distance between the two centers is therefore

30 + 30 = 60\,\mathrm{mm}

. Adding a dimension between those centers as driving would conflict with the existing constraints — Fusion 360 would reject it or flag an over-constrained sketch. Instead, you apply it as driven, and it correctly reports

60\,\mathrm{mm}

. Answer A is correct. Answer B is wrong on two counts: it claims the dimension is driving (it can't be — the sketch is already fully constrained) and reports

30\,\mathrm{mm}

, which is only the offset of one center, not the full center-to-center distance. Answer C incorrectly subtracts the circle diameter from the offsets. The distance between centers has nothing to do with radius or diameter — those describe circle size, not center position. Answer D compounds two errors: it calls the dimension driving (incorrect) and adds radii to the center offsets, confusing center-to-center distance with an edge-to-edge or overlap measurement. Study tip: In Fusion 360, always check whether the sketch is already fully constrained before deciding driven vs. driving — a fully constrained sketch turns black, and any extra dimension must be driven.

Question 4

A line's midpoint is constrained coincident with the sketch origin. A driving dimension sets the line length to $40\,\mathrm{mm}$ , and a driving angular dimension sets its direction to $30^\circ$ relative to the horizontal sketch axis.

Which assessment of the line is correct?

The line is fully constrained because its midpoint fixes translation, its length fixes size, and its angle fixes rotation. (correct answer)
The line remains under-constrained because at least one endpoint must be coincident with fixed geometry.
The line is over-constrained because fixing its midpoint and length also determines its angular orientation.
The line remains under-constrained because an angular dimension can report rotation but cannot drive it.

Explanation: When evaluating whether a sketch element is fully constrained in Fusion 360, ask yourself three questions: Is translation fixed? Is size fixed? Is rotation fixed? A sketch entity needs all three degrees of freedom removed to be fully constrained. For this line, the midpoint constrained coincident with the sketch origin removes all translational freedom — the line cannot slide left, right, up, or down. The driving dimension of

40\,\mathrm{mm}

fixes the line's length, removing size freedom. The driving angular dimension of

30°

from the horizontal axis fixes rotational freedom. All three degrees of freedom are eliminated, making answer A correct. A midpoint constraint is a perfectly valid way to anchor a line's position; it doesn't require an endpoint to touch fixed geometry. B is wrong because it assumes only endpoint constraints can fix translation. In reality, any coincident constraint to fixed geometry — including one applied to the midpoint — fully removes translational freedom. There is no rule requiring an endpoint specifically. C describes a misconception about over-constraint: fixing a midpoint and length does not determine angular orientation. A

40\,\mathrm{mm}

line centered at the origin could still rotate freely to any angle without the angular dimension. Length and position are independent of rotation. D mischaracterizes how angular dimensions work. A driving angular dimension does actively constrain rotation — it is not merely a reference or reporting value. A driven (reference) dimension reports without constraining, but the question explicitly states this is a driving dimension. Study tip: Always categorize constraints by which degree of freedom they remove — translation, size, or rotation — and verify all three are covered before concluding a sketch is fully constrained.

Question 5

A fully constrained rectangular plate has its left edge fixed at the sketch origin and a driving width of $100\,\mathrm{mm}$ . A hole center is dimensioned $20\,\mathrm{mm}$ horizontally from the plate's right edge. A driven dimension from the left edge to the hole center reports $80\,\mathrm{mm}$ .

If the plate width changes to $120\,\mathrm{mm}$ while the right-edge offset remains driving, what should the driven dimension report?

It should report $80\,\mathrm{mm}$ because a driven dimension retains its original measured value after an edit.
It should report $140\,\mathrm{mm}$ because the right-edge offset is added to the new plate width.
It should report $120\,\mathrm{mm}$ because the hole center follows the full width of the plate.
It should report $100\,\mathrm{mm}$ because the hole remains $20\,\mathrm{mm}$ from the moving right edge. (correct answer)

Explanation: Whenever you see a question mixing driving and driven dimensions in Fusion 360, your first move should be to identify which dimension controls geometry and which one simply reports a measurement. A driving dimension actively constrains the sketch — changing it moves geometry. A driven dimension (shown in gray) is read-only; it passively reflects whatever the geometry currently measures. In this sketch, the hole center is positioned by a driving

20\,\mathrm{mm}

offset from the right edge. That means as the right edge moves, the hole moves with it, always staying

20\,\mathrm{mm}

away. When the plate width increases from

100\,\mathrm{mm}

120\,\mathrm{mm}

, the right edge shifts

20\,\mathrm{mm}

to the right, and the hole center follows it. The driven dimension — measuring from the left edge to the hole center — then recalculates:

120 - 20 = 100\,\mathrm{mm}

. So D is correct. A is wrong because a driven dimension never "freezes" at its old value — it updates live whenever the geometry changes. That's the whole point of a driven dimension. B is wrong because it adds the offset to the width (

120 + 20 = 140

), which has no geometric meaning here. The offset subtracts from the right edge, it doesn't stack on top of the total width. C is wrong because the hole center doesn't sit at the full plate width; it sits

20\,\mathrm{mm}

inward from the right edge, so reporting

120\,\mathrm{mm}

would place it on the edge itself. Study tip: Always ask "what is this dimension anchored to?" Driving dimensions own the anchor; driven dimensions just measure the result. Trace the constraint chain before predicting any value change.

Question 6

An imported sketch reports that it is fully constrained because every entity has a Fix/UnFix constraint. The profile must now be made parametrically editable so future design changes can be entered through dimensions.

Which workflow best preserves design intent while allowing dimensions to drive the profile?

Leave all Fix constraints in place and add driven dimensions so their displayed values can later be edited.
Delete every constraint, add driven dimensions for all measured values, and rely on the closed profile for stability.
Leave all Fix constraints in place and add driving dimensions, accepting any resulting over-constraint warnings.
Remove appropriate Fix constraints, apply geometric constraints, and add only the necessary driving dimensions. (correct answer)

Explanation: When working with imported sketches in Fusion 360, you need to understand the difference between Fix/UnFix constraints and proper parametric constraints. A Fix constraint pins geometry to absolute coordinates in space — it "freezes" entities rather than defining relationships between them. While this makes a sketch fully constrained, it creates a rigid, position-dependent profile that cannot adapt when dimensions change. True parametric design intent means geometry is driven by relationships and dimensions, not locked coordinates. The correct approach, D, is to remove the Fix constraints strategically, replace them with geometric constraints (such as coincident, parallel, perpendicular, or tangent), and then add only the driving dimensions needed to fully constrain the sketch. This preserves design intent because dimensions can be changed later and the geometry will update predictably based on its relationships — exactly what parametric modeling is for. Option A fails because driven dimensions are read-only display values; they report measurements but cannot drive geometry. Editing their values does nothing to the sketch, making this approach useless for future design changes. Option B is dangerous because deleting every constraint removes all geometric relationships, leaving only a closed profile loop with no stability. The profile may appear closed but will be under-constrained and unpredictably shift during edits. Option C creates over-constrained sketches, which Fusion 360 will flag as errors. Competing constraints — Fix positions alongside driving dimensions — conflict with each other and prevent reliable parametric updates. A useful rule of thumb: Fix constraints are a temporary workaround, not a parametric solution. On exam questions about parametric design, always look for the answer that combines geometric constraints with driving dimensions.

Question 7

A fully constrained vertical line passes through the sketch origin. A circle lies to the right of the line. Its center is constrained to the horizontal sketch axis, the circle is tangent to the vertical line, and a driving diameter dimension is $30\,\mathrm{mm}$ .

Which statement correctly describes a horizontal dimension of $15\,\mathrm{mm}$ from the vertical line to the circle center?

It is required as a driving dimension because tangency controls contact but does not locate the circle center.
It is redundant as a driving dimension and should be driven if the offset needs to be displayed. (correct answer)
It must replace the diameter dimension because center offsets and diameter dimensions cannot drive the same circle.
It remains independent because a tangent circle's center may move horizontally without changing its diameter.

Explanation: When working with sketch constraints in Fusion 360, you need to understand the difference between driving and driven dimensions, and how geometric constraints interact with dimensional ones. Here, the circle already has three pieces of information fully defining its position and size: (1) its center is locked to the horizontal axis, (2) it is tangent to the vertical line, and (3) its diameter is

30\,\mathrm{mm}

. Because the circle is tangent to the vertical line and has a radius of

15\,\mathrm{mm}

, its center must sit exactly

15\,\mathrm{mm}

to the right of that line — this is not a free variable. The sketch is already fully constrained without any horizontal dimension. This confirms B is correct. Adding a

15\,\mathrm{mm}

horizontal dimension would be redundant as a driving dimension — the geometry already enforces it. If you want to display that offset value, you can add it as a driven (reference) dimension, which reads the existing geometry without controlling it. A is wrong because tangency does locate the center — given a fixed radius and a fixed line, tangency forces the center to be exactly one radius away from the line. No additional locating dimension is needed. C is wrong because driving dimensions for size (diameter) and position (offset) are not mutually exclusive — they control different properties and coexist routinely in fully constrained sketches. D is wrong because the center cannot move horizontally without breaking the tangency constraint, so it is not independent. Study tip: When evaluating whether a dimension is redundant, count all active constraints — geometric constraints like tangency carry positional information just as dimensional ones do.

Question 8

A square profile is fully constrained. A circle inside its upper-right corner is tangent to both the top edge and the right edge. Its diameter is controlled by a driving dimension of $20\,\mathrm{mm}$ .

If the diameter dimension is deleted while both Tangent constraints remain, what is the resulting condition of the circle?

It remains fully constrained because tangency to two perpendicular edges uniquely determines both center and diameter.
It becomes under-constrained by a size degree of freedom because its radius may change while preserving both tangencies. (correct answer)
It becomes under-constrained by two translation degrees of freedom because tangent constraints do not locate a circle center.
It becomes over-constrained because the two Tangent constraints attempt to define incompatible circle diameters.

Explanation: When working with geometric constraints in Fusion 360 sketches, always ask yourself: how many degrees of freedom (DOF) does each constraint remove, and how many remain? A circle in 2D has three DOF: two for position (x, y of center) and one for size (radius). Tangency between a circle and a line means the circle's center sits exactly one radius away from that line — this removes one DOF per constraint. With two tangent constraints (top edge and right edge), the circle's center is locked relative to its own radius: the center must sit at coordinates

(r, -r)

from the corner. However, r itself remains free. As the radius grows or shrinks, the center simply slides diagonally, maintaining both tangencies perfectly. That's one remaining DOF — size — which is exactly what the deleted diameter dimension was controlling. So removing it leaves the circle under-constrained by one size degree of freedom, confirming B is correct. A is wrong because two tangencies to perpendicular edges do not pin the radius — they only link the center's position to whatever the radius happens to be. Size is never determined by tangency alone. C is wrong because the two tangent constraints do effectively locate the center (relative to radius), leaving zero translation DOF. The unconstrained DOF is size, not position. D is wrong because both tangency constraints are completely compatible — they impose no conflicting size requirements. Over-constraint would require redundant constraints defining the same DOF simultaneously. Study tip: In Fusion 360 sketches, tangency removes positional DOF, not size DOF. Always check whether a dimension is separately controlling size when a constraint question involves circles.

Question 9

A sketch contains four lines forming a rectangle. Adjacent endpoints are coincident, the top and bottom lines are horizontal, and the side lines are vertical. One corner is coincident with the sketch origin. Driving dimensions set the width to $80\,\mathrm{mm}$ and the height to $40\,\mathrm{mm}$ . A driven dimension reports the diagonal length.

What happens to the sketch's constraint status if the driven diagonal dimension is deleted?

It remains fully constrained because the diagonal dimension only reported a value determined by the existing geometry. (correct answer)
It becomes under-constrained because deleting the diagonal dimension allows the rectangle to shear while preserving its width and height.
It becomes under-constrained because every fully constrained closed profile must retain dimensions for both sides and one diagonal.
It becomes over-constrained because the origin coincidence now conflicts with the width and height dimensions on the rectangle.

Explanation: When working with sketch constraints in Fusion 360, the key question to ask is: does this element add information that determines geometry, or does it merely report geometry already determined by other constraints? That distinction separates driving dimensions from driven dimensions. In this rectangle, the geometry is already fully pinned down: the origin coincidence fixes one corner's position, horizontal and vertical constraints fix the orientation of all four lines, the coincident constraints at adjacent corners close the shape, and the two driving dimensions fix the width at

80\,\mathrm{mm}

and height at

40\,\mathrm{mm}

. The diagonal length is therefore completely determined by the Pythagorean theorem —

\sqrt{80^2 + 40^2} \approx 89.44\,\mathrm{mm}

— before the driven dimension even exists. A driven dimension is a read-only reporter; it consumes zero degrees of freedom. Deleting it changes nothing about what the sketch knows, so the sketch remains fully constrained. That confirms A is correct. B is wrong because deleting a driven dimension cannot introduce freedom — shearing would require removing an actual constraint like the vertical or horizontal line conditions. C is a false rule; no such requirement exists that a closed profile must include a diagonal dimension. Only degrees of freedom matter, not a checklist of dimension types. D gets the direction of over-constraint backward and invents a conflict that doesn't exist — the origin coincidence and the width/height dimensions are entirely compatible and were already coexisting before the driven dimension was added. The study tip: always identify whether a dimension is driving or driven before predicting its effect on constraint status. Driven dimensions are display-only and are invisible to the constraint solver.

Question 10

A center-point arc has its center coincident with the sketch origin. Its radius is controlled by a driving dimension of $25\,\mathrm{mm}$ . The arc's start point is constrained to the positive horizontal sketch axis, but its end point remains free to move around the circle.

Which additional dimension most directly removes the arc's remaining degree of freedom without redundantly defining its radius?

A second radius dimension applied between the center and the arc's unconstrained end point.
A horizontal distance dimension from the origin to the arc's already constrained start point.
A driving angle dimension defining the included angle from the start radius to the end radius. (correct answer)
A driven arc-length dimension that reports the current distance between the arc's two end points.

Explanation: When working with sketch constraints in Fusion 360, think in terms of degrees of freedom (DOF). Every sketch element starts with some number of free movements, and your goal is to fully constrain it using dimensions and constraints — without over-defining it. An arc has four DOFs: center position (x, y), radius, and angular span. Here, the center is fixed at the origin (eliminating x and y), the radius is set to

25\,\mathrm{mm}

(eliminating the radial DOF), and the start point is locked to the positive horizontal axis (fixing the start angle). That leaves exactly one DOF: where the end point sits angularly around the circle. A driving angle dimension — the included angle from the start radius to the end radius — directly captures that single remaining freedom, fully constraining the arc. That makes C the correct answer. A is wrong because the distance from the center to the end point is the radius, which is already defined as

25\,\mathrm{mm}

. Applying it again would create a redundant, over-defined constraint — Fusion 360 would flag it as a conflict. B is wrong because the start point is already constrained to the horizontal axis at radius

25\,\mathrm{mm}

, meaning its position is fully determined. Dimensioning it again adds no new information and is also redundant. D is wrong because a driven dimension is computed from existing geometry — it reports a value but does not drive or constrain anything. It cannot remove a DOF. Study tip: Always count DOFs before adding a dimension. Ask yourself, "What motion is still possible?" — then pick the dimension that directly eliminates only that motion.