Answer key · Unit 7

Geometry.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 5

Lesson 1 of 5 · About 25 minutes

Angles and triangles

Angles on a straight line and across a crossing, angles between parallel lines, and why the angles of every triangle add up to 180 degrees.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Angles on a straight line
Practice · Calculate
Question
A ray leaves a point on a straight line and makes an angle of $65^\circ$ ⁠ with it. How many degrees are in the angle beside it, the one marked with a question mark?

Correct answer
115 ± 0.5
Setup, choices and feedback
The Scene at this step
Angle
65 °fixed
Angle A
60 °defaulthidden
Angle B
70 °defaulthidden
Given / Worked out
Given
Other settings
Mode/nodes/board/props/mode = "angles"
Other settings
Lines/nodes/board/props/lines = "straight"
Other settings
Unknown/nodes/board/props/unknown = "supplement"
Accepted within
± 0.5
Computed from the Scene at this step’s setting
Angles: 180° − turn, its neighbour on a straight line/nodes/board/outputs/supplement
Wrong answers it recognises
- 25 ± 0.5
  Misconception: Thinks angles on a straight line add up to 90°, as in a right angle.
  Feedback from the catalogue
  A straight line is half a turn, $180^\circ$ ⁠, not a right-angled corner. Two angles that together make a straight line add up to $180^\circ$ ⁠; it is two angles that make a right-angled corner that add up to $90^\circ$ ⁠.
Feedback for a right answer
$180^\circ - 65^\circ = 115^\circ$ ⁠. Two angles that together make a straight line always add up to $180^\circ$ ⁠. Choose Worked out under Look inside to see it.

Hint
Together the two angles fill the whole straight line, half a turn.

Worked solution
1. A straight line is half a turn: $180^\circ$ ⁠.
2. The given angle and the one beside it fill it together: $65^\circ + x = 180^\circ$ ⁠.
3. So $x = 180^\circ - 65^\circ = 115^\circ$ ⁠.
If you got $25^\circ$ ⁠, you made it up to $90^\circ$ ⁠, like the corner of a page. A straight line is two such corners, $180^\circ$ ⁠.
Step 2 of 6
Open this step
Vertically opposite angles
Practice · Predict, then look
Question
Two straight lines cross, and one of the angles they make is $40^\circ$ ⁠. How big is the angle opposite it, across the crossing point?

Correct answer
- Correct $40^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Angle
40 °fixed
Angle A
60 °defaulthidden
Angle B
70 °defaulthidden
Given / Worked out
Givenfixed
Other settings
Mode/nodes/board/props/mode = "angles"
Other settings
Lines/nodes/board/props/lines = "crossing"
Other settings
Unknown/nodes/board/props/unknown = "opposite"
Choices
- Correct $40^\circ$ ⁠
- Wrong $140^\circ$ ⁠
  Misconception: Thinks vertically opposite angles add up to 180° instead of being equal.
  Feedback from the catalogue
  An angle and its neighbour on the same straight line add up to $180^\circ$ ⁠. The angle opposite it, across the crossing point, is equal to it: both make up the same neighbour to $180^\circ$ ⁠.
- Wrong $50^\circ$ ⁠
  Misconception: Thinks angles on a straight line add up to 90°, as in a right angle.
  Feedback from the catalogue
  A straight line is half a turn, $180^\circ$ ⁠, not a right-angled corner. Two angles that together make a straight line add up to $180^\circ$ ⁠; it is two angles that make a right-angled corner that add up to $90^\circ$ ⁠.
- Wrong $320^\circ$ ⁠
  Feedback
  $320^\circ$ ⁠ is the rest of the full turn around the point: the other three angles together, not one of them.
Once the prediction is locked in, the Scene changes
Given / Worked out
Worked outfixed
The Scene then shows: 40 · Angles: the vertically opposite angle, equal to turn/nodes/board/outputs/opposite
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Vertically opposite angles are equal: $40^\circ$ ⁠. Each of them sits on a straight line with the same $140^\circ$ ⁠ angle, so both are $180^\circ - 140^\circ$ ⁠.

Worked solution
1. The $40^\circ$ ⁠ angle and the one beside it sit on a straight line: $180^\circ - 40^\circ = 140^\circ$ ⁠.
2. The angle opposite the $40^\circ$ ⁠ sits beside the same $140^\circ$ ⁠ angle, on the other line.
3. So it too is $180^\circ - 140^\circ = 40^\circ$ ⁠: vertically opposite angles are equal.
If you chose $140^\circ$ ⁠, that is the angle beside the $40^\circ$ ⁠, not the one opposite it.
Step 3 of 6
Open this step
Parallel lines
Practice · Calculate
Question
Two parallel lines are cut by a third straight line. An angle of $70^\circ$ ⁠ is marked at the lower line. How big is the angle marked with a question mark at the upper line, on the other side of the transversal?

Correct answer
70 ± 0.5
Setup, choices and feedback
The Scene at this step
Angle
70 °fixed
Angle A
60 °defaulthidden
Angle B
70 °defaulthidden
Given / Worked out
Given
Other settings
Mode/nodes/board/props/mode = "angles"
Other settings
Lines/nodes/board/props/lines = "parallel"
Other settings
Pair/nodes/board/props/pair = "alternate"
Other settings
Unknown/nodes/board/props/unknown = "alternate"
Accepted within
± 0.5
Computed from the Scene at this step’s setting
Angles: the alternate angle at the second parallel line/nodes/board/outputs/alternate
Wrong answers it recognises
- 110 ± 0.5
  Misconception: Thinks alternate angles between parallel lines add up to 180° instead of being equal.
  Feedback from the catalogue
  Alternate angles, on opposite sides of the transversal and between the two parallel lines, are equal. The ones that add up to $180^\circ$ ⁠ are the co-interior angles, on the same side of the transversal.
Feedback for a right answer
$70^\circ$ ⁠. Alternate angles between parallel lines are equal. Choose Worked out under Look inside: the lower angle slides up the transversal and turns half a turn onto the marked one.

Hint
In your mind, slide the $70^\circ$ ⁠ angle up the transversal to the upper line without turning it. What sits opposite it there?

Worked solution
1. The lines are parallel, so the transversal crosses both at the same slant: the corresponding angle at the upper line, on the same side and facing the same way, is also $70^\circ$ ⁠.
2. The marked angle is opposite that corresponding angle, across the crossing point: they are vertically opposite.
3. Vertically opposite angles are equal, so the marked angle is $70^\circ$ ⁠.
If you got $110^\circ$ ⁠, you made the two add up to $180^\circ$ ⁠ as if they sat side by side on one line. They are on opposite sides of the transversal, and equal.
Step 4 of 6
Open this step
The angles of a triangle
Practice · Predict, then look
Question
Tear the three corners off the triangle and lay them side by side, their points together. How many degrees will they make together?

Correct answer
- Correct $180^\circ$ ⁠: a straight line
Setup, choices and feedback
The Scene at this step
Angle
60 °defaulthidden
Angle A
50 °fixed
Angle B
60 °fixed
Given / Worked out
Givenfixed
Other settings
Unknown/nodes/board/props/unknown = "angleC"
Choices
- Correct $180^\circ$ ⁠: a straight line
- Wrong $360^\circ$ ⁠: a full turn
  Misconception: Thinks the angles of a triangle add up to 360°.
  Feedback from the catalogue
  $360^\circ$ ⁠ is a full turn, or the angles of a quadrilateral. The three corners of a triangle line up into a straight line, half a turn: $180^\circ$ ⁠.
- Wrong It depends on the triangle’s size and shape
  Misconception: Thinks the angle sum of a triangle depends on its size or shape.
  Feedback from the catalogue
  A bigger triangle seems to hold “more angle”, but an angle measures an opening, not a length. In every triangle, long or narrow, the corners meet in a straight line: $180^\circ$ ⁠.
- Wrong $90^\circ$ ⁠: a right angle
  Feedback
  Two of this triangle’s angles on their own, $50^\circ$ ⁠ and $60^\circ$ ⁠, already make more than $90^\circ$ ⁠.
Once the prediction is locked in, the Scene changes
Given / Worked out
Worked outfixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The three corners meet in a straight line: $50^\circ + 60^\circ + 70^\circ = 180^\circ$ ⁠. It works in every triangle, long or narrow, and in the next step you change an angle yourself.

Worked solution
1. Through the vertex $C$ ⁠ runs a straight line parallel to the base $AB$ ⁠.
2. The corner at $A$ ⁠ slides along the side $AC$ ⁠ to $C$ ⁠ without turning, and fits between the parallel line and $AC$ ⁠ carried on: corresponding angles.
3. The corner at $B$ ⁠ slides along $BC$ ⁠ in the same way.
4. The corner at $C$ ⁠ turns half a turn and fits exactly between them: vertically opposite angles.
5. The three corners fill the straight line, so $A + B + C = 180^\circ$ ⁠, in every triangle.
If you chose $360^\circ$ ⁠, that is a full turn around a point, or the angles of a quadrilateral. In a triangle they fill only half a turn.
Step 5 of 6
Open this step
A right triangle
Practice · Reach the target
Question
Angle A is locked at $35^\circ$ ⁠. Set Angle B so that the angle at $C$ ⁠ is a right angle, $90^\circ$ ⁠, then press Check.

Goal
- 90 ± 0.5
  At the start of the step: 75
Setup, choices and feedback
The Scene at this step
Angle
60 °defaulthidden
Angle A
35 °fixed
Angle B
70 °
Given / Worked out
Givenfixed
Other settings
Unknown/nodes/board/props/unknown = "angleC"
Read from the Scene at
Triangle: 180° − A − B (zero or less when there is no triangle)/nodes/board/outputs/angleC
One setting that reaches it
Angle B
55 °
Feedback for a right answer
$35^\circ + 55^\circ + 90^\circ = 180^\circ$ ⁠. In a right triangle the other two angles always add up to $90^\circ$ ⁠.

Feedback for a wrong answer
Not yet. The angle at $C$ ⁠ is what is left of $180^\circ$ ⁠ after $A$ ⁠ and $B$ ⁠.

Hint
The right angle takes $90^\circ$ ⁠ of the $180^\circ$ ⁠. How much is left for $A$ ⁠ and $B$ ⁠ together?

Worked solution
1. The angles of a triangle add up to $180^\circ$ ⁠, and the angle at $C$ ⁠ must be $90^\circ$ ⁠.
2. That leaves $180^\circ - 90^\circ = 90^\circ$ ⁠ for $A$ ⁠ and $B$ ⁠ together.
3. $A = 35^\circ$ ⁠, so $B = 90^\circ - 35^\circ = 55^\circ$ ⁠.
4. Set Angle B to $55^\circ$ ⁠. Check: $35^\circ + 55^\circ + 90^\circ = 180^\circ$ ⁠.
Step 6 of 6
Open this step
The exterior angle
Practice · Calculate
Question
In the triangle $A = 45^\circ$ ⁠ and $B = 75^\circ$ ⁠. The side $AB$ ⁠ is carried on past $B$ ⁠. How big is the exterior angle this makes at $B$ ⁠, between the extension and the side $BC$ ⁠?

Correct answer
105 ± 0.5
Setup, choices and feedback
The Scene at this step
Angle
60 °defaulthidden
Angle A
45 °fixed
Angle B
75 °fixed
Given / Worked out
Given
Other settings
Exterior/nodes/board/props/exterior = true
Other settings
Unknown/nodes/board/props/unknown = "exteriorB"
Accepted within
± 0.5
Computed from the Scene at this step’s setting
Triangle: the exterior angle at B, 180° − B = A + C/nodes/board/outputs/exteriorB
Wrong answers it recognises
- 75 ± 0.5
  Misconception: Takes the exterior angle to be the interior angle beside it.
  Feedback from the catalogue
  The exterior angle lies outside the triangle, between a side and the next side carried on. It sits on a straight line with the interior angle, so it equals $180^\circ$ ⁠ minus that angle, and also the sum of the other two interior angles.
Feedback for a right answer
$180^\circ - 75^\circ = 105^\circ$ ⁠, which is exactly $45^\circ + 60^\circ$ ⁠: the two interior angles that are not next to it. Choose Worked out under Look inside: the corners at $A$ ⁠ and $C$ ⁠ fill the exterior angle.

Hint
The exterior angle and the angle $B$ ⁠ sit together on a straight line.

Worked solution
1. The exterior angle and the angle $B$ ⁠ sit side by side on the straight line: $180^\circ - 75^\circ = 105^\circ$ ⁠.
2. Another way: the third angle is $C = 180^\circ - 45^\circ - 75^\circ = 60^\circ$ ⁠.
3. The exterior angle equals the sum of the two interior angles far from it: $45^\circ + 60^\circ = 105^\circ$ ⁠.
If you answered $75^\circ$ ⁠, you took the interior angle $B$ ⁠ itself. The exterior angle lies outside the triangle, between the side $BC$ ⁠ and $AB$ ⁠ carried on.

Lesson 2 of 5 · About 25 minutes

Area

Area counts squares: a rectangle, then a parallelogram, a triangle and a trapezoid cut and rearranged, and a circle unrolled into a strip.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Counting squares
Practice · Reach the target
Question
Set Base and Height so that the rectangle covers exactly $24$ ⁠ squares, then press Check.

Goal
- 24 ± 0.01
  At the start of the step: 15
Setup, choices and feedback
The Scene at this step
Base
5
Height
3
Radius
3defaulthidden
As drawn / Rearranged
Rearranged
Read from the Scene at
Triangle and similar: the triangle's area; area: the shape's; NaN otherwise/nodes/board/outputs/area
One setting that reaches it
Base
8
Height
3
Feedback for a right answer
$24$ ⁠ squares: for example $6 \times 4$ ⁠, $8 \times 3$ ⁠ or $4 \times 6$ ⁠. Each square is one unit of area, and a rectangle covers base times height of them.

Feedback for a wrong answer
Not $24$ ⁠ yet. Drag Look inside and count the rows.

Hint
A rectangle is equal rows of squares. Which two numbers multiply to $24$ ⁠?

Worked solution
1. The number of rows in a rectangle is its height, and each row holds as many squares as its base.
2. So a rectangle’s area is base times height.
3. You need two numbers whose product is $24$ ⁠: for example $6 \times 4 = 24$ ⁠.
4. Set Base to $6$ ⁠ and Height to $4$ ⁠ (or $8$ ⁠ and $3$ ⁠, or $4$ ⁠ and $6$ ⁠).
Step 2 of 6
Open this step
Perimeter and area
Practice · Calculate
Question
The rectangle’s base is $7$ ⁠ and its height $3$ ⁠. What is its perimeter, the length of the line around it?

Correct answer
20 ± 0.01
Setup, choices and feedback
The Scene at this step
Base
7fixed
Height
3fixed
Radius
3defaulthidden
As drawn / Rearranged
As drawn
Other settings
Unknown/nodes/board/props/unknown = "perimeter"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Triangle and similar: a + b + c; area: the shape's perimeter (a circle's circumference); NaN otherwise/nodes/board/outputs/perimeter
Wrong answers it recognises
- 21 ± 0.01
  Misconception: Confuses perimeter and area.
  Feedback from the catalogue
  The perimeter is a length: how much fence goes round. The area counts squares: how much grass goes inside. A $7 \times 3$ ⁠ rectangle has perimeter $20$ ⁠ and area $21$ ⁠.
Feedback for a right answer
$7 + 3 + 7 + 3 = 2 \times (7 + 3) = 20$ ⁠. The perimeter is a length, while the area $7 \times 3 = 21$ ⁠ counts squares: two different things.

Hint
Walk around it: base, height, base, height.

Worked solution
1. The perimeter is the whole way round: two bases and two heights.
2. $7 + 3 + 7 + 3 = 20$ ⁠, or in short $2 \times (7 + 3) = 20$ ⁠.
If you got $21$ ⁠, you worked out the area, $7 \times 3$ ⁠: how many squares are inside, not how long the line around them is.
Step 3 of 6
Open this step
A parallelogram
Practice · Predict, then look
Question
The parallelogram has base $6$ ⁠ and height $4$ ⁠. Its slanted side is longer than the height, about $4.47$ ⁠. What is its area?

Correct answer
- Correct $24$ ⁠: base times height
Setup, choices and feedback
The Scene at this step
Base
6fixed
Height
4fixed
Radius
3defaulthidden
As drawn / Rearranged
As drawnfixed
Other settings
Shape/nodes/board/props/shape = "parallelogram"
Other settings
Offset/nodes/board/props/offset = 2
Other settings
Unknown/nodes/board/props/unknown = "area"
Choices
- Correct $24$ ⁠: base times height
- Wrong About $26.8$ ⁠: base times the slanted side
  Misconception: Multiplies the base by the slanted side instead of the height.
  Feedback from the catalogue
  The height is the distance at right angles from the base to the opposite side, like the dashed line, not the length of the slanted side. A parallelogram leaning further gets lower and its area shrinks, even with sides of the same length.
- Wrong $12$ ⁠: half of base times height
  Feedback
  The half belongs to a triangle. The parallelogram is covered all over, like a rectangle.
Once the prediction is locked in, the Scene changes
As drawn / Rearranged
Rearrangedfixed
The Scene then shows: 24 · Triangle and similar: the triangle's area; area: the shape's; NaN otherwise/nodes/board/outputs/area
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Cut off the triangle left of the height line and move it to the right-hand side: the parallelogram becomes a $6 \times 4$ ⁠ rectangle, with nothing added or lost. The area is $24$ ⁠.

Worked solution
1. The dashed line is the height: the distance at right angles from the base to the side opposite it, $4$ ⁠.
2. Cut along it the triangle on the left and move it right, exactly the length of the base, $6$ ⁠.
3. You get a rectangle with base $6$ ⁠ and height $4$ ⁠.
4. So the area is $6 \times 4 = 24$ ⁠.
If you chose $26.8$ ⁠, you multiplied by the slanted side. The height is always measured at right angles to the base, not along the side.
Step 4 of 6
Open this step
A triangle
Practice · Calculate
Question
The triangle has base $8$ ⁠ and height $5$ ⁠. What is its area?

Correct answer
20 ± 0.01
Setup, choices and feedback
The Scene at this step
Base
8fixed
Height
5fixed
Radius
3defaulthidden
As drawn / Rearranged
As drawn
Other settings
Shape/nodes/board/props/shape = "triangle"
Other settings
Offset/nodes/board/props/offset = 3
Other settings
Unknown/nodes/board/props/unknown = "area"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Triangle and similar: the triangle's area; area: the shape's; NaN otherwise/nodes/board/outputs/area
Wrong answers it recognises
- 40 ± 0.01
  Misconception: Leaves out the ½ in the area of a triangle or trapezoid (or a triangular base).
  Feedback from the catalogue
  A triangle is half a parallelogram: two copies of it, one upside down, fill one. So a triangle’s area is base times height divided by $2$ ⁠. A trapezoid too is half a parallelogram whose base is its two bases added.
Feedback for a right answer
$8 \times 5 \div 2 = 20$ ⁠. Two copies of the triangle make a parallelogram with base $8$ ⁠ and height $5$ ⁠, area $40$ ⁠, and the triangle is half of it.

Hint
Two such triangles, one of them upside down, make a parallelogram. Drag Look inside.

Worked solution
1. Turn a copy of the triangle half a turn about the middle of its right-hand side.
2. The triangle and the copy make a parallelogram: base $8$ ⁠, height $5$ ⁠, area $8 \times 5 = 40$ ⁠.
3. The triangle is half of the parallelogram: $40 \div 2 = 20$ ⁠.
If you got $40$ ⁠, you forgot to divide by $2$ ⁠: that is the area of two triangles.
Step 5 of 6
Open this step
A trapezoid
Practice · Calculate
Question
The trapezoid has two parallel bases, $8$ ⁠ at the bottom and $4$ ⁠ at the top, and the height between them is $3$ ⁠. What is its area?

Correct answer
18 ± 0.01
Setup, choices and feedback
The Scene at this step
Base
8fixed
Height
3fixed
Radius
3defaulthidden
As drawn / Rearranged
As drawn
Other settings
Shape/nodes/board/props/shape = "trapezoid"
Other settings
Top/nodes/board/props/top = 4
Other settings
Offset/nodes/board/props/offset = 1
Other settings
Unknown/nodes/board/props/unknown = "area"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Triangle and similar: the triangle's area; area: the shape's; NaN otherwise/nodes/board/outputs/area
Wrong answers it recognises
- 36 ± 0.01
  Misconception: Leaves out the ½ in the area of a triangle or trapezoid (or a triangular base).
  Feedback from the catalogue
  A triangle is half a parallelogram: two copies of it, one upside down, fill one. So a triangle’s area is base times height divided by $2$ ⁠. A trapezoid too is half a parallelogram whose base is its two bases added.
Feedback for a right answer
$(8 + 4) \times 3 \div 2 = 18$ ⁠. The trapezoid and an upside-down copy make a parallelogram with base $8 + 4 = 12$ ⁠, and the trapezoid is half of it.

Hint
Here too an upside-down copy completes a parallelogram. What is its base?

Worked solution
1. Turn a copy of the trapezoid half a turn about the middle of its right-hand side.
2. Together they make a parallelogram with base $8 + 4 = 12$ ⁠ and height $3$ ⁠: area $12 \times 3 = 36$ ⁠.
3. The trapezoid is half of it: $36 \div 2 = 18$ ⁠.
If you got $36$ ⁠, you forgot to divide by $2$ ⁠. The two bases are added, not multiplied.
Step 6 of 6
Open this step
A circle
Practice · Calculate
Question
The circle’s radius is $3$ ⁠. What is its area? Round to two decimal places.

Correct answer
28.27 ± 0.05
Setup, choices and feedback
The Scene at this step
Base
6defaulthidden
Height
4defaulthidden
Radius
3fixed
As drawn / Rearranged
As drawn
Other settings
Shape/nodes/board/props/shape = "circle"
Other settings
Unknown/nodes/board/props/unknown = "area"
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Triangle and similar: the triangle's area; area: the shape's; NaN otherwise/nodes/board/outputs/area
Wrong answers it recognises
- 18.85 ± 0.05
  Misconception: Uses the circumference 2πr for the area of a circle.
  Feedback from the catalogue
  $2\pi r$ ⁠ is the circumference, the length of the line round the circle. The area is $\pi r^2$ ⁠: unroll the circle into a strip half the circumference wide, $\pi r$ ⁠, and $r$ ⁠ tall.
- 113.1 ± 0.1
  Misconception: Puts the diameter where the formula needs the radius.
  Feedback from the catalogue
  The radius is the distance from the centre to the edge, half the diameter. In $\pi r^2$ ⁠ it is the radius that is squared: with the diameter the answer is $4$ ⁠ times too big.
Feedback for a right answer
$\pi \times 3^2 = 9\pi \approx 28.27$ ⁠. The slices make almost a rectangle $\pi r$ ⁠ wide and $r$ ⁠ tall, so the circle’s area is $\pi r^2$ ⁠.

Hint
Drag Look inside: the slices line up into a shape whose width is half the circumference, $\pi r$ ⁠, and whose height is the radius $r$ ⁠.

Worked solution
1. Cut the circle into thin slices and lay them alternately, one point down and the next point up.
2. You get almost a rectangle: its height is the radius, $r = 3$ ⁠, and its width half the circumference, $\pi r = 3\pi$ ⁠.
3. The area: $\pi r \times r = \pi r^2 = 9\pi \approx 28.27$ ⁠.
If you got $18.85$ ⁠, you worked out the circumference, $2\pi r$ ⁠, which is a length. If you got $113.1$ ⁠, you used the diameter $6$ ⁠ in place of the radius.

Lesson 3 of 5 · About 25 minutes

Pythagoras' theorem

In a right triangle the squares on the two legs exactly fill the square on the hypotenuse: how to find a missing side, and how to tell whether an angle is right.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Squares on the sides
Practice · Predict, then look
Question
A square stands on each side of the right triangle. The squares on $a$ ⁠ and $b$ ⁠ are cut into pieces and slid into the square on $c$ ⁠. What will happen?

Correct answer
- Correct They fill it exactly
Setup, choices and feedback
The Scene at this step
Side a
3fixed
Side b
4fixed
Corner angle
90 °fixed
Squares / Rearranged
Squaresfixed
Choices
- Correct They fill it exactly
- Wrong Part of it stays empty
  Feedback
  Count the squares: $9 + 16 = 25$ ⁠, exactly the area of the square on $c$ ⁠.
- Wrong They are too big and overlap
  Feedback
  The pieces fit without overlapping. That only happens when the angle between $a$ ⁠ and $b$ ⁠ is not a right angle.
Once the prediction is locked in, the Scene changes
Squares / Rearranged
Rearrangedfixed
The Scene then shows: 25 · Pythagoras: c², equal to a² + b² only at a right angle/nodes/board/outputs/squareC
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Exactly: $a^2 + b^2 = 9 + 16 = 25 = c^2$ ⁠. This is Pythagoras’ theorem: in a right triangle, the areas of the squares on the two legs add up to the area of the square on the hypotenuse.

Worked solution
1. The square on $a$ ⁠ has area $a^2 = 3^2 = 9$ ⁠, and the one on $b$ ⁠ has $b^2 = 4^2 = 16$ ⁠.
2. Together: $9 + 16 = 25$ ⁠.
3. The square on $c$ ⁠ has area $c^2 = 25$ ⁠, so the pieces fill it exactly.
4. In a right triangle always $a^2 + b^2 = c^2$ ⁠, where $c$ ⁠ is the hypotenuse, the side opposite the right angle.
Step 2 of 5
Open this step
Finding the hypotenuse
Practice · Calculate
Question
The legs are $a = 6$ ⁠ and $b = 8$ ⁠. How long is the hypotenuse $c$ ⁠?

Correct answer
10 ± 0.01
Setup, choices and feedback
The Scene at this step
Side a
6fixed
Side b
8fixed
Corner angle
90 °fixed
Squares / Rearranged
Squares
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Pythagoras: the side c opposite the corner, by the law of cosines/nodes/board/outputs/sideC
Wrong answers it recognises
- 14 ± 0.01
  Misconception: Thinks the hypotenuse is the sum of the two legs.
  Feedback from the catalogue
  The straight way is shorter than going round the corner: the hypotenuse is always shorter than the two legs together. It is the squares of the sides that add, $a^2 + b^2 = c^2$ ⁠, and then you take the square root.
- 100 ± 0.01
  Misconception: Gives c² as the length of the hypotenuse.
  Feedback from the catalogue
  $a^2 + b^2$ ⁠ is the area of the square on the hypotenuse, not its length. The length is the side of that square: the square root of the area.
Feedback for a right answer
$c^2 = 36 + 64 = 100$ ⁠, so $c = \sqrt{100} = 10$ ⁠.

Hint
The board shows $c^2$ ⁠. Which number squared gives it?

Worked solution
1. By Pythagoras’ theorem $c^2 = a^2 + b^2$ ⁠.
2. Substitute: $c^2 = 6^2 + 8^2 = 36 + 64 = 100$ ⁠.
3. $c$ ⁠ is the number whose square is $100$ ⁠: $c = \sqrt{100} = 10$ ⁠.
If you answered $14$ ⁠, you added the sides themselves, $6 + 8$ ⁠. If you answered $100$ ⁠, you stopped at $c^2$ ⁠ and forgot the square root.
Step 3 of 5
Open this step
When the root is not whole
Practice · Calculate
Question
The legs are $a = 2$ ⁠ and $b = 3$ ⁠. How long is the hypotenuse $c$ ⁠? Round to two decimal places.

Correct answer
3.606 ± 0.01
Setup, choices and feedback
The Scene at this step
Side a
2fixed
Side b
3fixed
Corner angle
90 °fixed
Squares / Rearranged
Squares
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Pythagoras: the side c opposite the corner, by the law of cosines/nodes/board/outputs/sideC
Wrong answers it recognises
- 5 ± 0.01
  Misconception: Thinks the hypotenuse is the sum of the two legs.
  Feedback from the catalogue
  The straight way is shorter than going round the corner: the hypotenuse is always shorter than the two legs together. It is the squares of the sides that add, $a^2 + b^2 = c^2$ ⁠, and then you take the square root.
- 13 ± 0.01
  Misconception: Gives c² as the length of the hypotenuse.
  Feedback from the catalogue
  $a^2 + b^2$ ⁠ is the area of the square on the hypotenuse, not its length. The length is the side of that square: the square root of the area.
Feedback for a right answer
$c = \sqrt{13} \approx 3.61$ ⁠. Most hypotenuses are not whole numbers; $3$ ⁠, $4$ ⁠, $5$ ⁠ is a special case.

Hint
$c^2 = 4 + 9$ ⁠. Its square root lies between $3$ ⁠ and $4$ ⁠.

Worked solution
1. $c^2 = 2^2 + 3^2 = 4 + 9 = 13$ ⁠.
2. $c = \sqrt{13}$ ⁠. Check: $3^2 = 9$ ⁠ and $4^2 = 16$ ⁠, so $c$ ⁠ lies between $3$ ⁠ and $4$ ⁠.
3. On a calculator: $\sqrt{13} \approx 3.61$ ⁠.
If you answered $13$ ⁠, that is $c^2$ ⁠, not $c$ ⁠. If you answered $5$ ⁠, you added $2 + 3$ ⁠: the hypotenuse is always shorter than the two legs put together.
Step 4 of 5
Open this step
Finding a leg
Practice · Reach the target
Question
Side a is locked at $5$ ⁠. Set Side b so that the hypotenuse is $c = 13$ ⁠, that is $c^2 = 169$ ⁠, then press Check.

Goal
- 13 ± 0.01
  At the start of the step: 6.403
Setup, choices and feedback
The Scene at this step
Side a
5fixed
Side b
4
Corner angle
90 °fixed
Squares / Rearranged
Squares
Read from the Scene at
Pythagoras: the side c opposite the corner, by the law of cosines/nodes/board/outputs/sideC
One setting that reaches it
Side b
12
Feedback for a right answer
$b = 12$ ⁠: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ ⁠. To find a leg you subtract: $b^2 = c^2 - a^2$ ⁠.

Feedback for a wrong answer
Not yet. Watch the sum at the top of the board: $c^2$ ⁠ has to reach $169$ ⁠.

Hint
$b^2 = c^2 - a^2$ ⁠. What is that, and what is its square root?

Worked solution
1. By Pythagoras $a^2 + b^2 = c^2$ ⁠, so $b^2 = c^2 - a^2$ ⁠.
2. Substitute: $b^2 = 13^2 - 5^2 = 169 - 25 = 144$ ⁠.
3. $b = \sqrt{144} = 12$ ⁠.
4. Set Side b to $12$ ⁠. Check: $25 + 144 = 169$ ⁠.
Step 5 of 5
Open this step
Is the angle right?
Practice · Choose
Question
In a triangle the sides $5$ ⁠ and $12$ ⁠ meet at a corner, and the third side is $14$ ⁠. Is the angle between them a right angle?

Correct answer
- Correct No, it is obtuse, more than $90^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Side a
5fixed
Side b
12fixed
Corner angle
90 °
Squares / Rearranged
Squares
Choices
Shown in a random order.
- Correct No, it is obtuse, more than $90^\circ$ ⁠
- Wrong Yes, every triangle has $a^2 + b^2 = c^2$ ⁠
  Misconception: Thinks a² + b² = c² holds in every triangle.
  Feedback from the catalogue
  $a^2 + b^2 = c^2$ ⁠ holds only when the angle between $a$ ⁠ and $b$ ⁠ is a right angle. When it is wider, $c^2$ ⁠ is more than $a^2 + b^2$ ⁠; when it is sharper, less. Try Corner angle on the board.
- Wrong No, it is acute, less than $90^\circ$ ⁠
  Feedback
  A sharper corner brings the ends of the sides $5$ ⁠ and $12$ ⁠ closer together, and the third side shrinks below $13$ ⁠. Here it is longer.
Feedback for a right answer
At a right angle the third side would be $\sqrt{25 + 144} = 13$ ⁠. It is $14$ ⁠, longer, so the corner is wider than $90^\circ$ ⁠: $14^2 = 196 > 169$ ⁠. Set Corner angle to about $103^\circ$ ⁠ and you will see $c^2 \approx 196$ ⁠.

Hint
If the angle were right, how long would the third side be? Try it on the board.

Worked solution
1. If the angle were right: $c^2 = 5^2 + 12^2 = 25 + 144 = 169$ ⁠, so $c = 13$ ⁠.
2. Here $c = 14$ ⁠, and $c^2 = 196$ ⁠, more than $169$ ⁠.
3. A wider corner pushes the ends of the sides apart and $c$ ⁠ grows; a sharper one brings them together and $c$ ⁠ shrinks.
4. So the angle is obtuse. Pythagoras’ theorem also works backwards: $a^2 + b^2 = c^2$ ⁠ exactly only when the angle is right.
If you chose “yes”, you assumed the theorem holds in every triangle. It holds only in a right triangle.

Lesson 4 of 5 · About 20 minutes

Similarity

Similar triangles: the same angles, every side multiplied by the same number, and the area by its square. And at the end, measuring a tree by its shadow.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
The same shape at another size
Practice · Predict, then look
Question
A copy of the triangle will appear on the right, with every side twice as long. What will its angles be?

Correct answer
- Correct The same angles as the original
Setup, choices and feedback
The Scene at this step
Scale factor
2fixed
Side AB
4fixed
Angle A
50 °fixed
Original / Scaled copy
Originalfixed
Other settings
Show/nodes/board/props/show = "both"
Choices
- Correct The same angles as the original
- Wrong Doubled, like the sides
  Misconception: Thinks enlarging a triangle enlarges its angles.
  Feedback from the catalogue
  An enlargement multiplies distances but leaves directions alone, and an angle is the opening between two directions. Similar triangles have equal angles; only the sides are multiplied.
- Wrong Smaller, because the triangle is stretched
  Feedback
  The triangle is not stretched one way: it grows in the same ratio in every direction, so the shape stays the same shape.
Once the prediction is locked in, the Scene changes
Original / Scaled copy
Scaled copyfixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The same angles: $50^\circ$ ⁠, $60^\circ$ ⁠ and $70^\circ$ ⁠. Similar triangles are the same shape at another size: equal angles, and every side multiplied by the same number, the scale factor.

Worked solution
1. Enlarging by $2$ ⁠ doubles every distance but does not change the direction of any side.
2. An angle is the difference between the directions of two sides, so it does not change.
3. The copy is a triangle with the same angles, $50^\circ$ ⁠, $60^\circ$ ⁠ and $70^\circ$ ⁠, and doubled sides.
If you chose “doubled”, notice that doubled angles would add up to $360^\circ$ ⁠, while the angles of every triangle add up to $180^\circ$ ⁠.
Step 2 of 5
Open this step
Multiplying the sides
Practice · Calculate
Question
In the original triangle $AB = 4$ ⁠ and $BC = 3$ ⁠. In the similar copy $A'B' = 6$ ⁠. How long is $B'C'$ ⁠?

Correct answer
4.5 ± 0.01
Setup, choices and feedback
The Scene at this step
Scale factor
1.5fixed
Side AB
4fixed
Angle A
36.87 °fixed
Original / Scaled copy
Original
Other settings
Angle b/nodes/board/props/angleB = 90
Other settings
Unknown/nodes/board/props/unknown = "scaledSideA"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Similar: k · BC/nodes/board/outputs/scaledSideA
Wrong answers it recognises
- 5 ± 0.01
  Misconception: Adds the same amount to every side instead of multiplying by the scale factor.
  Feedback from the catalogue
  In similar figures every side is multiplied by the same number, the scale factor. Adding the same length to every side changes the shape: a short side grows relatively more than a long one.
Feedback for a right answer
The scale factor is $k = 6 \div 4 = 1.5$ ⁠, so $B'C' = 1.5 \times 3 = 4.5$ ⁠.

Hint
By what factor did $AB$ ⁠ grow? Every side grows by the same factor.

Worked solution
1. The scale factor is the ratio of matching sides: $k = \frac{A'B'}{AB} = \frac{6}{4} = 1.5$ ⁠.
2. Every side of the copy is $k$ ⁠ times as long: $B'C' = k \times BC = 1.5 \times 3 = 4.5$ ⁠.
3. Check: $\frac{4.5}{3} = 1.5$ ⁠, the same ratio.
If you answered $5$ ⁠, you added $2$ ⁠ to every side, as $AB$ ⁠ grew from $4$ ⁠ to $6$ ⁠. Similar figures multiply, they do not add: a short side grows less than a long one.
Step 3 of 5
Open this step
Finding the scale factor
Practice · Reach the target
Question
In the original triangle $AB = 4$ ⁠. Set the Scale factor so that $A'B' = 10$ ⁠ in the copy, then press Check.

Goal
- 10 ± 0.01
  At the start of the step: 4
Setup, choices and feedback
The Scene at this step
Scale factor
1
Side AB
4fixed
Angle A
50 °fixed
Original / Scaled copy
Scaled copy
Read from the Scene at
Similar: k · AB/nodes/board/outputs/scaledSideC
One setting that reaches it
Scale factor
2.5
Feedback for a right answer
$k = 10 \div 4 = 2.5$ ⁠. The scale factor is a side of the copy divided by the matching side of the original.

Feedback for a wrong answer
Not yet. $A'B'$ ⁠ is $k$ ⁠ times $4$ ⁠.

Hint
$4 \times k = 10$ ⁠. By what factor?

Worked solution
1. In the copy every side is $k$ ⁠ times as long: $A'B' = k \times AB$ ⁠.
2. $10 = k \times 4$ ⁠, so $k = \frac{10}{4} = 2.5$ ⁠.
3. Set the Scale factor to $2.5$ ⁠.
Step 4 of 5
Open this step
The area grows faster
Practice · Predict, then look
Question
The original triangle’s area is $6$ ⁠, and every side of the copy is twice as long. By what factor does the area grow?

Correct answer
- Correct By $4$ ⁠
Setup, choices and feedback
The Scene at this step
Scale factor
2fixed
Side AB
4fixed
Angle A
36.87 °fixed
Original / Scaled copy
Originalfixed
Other settings
Angle b/nodes/board/props/angleB = 90
Other settings
Area readout/nodes/board/props/areaReadout = true
Other settings
Unknown/nodes/board/props/unknown = "scaledArea"
Choices
- Wrong By $2$ ⁠
  Misconception: Thinks area grows by the scale factor k rather than k².
  Feedback from the catalogue
  Area spreads in two directions, and an enlargement by $k$ ⁠ stretches both: the base by $k$ ⁠ and the height by $k$ ⁠. So the area grows by $k^2$ ⁠: by $4$ ⁠ when you double.
- Correct By $4$ ⁠
- Wrong By $8$ ⁠
  Feedback
  Volume, measured in three directions, grows by $8 = 2^3$ ⁠. Area is measured in two.
Once the prediction is locked in, the Scene changes
Original / Scaled copy
Scaled copyfixed
The Scene then shows: 4 · Similar: k²/nodes/board/outputs/areaRatio
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
By $4 = 2^2$ ⁠: the area of $6$ ⁠ becomes $24$ ⁠. Both the base and the height grow by $2$ ⁠, so the area grows by $2 \times 2$ ⁠.

Worked solution
1. A triangle’s area is base times height divided by $2$ ⁠.
2. In the copy the base is twice as long and the height twice as tall.
3. So the area grows by $2 \times 2 = 2^2 = 4$ ⁠: from $6$ ⁠ to $24$ ⁠.
4. In general, at a scale factor $k$ ⁠ the area grows by $k^2$ ⁠.
If you chose $2$ ⁠, you doubled only one direction. Area spreads in two directions, and both grow.
Step 5 of 5
Open this step
By the shadow
Practice · Calculate
Question
A pole $2$ ⁠ metres tall casts a shadow $3$ ⁠ metres long. At the same moment a tree casts a shadow $12$ ⁠ metres long. The sun’s rays meet the ground at the same angle, so the triangles are similar. How tall is the tree, in metres?

Correct answer
8 ± 0.01
Setup, choices and feedback
The Scene at this step
Scale factor
4fixed
Side AB
3fixed
Angle A
33.69 °fixed
Original / Scaled copy
Original
Other settings
Angle b/nodes/board/props/angleB = 90
Other settings
Unknown/nodes/board/props/unknown = "scaledSideA"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Similar: k · BC/nodes/board/outputs/scaledSideA
Wrong answers it recognises
- 11 ± 0.01
  Misconception: Adds the same amount to every side instead of multiplying by the scale factor.
  Feedback from the catalogue
  In similar figures every side is multiplied by the same number, the scale factor. Adding the same length to every side changes the shape: a short side grows relatively more than a long one.
Feedback for a right answer
The shadow is $12 \div 3 = 4$ ⁠ times as long, so the tree is also $4$ ⁠ times as tall as the pole: $4 \times 2 = 8$ ⁠ metres.

Hint
How many times longer is the tree’s shadow than the pole’s?

Worked solution
1. In the small triangle the shadow is $AB = 3$ ⁠ and the pole $BC = 2$ ⁠. In the large one the shadow is $A'B' = 12$ ⁠ and the tree $B'C'$ ⁠.
2. The scale factor: $k = \frac{12}{3} = 4$ ⁠.
3. The tree’s height: $B'C' = 4 \times 2 = 8$ ⁠ metres.
If you answered $11$ ⁠, you added to the height the $9$ ⁠ metres the shadow grew by. In similar figures every length is multiplied by the same number.

Lesson 5 of 5 · About 20 minutes

Volume

Volume counts unit cubes: the area of the base times the height, for a box, a prism and a cylinder, and why a cube twice as big holds eight times as much.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Counting cubes
Practice · Calculate
Question
A box is $4$ ⁠ long, $3$ ⁠ wide and $2$ ⁠ tall. How many unit cubes fill it?

Correct answer
24 ± 0.01
Setup, choices and feedback
The Scene at this step
Length
4fixed
Width
3fixed
Height
2fixed
Base / Built up
Base
Other settings
Unknown/nodes/board/props/unknown = "volume"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Solid: base area × height/nodes/board/outputs/volume
Wrong answers it recognises
- 9 ± 0.01
  Misconception: Adds the length, width and height to get the volume.
  Feedback from the catalogue
  Volume counts cubes: one layer covers the base, length times width cubes, and there are as many layers as the height. So you multiply the three measurements, not add them.
- 52 ± 0.01
  Misconception: Gives the surface area (or the curved surface) for the volume.
  Feedback from the catalogue
  The surface area is the wrapping, the cardboard or paper round the outside, counted in squares. The volume is what is inside, counted in cubes: base area times height.
Feedback for a right answer
One layer holds $4 \times 3 = 12$ ⁠ cubes, and there are two layers: $12 \times 2 = 24$ ⁠. That is the volume: the number of unit cubes that fill the solid.

Hint
How many cubes are in the bottom layer, and how many such layers are there?

Worked solution
1. The bottom layer covers the base: $4 \times 3 = 12$ ⁠ cubes.
2. The height is $2$ ⁠, so there are two such layers.
3. The volume: $12 \times 2 = 24$ ⁠ unit cubes.
If you answered $9$ ⁠, you added the three measurements, $4 + 3 + 2$ ⁠. If you answered $52$ ⁠, you worked out the surface area, the cardboard that wraps the box, not what is inside it.
Step 2 of 5
Open this step
Layer on layer
Practice · Reach the target
Question
The box’s base is $4 \times 3$ ⁠. Set Height so that the volume is $36$ ⁠, then press Check.

Goal
- 36 ± 0.01
  At the start of the step: 12
Setup, choices and feedback
The Scene at this step
Length
4fixed
Width
3fixed
Height
1
Base / Built up
Basefixed
Other settings
Unknown/nodes/board/props/unknown = "volume"
Read from the Scene at
Solid: base area × height/nodes/board/outputs/volume
One setting that reaches it
Height
3
Feedback for a right answer
$12 \times 3 = 36$ ⁠: three layers of $12$ ⁠. Volume is the area of the base times the height.

Feedback for a wrong answer
Not $36$ ⁠ yet. How many layers of $12$ ⁠ do you need?

Hint
Each layer holds $12$ ⁠ cubes. How many layers make $36$ ⁠?

Worked solution
1. The base area: $4 \times 3 = 12$ ⁠, that is $12$ ⁠ cubes in each layer.
2. Volume = base area $\times$ ⁠ height, so $12 \times h = 36$ ⁠.
3. $h = 36 \div 12 = 3$ ⁠.
4. Set Height to $3$ ⁠.
Step 3 of 5
Open this step
A prism
Practice · Calculate
Question
The prism’s base is a right triangle with legs $4$ ⁠ and $3$ ⁠, and the prism is $5$ ⁠ tall. What is its volume?

Correct answer
30 ± 0.01
Setup, choices and feedback
The Scene at this step
Length
4fixed
Width
3fixed
Height
5fixed
Base / Built up
Base
Other settings
Solid/nodes/board/props/solid = "prism"
Other settings
Unknown/nodes/board/props/unknown = "volume"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Solid: base area × height/nodes/board/outputs/volume
Wrong answers it recognises
- 60 ± 0.01
  Misconception: Leaves out the ½ in the area of a triangle or trapezoid (or a triangular base).
  Feedback from the catalogue
  A triangle is half a parallelogram: two copies of it, one upside down, fill one. So a triangle’s area is base times height divided by $2$ ⁠. A trapezoid too is half a parallelogram whose base is its two bases added.
Feedback for a right answer
The base area is $4 \times 3 \div 2 = 6$ ⁠, and the volume $6 \times 5 = 30$ ⁠. The same rule as for a box: base area times height.

Hint
First the base area: a triangle, half of base times height.

Worked solution
1. The base is a triangle: $4 \times 3 \div 2 = 6$ ⁠.
2. Each layer $1$ ⁠ tall holds a volume of $6$ ⁠.
3. There are $5$ ⁠ layers: $6 \times 5 = 30$ ⁠.
If you got $60$ ⁠, you forgot to halve the base area: that is the volume of a whole box, $4 \times 3 \times 5$ ⁠, and the prism is half of it.
Step 4 of 5
Open this step
A cylinder
Practice · Calculate
Question
The cylinder has radius $3$ ⁠ and height $4$ ⁠. What is its volume? Round to one decimal place.

Correct answer
113.1 ± 0.1
Setup, choices and feedback
The Scene at this step
Length
4defaulthidden
Width
3defaulthidden
Height
4fixed
Base / Built up
Base
Other settings
Solid/nodes/board/props/solid = "cylinder"
Other settings
Radius/nodes/board/props/radius = 3
Other settings
Unknown/nodes/board/props/unknown = "volume"
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Solid: base area × height/nodes/board/outputs/volume
Wrong answers it recognises
- 75.4 ± 0.1
  Misconception: Gives the surface area (or the curved surface) for the volume.
  Feedback from the catalogue
  The surface area is the wrapping, the cardboard or paper round the outside, counted in squares. The volume is what is inside, counted in cubes: base area times height.
- 452.4 ± 0.2
  Misconception: Puts the diameter where the formula needs the radius.
  Feedback from the catalogue
  The radius is the distance from the centre to the edge, half the diameter. In $\pi r^2$ ⁠ it is the radius that is squared: with the diameter the answer is $4$ ⁠ times too big.
Feedback for a right answer
The base area is $\pi \times 3^2 = 9\pi \approx 28.27$ ⁠, and the volume $9\pi \times 4 = 36\pi \approx 113.1$ ⁠.

Hint
The base is a circle: $\pi r^2$ ⁠. How many such layers are there?

Worked solution
1. The base area, a circle of radius $3$ ⁠: $\pi r^2 = 9\pi \approx 28.27$ ⁠.
2. The cylinder is $4$ ⁠ such layers: $9\pi \times 4 = 36\pi$ ⁠.
3. $36\pi \approx 113.1$ ⁠.
If you got $75.4$ ⁠, you worked out the curved surface, $2\pi r h$ ⁠, the sheet that rolls round the cylinder. If you got $452.4$ ⁠, you used the diameter $6$ ⁠ in place of the radius.
Step 5 of 5
Open this step
A cube twice as big
Practice · Predict, then look
Question
A cube with edge $1$ ⁠ holds one unit cube. Double every edge, to $2$ ⁠. How many unit cubes will the new cube hold?

Correct answer
- Correct $8$ ⁠
Setup, choices and feedback
The Scene at this step
Length
2fixed
Width
2fixed
Height
2fixed
Base / Built up
Basefixed
Other settings
Unknown/nodes/board/props/unknown = "volume"
Choices
- Wrong $2$ ⁠
  Misconception: Thinks doubling every edge doubles the volume.
  Feedback from the catalogue
  Doubling every edge stretches three directions: length, width and height. Each of them doubles the number of cubes, so the volume grows by $2 \times 2 \times 2 = 8$ ⁠.
- Wrong $4$ ⁠
  Misconception: Thinks volume grows by k² when every length grows by k.
  Feedback from the catalogue
  $k^2$ ⁠ fits area, which spreads in two directions. Volume also has a height that grows too, so volume grows by $k^3$ ⁠: by $8$ ⁠ when you double.
- Correct $8$ ⁠
- Wrong $6$ ⁠
  Feedback
  $6$ ⁠ is the number of faces of a cube, not the number of cubes inside it.
Once the prediction is locked in, the Scene changes
Base / Built up
Built upfixed
The Scene then shows: 8 · Solid: base area × height/nodes/board/outputs/volume
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$2 \times 2 \times 2 = 2^3 = 8$ ⁠: two layers of $4$ ⁠. When every length grows by $k$ ⁠, area grows by $k^2$ ⁠ and volume by $k^3$ ⁠.

Worked solution
1. The new base is $2 \times 2 = 4$ ⁠ cubes.
2. The height is $2$ ⁠: two layers of $4$ ⁠.
3. $4 \times 2 = 8 = 2^3$ ⁠.
4. Volume grows by $k^3$ ⁠, because it is measured in three directions and each of them grows by $k$ ⁠.
If you chose $4$ ⁠, you doubled the base and forgot that the height doubled too. If you chose $2$ ⁠, you doubled one direction only.

Across the unit · 21 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Angles and triangles

Angles on a straight line

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Vertically opposite angles

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Parallel lines

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The angles of a triangle

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A right triangle

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The exterior angle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Area

Counting squares

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Perimeter and area

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A parallelogram

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A triangle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A trapezoid

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A circle