Area 4 square units, perimeter 8 units. Students who say 9 are counting pegs, not squares: a 2 × 2 square has 9 pegs on and inside it but only 4 unit squares. Turn on ▦ Grid to show the squares.

Area 4: half of the 4 × 2 rectangle that fits around it. Predict works with every measurement, so you can make any of them a prediction first.

It stays 6 (½ × 4 × 3) wherever the top corner is on that row, because the base and the height stay the same. Even when the triangle leans far over and becomes obtuse, the height is still 3.

From (3, 3): Square; Kite at (4, 4); Arrowhead at (0, 0); Right-angled trapezium at (2, 3), (3, 2), (3, 4) and (4, 3); Quadrilateral at (2, 4) and (4, 2); and a right-angled triangle at (2, 2), (1, 4) and (4, 1), where the corner lands in line with a side so the shape has only three corners. Many other pegs make the band cross itself.

Reflect copies it in the vertical line through the centre peg, so the image touches the original along the edge from (2, 0) to (2, 1). ⟲ 90° turns a copy a quarter turn anticlockwise about the centre peg (2, 2), so the L lands on its side along the bottom of the board. Copies always come in the next colour, so the original is easy to see.

Only 3 × 4 (perimeter 14) and 2 × 6 (perimeter 16) fit: 1 × 12 needs 13 pegs in a row. For a fixed area, the closer to a square, the smaller the perimeter.

The box is 4 × 3 = 12 and the triangle is exactly half of it, 6. A right-angled triangle with shorter sides 5 and 3 has area 7.5. Halving the rectangle is where ½ × base × height comes from.

All three are possible: (0, 0), (2, 0), (0, 2) has area 2, (0, 0), (3, 0), (0, 2) has area 3 and (0, 0), (2, 1), (1, 3) has area 2.5. On a square grid every area is a whole number or a half: Pick’s theorem shows why. The largest triangle on 5 × 5 has area 8, half of the whole 4 × 4 board.

The corner triangles are 1.5, 1.5 and 4, so the outside is 7 and the triangle is 12 − 7 = 5. This “surround and subtract” method works for any shape on a grid.

It is a square with sides √5. Area: box 9 − four triangles of 1 = 5, and (√5)² = 5. Each side goes 2 along and 1 up, so its length squared is 2² + 1² = 5: Pythagoras appearing on its own.

Eight areas: 1, 2, 4, 5, 8, 9, 10 and 16. A square whose side goes a along and b up has area a² + b², and here a + b can be at most 4. 3, 6, 7, 11, 12, 13, 14 and 15 are missing: 13 = 2² + 3² needs a 6 × 6 board, and 3, 6, 7, 11, 12, 14 and 15 are not the sum of two square numbers at all.

It is the hypotenuse of a right-angled triangle with shorter sides 2 and 1, so its length is √(2² + 1²) = √5 ≈ 2.24. Three along and one up is √10 ≈ 3.16. √ Exact keeps the surd; decimals are rounded.

5 exactly, since 3² + 4² = 25. Then √2, √8 = 2√2 and √10. The 3-4-5 triangle is worth remembering: it is the smallest right-angled triangle with three whole-number sides.

14 different lengths: 1, √2, 2, √5, 2√2, 3, √10, √13, 4, 3√2, √17, 2√5, 5 and 4√2. They are the square roots of a² + b² for 0 ≤ a ≤ b ≤ 4. Most lengths come from two pegs, (a, b) and (b, a), which are reflections of each other in the diagonal.

The sides are 4, 2√2 and 2√2. 4 + 4√2 ≈ 9.66 can’t be simplified, because 4 and 4√2 are unlike terms, just as 4 + 4x can’t. The decimal is only an approximation.

Only one length works: 5, from 5 along (or 5 up) and from 3 along, 4 up (or 4 along, 3 up). 25 is the only number up to 50 here that is the sum of two squares in two genuinely different ways (0² + 5² and 3² + 4²).

They always total 180°. The angles are rounded to 1 decimal place, so a hand-added total can come to 179.9° or 180.1°; the card beside the board shows the exact angle sum. Talk about why rounding causes this.

For example (0, 1), (2, 0), (3, 2): the sides from (2, 0) go (−2, 1) and (1, 2), which are perpendicular, so the tool calls it a right-angled isosceles triangle with sides √5, √5 and √10.

The right angle is at (2, 0). Proof by Pythagoras: the sides are √5, √5 and √10, and 5 + 5 = 10. Or by gradients: −½ and 2 multiply to −1. Turning one side through 90° turns (1, 2) into (−2, 1).

It is impossible on a square grid: every grid triangle has an area that is a whole number or a half, but an equilateral triangle with side √n has area (√3 ÷ 4) × n, which is never a whole number or a half. On the isometric board it is easy.

For example (0, 0), (2, 1), (1, 2) has sides √5, √5 and √2, with its line of symmetry along the diagonal y = x. Some students only accept isosceles triangles that “stand on” their odd side.

The rhombus that isn’t a square is usually hardest: all four sides must be equal but slanted, for example (0, 1), (2, 0), (4, 1), (2, 2) with sides √5. The tool names each shape by its most specific name, so a square is never called a rectangle.

The angles are 29.7°, 233.1°, 29.7° and 67.4° (to 1 decimal place), which total 360°. As the corner rises the reflex angle gets smaller: at (2, 2) it is exactly 270°.

The midpoints always make a parallelogram (Varignon’s theorem), whatever the red quadrilateral. Here its sides are 5 and √20. Its area is always half the red area: 22 against 44 here.

360°, 540°, 720°, and 1080° for an octagon: add 180° for every extra side, (n − 2) × 180°. Concave shapes follow the same rule once their reflex angles are counted.

Moving only one corner breaks a pair of parallel sides, so move two: slide the top edge sideways to (2, 3) and (6, 3). Area stays 8, since the base 4 and height 2 are unchanged. Parallelograms on the same base and between the same parallels have equal area.

Square: 4 lines, order 4. Rectangle and rhombus: 2 lines, order 2. Kite: 1 line, order 1. Parallelogram: no lines, but order 2. The parallelogram surprises students who think “symmetry” only means mirror lines.

A parallelogram such as (0, 0), (3, 0), (4, 2), (1, 2) has order 2 and no lines. Order 4 with no lines needs a pinwheel, for example (0, 0), (1, 1), (6, 0), (5, 1), (6, 6), (5, 5), (0, 6), (1, 5): one bent arm turned a quarter turn four times about the centre.

Reflecting in the y-axis changes the sign of x: (1, 1) → (−1, 1), (4, 1) → (−4, 1). Reflecting in the x-axis changes the sign of y. Students often mix up which axis is which, so ask them to predict before pressing.

(x, y) → (−y, x) for 90° anticlockwise: (4, 1) → (−1, 4). Two turns give (−x, −y), a half turn. ⟳ 90° (clockwise) gives (y, −x).

Congruent shapes have the same three side lengths (here 2, 3 and √13) and the same angles, wherever they are and whichever way round. A reflected copy is still congruent, which some students doubt.

Only an equilateral triangle has three lines of symmetry, and that is impossible on a square grid. On the isometric board it is easy, and so is a regular hexagon (6 lines, order 6).

I = 2 and B = 10, so 2 + 5 − 1 = 6, which is the area. This is Pick’s theorem: for any shape with corners on a square grid and no holes, Area = I + B/2 − 1. With all three on, the card shows the check.

I = 5 and B = 3, so the area is 5 + 1.5 − 1 = 5.5. With no horizontal sides, Pick is quicker than box it here (box 12 − 2 − 1.5 − 3 = 5.5).

With I = 0 the formula becomes A = B/2 − 1. A triangle with only its three corners on the grid (B = 3) has area ½, however long and thin it is, for example (0, 0), (1, 0), (5, 1).

For example the 2 × 2 square (I = 1, B = 8) and the 1 × 4 rectangle (I = 0, B = 10). Keeping the area fixed, each extra interior peg costs two boundary pegs, because A = I + B/2 − 1.

The whole board, area 16, with I = 9 and B = 16 (9 + 8 − 1 = 16). A zigzag such as (0, 0), (4, 0), (4, 1), (1, 1), (1, 2), (4, 2), (4, 3), (1, 3), (1, 4), (0, 4) has B = 22 but I = 0, so its area is only 10: big boundaries don’t mean big areas.

No: this hexagon has I = 2, B = 8 and area 10 small triangles. The rule on a triangular grid is A = 2I + B − 2 (4 + 8 − 2 = 10), because each small triangle is half of a square-grid unit. The card shows the check when Area, I and B are all on.

The corners are (1, 1), (4, 1) and (1, 3). The classic error is writing (y, x): ask “along the corridor, then up the stairs”.

The length is 7: from −3 to 4 is 4 − (−3) = 7. The student used 3 instead of −3. Counting the gaps between pegs on the board settles it.

Both are 5: 4 across and 3 up, so √(4² + 3²) = √25. The distance formula is Pythagoras on the differences in coordinates, and negative coordinates make no difference to the method.

Gradient = up ÷ along: 2/4 = 1/2. A vertical band has an “along” of 0, so up ÷ 0 is undefined. Students often think a horizontal band (m = 0) has no gradient.

Sides that meet at a right angle have gradients that multiply to −1: here 1/2 and −2. Parallel sides have equal gradients.

The midpoint is (0, 1), the mean of the x-coordinates and the mean of the y-coordinates. The full band is 10 long and the half band is 5.

Every 4th: equilateral triangle. Every 3rd: square. Every 2nd: regular hexagon. Every peg: regular dodecagon. These are the divisors of 12 (12 ÷ 3, 12 ÷ 4…). Every 5th gives a 12-pointed star that crosses itself, so the tool calls it a crossed shape.

It is always 90°: the angle in a semicircle is a right angle, one of the circle theorems. The other two angles change but always add up to 90°.

Every angle is a multiple of 15°: an angle at the edge is half the angle at the centre, and each gap between pegs is 30° at the centre. Pegs 1, 4 and 9 cut the circle into gaps of 3, 5 and 4, so the angles are 45°, 75° and 60°.

Opposite angles always add up to 180° (a cyclic quadrilateral). Here the angles are 105°, 75°, 75° and 105°, so each opposite pair makes 180°. Drag any corner to another peg: the angles change but the opposite pairs still add to 180°.

The angle at the centre (90°) is twice the angle at the edge (45°). Move the blue corner to any peg on the same side, and it stays 45°, because angles in the same segment are equal.

A side of n pegs makes (n − 1)² small triangles: 4, 9, 16 and 25 for 3, 4, 5 and 6 pegs. The square numbers appear because each row of the triangle holds the next odd number of small triangles: 1 + 3 + 5 + 7.

Side 1: 6 small triangles. Side 2: 24. Doubling the lengths multiplies the area by 4, for hexagons just as for squares.

For example: start at a peg, put the second corner 1½ along and one row up, and the third corner two rows straight above the first. Its sides are √3 and its area is 3 small triangles, so it is one third of the hexagon around it.

60° and 120° follow the grid lines. 90° and 30° need a band that goes one along and one up-right (length √3). 45° is impossible here, just as 60° is impossible on a square grid.

It is a square: four equal sides (2√2) and four right angles. A shape’s name doesn’t depend on which way up it is. Press ⟲ 90° or tilt your head: nothing about it changes.

No. The 1 × 6 rectangle has perimeter 14 and area 6; the 3 × 3 square has perimeter 12 and area 9. Area and perimeter measure different things, and either can grow while the other shrinks.

The area goes from 2 to 8: doubling the lengths multiplies the area by 2 × 2 = 4. The perimeter does double (6 to 12). The same happens for any shape.

They counted the 16 pegs on and inside the square. Area counts the 9 unit squares between the pegs. Pick’s theorem links the two: 4 interior + 12 ÷ 2 boundary − 1 = 9.

It is 3√2 ≈ 4.24. Each diagonal step is √2, not 1, because it is the hypotenuse of a 1 × 1 right-angled triangle. Counting steps only works for horizontal and vertical lines.

No mistake: the angles are 63.43…°, 63.43…° and 53.13…°, which do add to exactly 180°, but each is rounded to 1 decimal place. Rounding each value first loses a little, which is why you should round at the end.

The only other rectangle is 3 × 6 (area 18, perimeter 18). For a triangle, the right-angled triangle with sides 6, 8 and 10 has area 24 and perimeter 24. The 5-12-13 triangle (area and perimeter 30) needs a bigger board.

50 squares. Upright: 16 + 9 + 4 + 1 = 30. Tilted: 9 + 4 + 4 + 1 + 1 + 1 = 20 (a side going 1 along and 1 up fits in 9 places; 2 and 1, or 1 and 2, fit in 4 places each; 3 and 1, 1 and 3, and 2 and 2 fit in 1 place each). A good test of working systematically.

There are 29 different triangles on a 4 × 4 board (16 pegs). Two triangles are congruent when their three lengths match, so comparing exact lengths is the quickest test.

Three lines needs a shape like an equilateral triangle, which is possible on the isometric, triangle and circle boards (every 4th peg of 12) but never on a square grid. Four lines needs a square board or the circle board.

Good questions start with a prediction, need the tool to answer, and can be checked. The link carries the board, its bands and the measurements that are switched on.