When you mark every point that is exactly 4 cm from P and join them, they form a perfectly round circle of radius 4 cm, with P at the centre.

All points 4 cm from P form a circle

Since the wave is a half circle sitting exactly on the centre of AB, its diameter equals half of AB.

Radius to be set in the compass = 2 cm

Length of AX = 4 cm (X is the centre of the semicircle, the point where the compass tip is placed).

Here the same idea is used on a shorter central line, giving three neat waves instead of one full wave-pair:

Wave drawn on a differently sized central line

Both shallow waves become identical only when the two compass centres are placed at equal distances from the line, mirrored on opposite sides. A little trial and error with the compass radius gets both arcs to match:

Shallow, identical waves smaller than a half circle

Going around the square in order gives S → P → Q → R (or its reverse). Checking each option:

• SPQR — follows the order around the square ✔
• RSPQ — same cyclic order, starting from R ✔
• QRSP — same cyclic order, starting from Q ✔
• PQSR — jumps from Q straight to S, skipping across the square instead of along a side ✘

PQSR is not a valid name for the given square.

Draw the rectangle first, using its dot-to-dot corners as a guide. Then, leaving exactly one dot of gap diagonally from each corner of the rectangle, draw a small square in each of the four corner regions. Keeping the same gap and the same square size on all four corners is what makes the arrangement look symmetrical.

Checking side lengths and angles using the dot-grid positions (each figure’s corners can be traced as steps of dots, e.g. 3 dots across and 1 dot up):

• A — all four sides equal, all angles 90° → this is a square.
• B — sides are not all equal (a taller, thinner rhombus) → not a square.
• C — sides are not all equal → not a square.
• D — a small tilted rectangle, sides unequal → not a square.

Only A is a square.

Think — answer: Yes. Counting how many dot-steps each side moves across and up/down tells us the side lengths (using the same ‘steps’ pattern rotated 90° confirms right angles), so we can reason this out from the grid positions alone, without a ruler or protractor.

Three rotated quadrilaterals drawn with all corners on dots — each was verified to have equal opposite sides and right angles (for the rectangles) or all four equal sides and right angles (for the squares):

Three rotated squares/rectangles on a dot grid

Yes — a table with these three columns is exactly the shorthand needed: each row records one experiment instead of writing a full sentence.

Whenever X and Y are the same distance from A and B, the length XY stays constant and equal to AB, no matter what that equal distance is.

(i) \( XY = AB \) — the segment XY is always exactly as long as AB.

(ii) The 4-sided figure ABYX is a rectangle — because AX and BY are equal and parallel (both vertical), which automatically makes AB and XY equal and parallel too.

When X sits at A and Y sits at C (or X at D and Y at B), the segment XY becomes a diagonal of the rectangle. So the farthest distance between X and Y equals the length of diagonal AC (or equivalently diagonal BD, since both diagonals of a rectangle are equal).

For two identical squares, the plan is exactly the same idea, only simpler:

1. Draw a rough sketch: rectangle ABCD with a middle line splitting it into two identical squares ABEF and FECD (choose any convenient side length for the square, e.g. \(AF = 4\text{ cm}\)).
2. Since the two parts must be identical squares, the long side must be exactly twice the short side: if \(AF = 4\text{ cm}\), then \(AC = 8\text{ cm}\).
3. Draw AF = 4 cm and a perpendicular at A. Using the compass (not a ruler) opened to AF’s length, step off the same length twice along the base to mark the two square widths, then complete the rectangle with perpendiculars.

This gives a rectangle whose sides are in the ratio 2 : 1, split by a middle line into two identical squares.

• Cannot be divided into two identical squares: Length = 4 cm, Breadth = 2.5 cm  (since \(4 \ne 2 \times 2.5\)). Try other pairs where length ≠ 2 × breadth.
• Cannot be divided into three identical squares: Length = 7 cm, Breadth = 2 cm  (since \(7 \ne 3 \times 2\)). Try other pairs where length ≠ 3 × breadth.

The rectangle must be constructed with all four sides equal — in other words, the rectangle must actually be a square. Only then does each diagonal split the 90° corner angles into two equal 45° parts.

From measuring several rectangles, two patterns hold every time:

• The two diagonals of a rectangle are always equal in length.
• A diagonal always splits each pair of opposite angles into two equal pairs of smaller angles (e.g. \(c = h\) and \(d = g\) in the labelled figure) — and when the rectangle is a square, a diagonal splits every 90° angle exactly in half, into two 45° angles.

A few measured examples make the pattern look true, but true certainty needs a general geometric reason (a proof) that works for every rectangle, not just the ones we happened to measure — for instance, using the fact that a rectangle’s diagonals form pairs of congruent triangles. In Grade 6 we build this confidence by testing many different rectangles and checking the pattern never breaks; formal proofs come in later grades.

Step 1 — Head: Draw a circle of a convenient radius for the head, then draw a short straight ‘neck’ line below it.

Step 2 — Body: Draw the rectangle for the body first with straight sides, leaving the top open.

Step 3 — Curved shoulders: To curve the top inward, the compass tip is placed below the rectangle’s top edge — roughly near the centre of the rectangle, a little below the top corners — and an arc is swept between the two top corners. Placing the tip lower makes the curve bulge downward (concave), matching the shoulder shape shown.

A bit of trial with the compass tip position (try a few points below the top edge and see which arc passes closest to both top corners) gets the curve looking right.

Mark the midpoint X of AB. Open the compass to a radius equal to half of AX (i.e. a quarter of AB):

1. With centre at the midpoint of AX, draw a semicircle bulging above the line, from A to X.
2. With centre at the midpoint of XB, draw a semicircle bulging below the line, from X to B.

For \(AB = 8\text{ cm}\): \(AX = 4\text{ cm}\), so each semicircle has radius 2 cm. See the worked answers to the ‘Figure it Out’ questions above for the completed waves.

Each eye is made of two shallow arcs (like the ‘smaller than a half circle’ wave from Q3 above) meeting at sharp points on the left and right — this is exactly the supporting-curve technique used for the shoulders of ‘A Person’.

Hint: points A and B are where the compass tip is placed for the upper and lower curve

Two support points, A (above the eye’s centre line) and B (below it), are chosen at equal distances from the centre line. With the compass tip at A, draw the shallow upper arc between the two corner points; with the tip at B (same radius), draw the shallow lower arc between the same two corners. Because A and B are placed symmetrically (mirror images of each other about the centre line), the two arcs come out identical, giving a symmetric almond eye shape. A dot (or small filled circle) is added in the middle for the pupil.

Repeat the same construction a little to the side for the second eye.

Rectangle ABCD, 4 cm × 6 cm

\(\angle A = \angle B = \angle C = \angle D = 90^\circ\)  → satisfies R2 (all angles 90°).

\(AB = CD = 4\text{ cm}\) and \(AD = BC = 6\text{ cm}\)  → satisfies R1 (opposite sides equal).

Rectangle PQRS, 10 cm × 2 cm

\(\angle P = \angle Q = \angle R = \angle S = 90^\circ\)  → satisfies R2.

\(PQ = SR = 10\text{ cm}\) and \(PS = QR = 2\text{ cm}\)  → satisfies R1.

No. Once all four angles of a 4-sided figure are fixed at 90°, the figure is forced to close up only when each pair of opposite sides is equal — that is precisely what makes it a rectangle (or a square). There’s no way to keep all angles at 90° and have unequal opposite sides; the shape simply wouldn’t close into a proper 4-sided figure.

Draw a rough diagram first with all the short sides tick-marked as equal (as shown), which shows that the rectangle’s long side must be exactly 3 times its short side.

1. Choose any convenient length for the square’s side, say \(4\text{ cm}\), and draw the first square using a ruler and perpendiculars.
2. Using the compass opened to that same side length (no need to re-measure with the ruler), step off two more equal widths along the base line.
3. Draw perpendiculars up from each new base point, and join the tops to complete a rectangle of size \(12\text{ cm} \times 4\text{ cm}\), automatically divided into 3 identical squares by the vertical lines.

Hint from the book: take the length of the rectangle to be three times its breadth.

Since the inner square must reach the full height of the rectangle to touch the top and bottom edges (as shown in the figure), its side length must equal the rectangle’s shorter side:

Side of square = 4 cm (same as the rectangle’s breadth).

For the square to be centred, the leftover width \((8 – 4 = 4\text{ cm})\) must be split equally on both sides:

Gap on each side = 2 cm.

1. Draw the 8 cm × 4 cm rectangle.
2. On the top and bottom sides, mark points 2 cm in from the left edge and 2 cm in from the right edge.
3. Join the corresponding top and bottom marks with vertical lines to complete the centred 4 cm square.

Build the staircase one square at a time, always starting the next square from the midpoint of the previous square’s top (or bottom) edge, as shown by the alignment in the figure:

1. Draw the first (largest) square using a ruler and set-square for right angles — side 7 cm.
2. Find the midpoint of its top edge; from there, draw the next square of side 5 cm rising up and to the right.
3. Repeat from the midpoint of that square’s top edge for the smallest 3 cm square.

Keeping every new square’s side flush with the midpoint mark (not the corner) is what keeps the staircase looking aligned exactly as shown.

Choose a convenient side for the big square (e.g. 9 cm) so it divides evenly into 3 equal smaller squares of 3 cm each.

1. Construct the large square (9 cm side) using perpendiculars.
2. Divide it into a 3 × 3 grid of 3 cm squares by marking off equal steps with the compass along each side and drawing parallel grid lines.
3. In each small square chosen for shading, draw one diagonal, then fill the triangular half with parallel hatching lines, matching the pattern shown (note that the shading direction alternates from square to square).

The centre of the circular hole must coincide with the centre of the square — the point where the square’s two diagonals cross.

1. Construct the square with your chosen side length.
2. Lightly draw both diagonals; their intersection point is the centre of the square.
3. Placing the compass tip exactly on that intersection point, draw a circle of any radius smaller than half the square’s side — this is the ‘hole’.

This combines the earlier ideas: dividing a square into 4 equal parts (like the rectangle-into-squares construction) and centring a circle in each part.

1. Construct the large square and divide it into 4 equal smaller squares using one horizontal and one vertical mid-line.
2. In each of the 4 small squares, draw its two diagonals to locate its own centre.
3. Draw a circle of the same small radius centred at each of those 4 points.

Each arc is a quarter-circle centred at one corner of the square, with radius equal to the square’s side length.

1. Construct the 8 cm square, label its corners.
2. Place the compass tip at one corner, open it to the full side length (8 cm), and draw an arc from one adjacent corner to the other (this arc bulges inward from the far side).
3. Repeat with the tip at each of the other 3 corners, using the same 8 cm radius each time.

Because all four arcs use the same radius (the side length) and are centred at the four corners, they bulge inward by exactly the same amount from each side, giving the symmetric pillow-like curve pattern shown.

1. Draw AB of any convenient (arbitrary) length.
2. Construct a perpendicular to AB at B (this is the line on which C will lie).
3. At A, draw a ray making a \(60^\circ\) angle with AB (using a protractor); this ray meets the perpendicular from B at point C.
4. Since \(\angle A = 90^\circ\) in a rectangle and one part is already \(60^\circ\), the remaining part is \(\angle DAC = 90^\circ – 60^\circ = 30^\circ\), matching the given split — draw a perpendicular to AB through A; D will lie on this line.
5. Method 1: draw a perpendicular to BC at C; it meets the vertical line through A at the fourth point D.
   Method 2: using the compass, mark D on the vertical line through A such that \(AD = BC\), then join CD.

This gives rectangle ABCD where diagonal AC splits \(\angle A\) into \(60^\circ\) and \(30^\circ\), and (by the equal-alternate-angle property explored on page 204) also splits the opposite \(\angle C\) into \(30^\circ\) and \(60^\circ\).

1. Draw the base \(DC = 5\text{ cm}\).
2. Construct a perpendicular line \(l\) to DC at point C.
3. The fourth point B must lie on line l and be exactly 7 cm from D — instead of guessing this point by trial and error with a ruler, draw an arc of radius 7 cm centred at D; where this arc crosses line \(l\) is exactly point B (every point on the arc is 7 cm from D, so their intersection guarantees both conditions at once).
4. Construct perpendiculars to DC (through D) and to BC (through B); where these two perpendiculars meet is the fourth point A.

Completed rectangle ABCD with DC = 5 cm and diagonal AC = 7 cm

Checking: \(\angle A = \angle B = \angle C = \angle D = 90^\circ\) and opposite sides are equal — ABCD satisfies both rectangle properties R1 and R2.

Rectangle ABCD with a 45°–45° diagonal split — all sides equal, tick-marked

Observation: when the diagonal splits each angle exactly in half (45° and 45°), all four sides turn out equal — the ‘rectangle’ is actually a square. This matches the earlier Explore answer: only a square has diagonals that bisect its angles equally.

The tricky part is the peaked roof: point A must be exactly 5 cm from both B and C at once — this is the same ‘equidistant point’ idea used throughout the chapter.

1. Draw the square base DE = 5 cm with the small 2 cm × 3 cm door rectangle marked inside it, and erect the two 5 cm walls DB and EC.
2. To locate the roof-peak A: with the compass opened to 5 cm, draw an arc centred at B (all points on it are 5 cm from B), then draw a second 5 cm arc centred at C (all points on it are 5 cm from C).
3. The point where the two arcs cross is the only point that is 5 cm from both B and C at once — that point is A.
4. Join A to B and A to C with straight lines to complete the roof’s triangular peak.
5. Finally, keeping the compass still at 5 cm, place the tip at A and draw the connecting arc from B to C — this is the curved underside of the roof.

The completed house:

The completed House construction

Yes — a rhombus. All four sides can be made equal while the angles are tilted away from 90° (two angles bigger, two smaller, always adding to 90°+90° pairs).

A rhombus — all four sides equal, but angles not 90°

It can be constructed using the same ‘two equidistant arcs’ method as the House roof: draw one side, then from each end swing an arc equal to the side length; the arcs’ intersection gives a third point that, together with a mirrored fourth point, completes a 4-sided figure with all sides equal but slanted angles — a rhombus, not a square.