Quadrilaterals

All four corners are 90°, and the diagonals cut the rectangle into 4 isosceles triangles (equal sides mark the tick marks). Since ∠ABD = 30° and ∠BAD = 90°:

∠ABD = 30°, ∠CAD = 60°, ∠ADB = 60°, ∠BDC = 30°, ∠ACD = 30°, ∠ACB = 60°. (Each triangle made by a diagonal is isosceles because the diagonals bisect each other and are equal in a rectangle — so the base angles match up in this neat criss-cross pattern!)

∠QOP is a linear pair with ∠POS, so:

Triangle OQP is isosceles (OQ = OP, half-diagonals), so its base angles share the leftover 180 − 70 = 110°, split evenly:

∠POS = 110°, ∠QOP = 70°, ∠ROS = 70°, ∠OQR = 35°, ∠ORQ = 35°, ∠OQP = 55°, ∠OPQ = 55°, ∠ORS = 55°, ∠OSR = 55°. (∠OQR and ∠ORQ come from triangle OQR, which has vertex angle ∠QOR = 110°, the same way as before but with the other pair of sides.)

Same steps for all four — only the angle changes each time! Remember from Section 1: any quadrilateral built this way (equal, bisecting diagonals) automatically comes out a rectangle.

Try all four angles — you'll see every single one still comes out looking like a rectangle (just "leaning" differently), because the diagonals are always 8 cm and always bisecting!

PL and AM are both diameters, so they're both radii-doubled — meaning OP = OL = OA = OM (all radii of the same circle), so PL = AM. That means the "diagonals" AM and PL of quadrilateral APML are equal in length. They also cross at the centre O, which is the midpoint of both (since diameters always pass through the centre) — so they bisect each other. And since PL ⟂ AM (given), they cross at 90°.

Equal diagonals that bisect each other AT 90° — that's exactly the recipe for a square! So APML is a square.

Since the sticks are equal and cross at their shared midpoint, their endpoints form a quadrilateral with equal, bisecting diagonals — and we already proved that ALWAYS makes a rectangle, no matter the angle between the sticks!

So the thread traces out a rectangle, and every corner of a rectangle is a perfect 90° — job done, with no paper or protractor needed!

No! That description is exactly a parallelogram, which doesn't have to have 90° angles — it could be "squished" at any angle and still have parallel, equal opposite sides.

(i) Parallelogram PRAE, ∠P = 40°:

(ii) Parallelogram PQRS, ∠P = 110°:

(iii) XWVU with diagonal XV, ∠XVU = 30° (and the tick marks show WV = VU, XW = XU, an isosceles-triangle pattern on each side of the diagonal):

(iv) OIEA with diagonal AI, ∠AEO=20° (tick marks show OI = OA and AE = IE):

Remember: a parallelogram's diagonals bisect each other, but don't have to be equal — so here they're different lengths (7 cm and 5 cm), unlike a rectangle.

A rhombus's diagonals bisect each other AT 90° — that's the key extra rule versus a plain parallelogram.

Every angle of an equilateral triangle is 60°. Gluing two of them together along a shared side ABCD gives a 4-sided shape with:

All 4 sides equal makes this a rhombus, with angles 60°, 120°, 60°, 120° going around!

A kite's diagonals cross at 90°, but only ONE diagonal gets bisected by the other (not both, unlike a rhombus).

In a trapezium, co-interior angles (on the same non-parallel side/leg) always add to 180°:

The remaining angles are 45° and 75°.

Since AB ∥ DC, ∠A and ∠D are co-interior angles on leg AD:

Because AD = BC, ABCD is an isosceles trapezium — so the angles at the ends of the same parallel side are equal:

The four angles are ∠A = 80°, ∠B = 80°, ∠C = 100°, ∠D = 100°.

(i) A shape that's both a kite and a parallelogram must be a rhombus (or the special case, a square) — that's the only place the Kite and Parallelogram circles overlap.

(ii) No — a general kite and a rectangle don't overlap on the diagram. The only quadrilateral that's both is again the square (a kite AND a rectangle at once), which already lives fully inside both the Rhombus and Rectangle regions.

(iii) No, not every kite is a rhombus — a kite only needs two DIFFERENT pairs of adjacent equal sides, while a rhombus needs all 4 sides equal. The correct relationship: every rhombus IS a kite, but most kites are not rhombuses (the Rhombus circle sits fully inside the Kite circle, not the other way round).

Using the geometry of the two overlapping rectangles (their 90° corners and the marked 30° angle at R), the diagonal reasoning works out so that ∠IOD = 30° — the same as the angle marked at R, carried through the rectangles' parallel sides and equal-diagonal properties.

A square's diagonals are equal, bisect each other, AND cross at exactly 90° — and a compass can make a perfect 90° without any protractor!

If the square's side is x, each half-side is x/2. Using the Pythagorean theorem on the little corner triangle:

All four sides of UVWX work out equal, and all four angles work out to 90° — UVWX is also a square! (just smaller and rotated 45°.)

Yes! 4 equal sides makes it a rhombus. In a rhombus, adjacent angles add to 180°, so one 90° angle forces the next one to also be 90° — and that cascades all the way around. All four angles become 90°, making it a square.

Yes! Split the dart shape into two triangles using a diagonal — each triangle's angles sum to 180°, and 180+180=360°, no matter how "dented" or unusual the quadrilateral's shape is.

(i) False — that only guarantees a rectangle; squares additionally need perpendicular diagonals.

(ii) True — the 4th angle is forced to 90° too, since all four must sum to 360°.

(iii) False — it could be a kite instead, which also has perpendicular diagonals.

(iv) False — the non-parallel sides (legs) being equal doesn't mean they're also parallel to each other.

(v) True — if opposite angles keep coming out equal, that forces the adjacent angles to be supplementary too, which is exactly what makes opposite sides parallel.

(vi) True — a quadrilateral with all 4 angles equal must have each one be 360 ÷ 4 = 90°, which is exactly the definition of a rectangle.

(vii) False — an isosceles trapezium only guarantees ONE pair of parallel sides (the "parallel bases") plus equal legs; the legs themselves don't have to be parallel to each other, so the other pair of opposite sides may not be parallel.

No, not every kite is a rhombus — but every rhombus IS a kite! A rhombus is just a special, extra symmetric kite where BOTH pairs of adjacent sides happen to be equal to each other too (all 4 sides equal), not just each pair separately.

You get a square! The two "diagonals" you made are equal in length, bisect each other (cross at the midpoint), AND cross at 90° — and we proved right back in Section 1 that equal + bisecting + perpendicular diagonals is EXACTLY the recipe for a square.

Now you get a kite! The diagonals are no longer equal (one is longer), but they still cross at 90° and one of them (the shorter, un-extended one) is still bisected by the other. That's exactly a kite's diagonal rule — only ONE diagonal gets cut in half by the other, not both — so the shape "opens up" into a kite shape instead of staying a square.

A rhombus! All four outer sides are 8 cm (every side of an equilateral triangle is equal), so all 4 sides of the new quadrilateral match — that's the rhombus rule. The angles come out 60°, 120°, 60°, 120° (each 60° corner stays alone, each 120° corner is two 60°s stacked together).

Joined along an 8 cm side: you get a kite — two adjacent 8 cm sides on top, two adjacent 6 cm sides on the bottom (two DIFFERENT pairs of equal adjacent sides, the kite rule).

Joined along the 6 cm side: you get a different quadrilateral where the 6 cm sides sit opposite each other in the middle — this one is just a general (non-special) quadrilateral, since matching sides aren't adjacent to each other the way a kite needs.

Since a scalene triangle has 3 DIFFERENT side lengths, you can glue the two triangle copies together along any of the 3 sides (6 cm, 9 cm, or 12 cm) — that's 3 different shared-side choices, and for each one the second triangle can also be flipped, giving a kite shape or a "twisted" one.

Joining them along the 9 cm side (with the triangles mirrored) gives the neat kite shown in the book — with two adjacent sides of 6 cm and two adjacent sides of 12 cm. Try the other shared sides yourself and see what you get — some give kites, others give general (non-special) quadrilaterals!

The creases form a square (rotated 45°, so it looks like a diamond)! Because the paper was folded into matching quarters first, the single triangular crease gets copied identically into all 4 quarters when unfolded — giving 4 equal sides, each meeting the next at a right angle, exactly like the CASE/UVWX midpoint-square you met in Section 5.

Nested diamonds (step 5): instead of one triangular crease at the corner, make SEVERAL parallel triangular creases at different distances from the corner (like folding the corner in, then folding the new corner in again, several times). Each fold-line creates one more nested diamond/square when unfolded.

Plain square outline (step 6): fold the quarter-paper so the crease runs parallel to one edge rather than diagonally across the corner — folding all 4 corners of the original sheet in this straight, edge-parallel way produces an ordinary square (not tilted into a diamond) when unfolded.

(b) 90°

(b) trapezium

Draw diagonal AC in quadrilateral ABCD where AB=CD and BC=AD. Triangles ABC and CDA share side AC, with AB=CD and BC=AD given — so by SSS, △ABC ≅ △CDA. This makes their matching angles equal, which turns out to be exactly the alternate-angle pairs needed to prove AB∥CD and BC∥AD — both pairs of opposite sides parallel means the shape is a parallelogram.

(a) Each rectangle's diagonals must be equal in length and bisect each other — that's a defining property of every rectangle.

(b) The 30° angle transfers through the shared point R using alternate/matching angle relationships between the two rectangles' sides — since rectangle sides are at fixed 90° angles to each other, the given 30° angle at R determines matching 30° angles elsewhere via parallel-line angle rules.

(c) Not necessarily as cleanly — plain parallelograms don't guarantee equal diagonals or 90° corners, so the same angle-transfer logic wouldn't be guaranteed to produce a clean 30° result without knowing more about the specific parallelogram's angles.