PSLE Paper Folding Heuristics: 10 Question Types on Creases, Flaps and Strips

Corner Folded Along a Straight Edge

propertyA fold makes equal angles on both sides of the crease; on a straight edge the two copies and the rest add to 180°

ABCD is a rectangle. P is a point on DC and Q a point on CB. The corner C is folded along the crease PQ and lands at C'. ∠ DPC' = 74^°. Find ∠ CPQ and ∠ C'QB.

The crease PQ mirrors the corner C to C’: the two angles at P beside the crease are equal.

The fold makes ∠ C'PQ = ∠ CPQ.

step 1 of 6

Folding is a reflection in the crease, so the angle between PC and PQ equals the angle between PC' and PQ. Those two equal angles and ∠ DPC' lie along the straight edge DC.

The fold makes ∠ C'PQ = ∠ CPQ.
DPC is a straight line: ∠ DPC' + ∠ C'PQ + ∠ QPC = 180^°.
2 × ∠ CPQ = 180^° − 74^° = 106^°, so ∠ CPQ = 53^°.
In triangle PCQ, ∠ PCQ = 90^° (corner of the rectangle), so ∠ PQC = 180^° − 90^° − 53^° = 37^°.
The fold also makes ∠ PQC' = ∠ PQC = 37^°.
CQB is a straight line: ∠ C'QB = 180^° − 37^° − 37^° = 106^°.

answer∠ CPQ = 53^°; ∠ C'QB = 106^°

Common pitfalls

Giving ∠ CPQ = 180^° − 74^° = 106^°: that is both copies together, and the crease sits in the middle of it.
Forgetting the fold copies the angle at Q too, and reading ∠ C'QB straight off the picture.

Corner Folded into a Right Angle

propertyAt a right-angled corner, the two copies of the fold angle and the leftover add to 90°

ABCD is a rectangle. The corner C is folded along the crease BE, with E on DC, and lands at C' inside the rectangle. ∠ ABC' = 26^°. Find ∠ CBE and ∠ DEC'.

The crease BE passes through the corner B; the fold mirrors C to C’.

The fold makes ∠ C'BE = ∠ CBE.

step 1 of 6

The crease BE passes through the corner B, whose angle is 90^°. The fold copies ∠ CBE to the other side of the crease, and what is left of the right angle is the given 26^°.

The fold makes ∠ C'BE = ∠ CBE.
At the corner B: ∠ ABC' + ∠ C'BE + ∠ EBC = 90^°.
2 × ∠ CBE = 90^° − 26^° = 64^°, so ∠ CBE = 32^°.
In triangle BCE: ∠ BEC = 180^° − 90^° − 32^° = 58^°.
The fold copies it: ∠ BEC' = 58^°.
DEC is a straight line: ∠ DEC' = 180^° − 58^° − 58^° = 64^°.

answer∠ CBE = 32^°; ∠ DEC' = 64^°

Common pitfalls

Using 180^° at B instead of the corner's 90^°.
Answering ∠ DEC' = 58^°: that is one copy; the angle asked is what is left of the straight edge after both.

Corner Folded onto the Opposite Edge

propertyThe folded edge becomes a transversal between the two parallel edges, so a Z appears

ABCD is a rectangle with AB parallel to DC. It is folded along the crease EF, with E on DC and F on AB, so that the corner A lands at A' on the edge DC and the corner D lands at D'. ∠ AFE = 62^°. Find ∠ A'FB and ∠ DA'D'.

A lands on the top edge at A’; the fold copies the 62° at F.

The fold makes ∠ A'FE = ∠ AFE = 62^°.

step 1 of 5

The fold copies ∠ AFE across the crease, which leaves ∠ A'FB on the straight edge. A'F then runs between the parallel edges, so an alternate angle appears at A'; the folded corner is still a right angle, and that alternate angle is part of it.

The fold makes ∠ A'FE = ∠ AFE = 62^°.
AFB is a straight line: ∠ A'FB = 180^° − 62^° − 62^° = 56^°.
AB ∥ DC and FA' crosses both: ∠ FA'D is alternate to ∠ A'FB, so ∠ FA'D = 56^°.
The corner A was a right angle between AB and AD; the fold keeps it: ∠ FA'D' = 90^°.
A'D lies inside that right angle: ∠ DA'D' = 90^° − 56^° = 34^°.

answer∠ A'FB = 56^°; ∠ DA'D' = 34^°

Common pitfalls

Losing the right angle: the corner A is still 90^° after the fold, now between A'F and A'D'.
Taking ∠ FA'D = 62^° by matching it with the given angle instead of with ∠ A'FB across the parallels.

Fold Along a Diagonal: the Hidden Isosceles Triangle

propertyAn alternate angle and a folded angle are both equal to the same angle, so two sides are equal

ABCD is a rectangle with AB parallel to DC. It is folded along the diagonal BD, so that C lands at C'. DC' crosses AB at E. ∠ CDB = 34^°. Find ∠ BED and ∠ ADE.

Z shape across the parallel edges: ∠ EBD = 34°.

AB ∥ DC: ∠ EBD is alternate to ∠ CDB, so ∠ EBD = 34^°.

step 1 of 5

Two things make the same angle: the parallel edges give ∠ ABD = ∠ CDB, and the fold gives ∠ EDB = ∠ CDB. So triangle BDE has two equal angles and is isosceles.

AB ∥ DC: ∠ EBD is alternate to ∠ CDB, so ∠ EBD = 34^°.
The fold along BD copies ∠ CDB: ∠ EDB = 34^°.
Triangle BDE has equal angles at B and D, so EB = ED.
∠ BED = 180^° − 34^° − 34^° = 112^°.
At the corner D: ∠ ADB = 90^° − 34^° = 56^°, so ∠ ADE = 56^° − 34^° = 22^°.

answer∠ BED = 112^°; ∠ ADE = 22^°

Common pitfalls

Assuming E is the midpoint of AB or that BE = EA: only EB = ED follows.
Answering ∠ ADE = 34^° by copying the fold angle once too often; ∠ ADE is the rest of the corner.

Folded Strip: the Overlap Triangle

propertyA strip folded across itself always makes an isosceles overlap triangle

A strip of paper with parallel edges is folded once along the crease PQ, with P on one edge and Q on the other. The crease makes an angle of 52^° with the edge at P. The folded part overlaps the rest in triangle PQR. Find ∠ PQR and ∠ PRQ.

The part beyond the crease folds back over the strip; the 52° at P is copied.

The fold makes ∠ RPQ = 52^°, a copy of the 52^° between the crease and the edge.

step 1 of 4

The fold copies the 52^° at P to the other side of the crease, and the parallel edges carry the same 52^° to Q as an alternate angle. Two equal angles, one isosceles triangle.

The fold makes ∠ RPQ = 52^°, a copy of the 52^° between the crease and the edge.
The two edges of the strip are parallel and PQ crosses both: ∠ PQR is alternate to the 52^° at P, so ∠ PQR = 52^°.
Triangle PQR has equal angles at P and Q: it is isosceles, with RP = RQ.
∠ PRQ = 180^° − 52^° − 52^° = 76^°.

answer∠ PQR = 52^°; ∠ PRQ = 76^°

Common pitfalls

Treating the overlap as a right-angled triangle because the strip has square ends.
Doubling the 52^° for ∠ PRQ: 104^° is the angle the folded edge makes with the original edge, outside the triangle.

Folded Strip: the Equilateral Overlap

propertyThe overlap is equilateral only when the crease makes 60° with the edge

A strip of paper with parallel edges is folded once along the crease PQ. The overlap triangle PQR turns out to be equilateral. Find the angle the crease makes with the edge at P, and the angle between the original edge and the folded edge at P.

The overlap is isosceles whatever the crease angle; here all three sides are equal.

In any such fold, ∠ RPQ and ∠ PQR both equal the crease angle at P.

step 1 of 5

The overlap triangle is always isosceles with its two equal angles each equal to the crease angle. For it to be equilateral those angles must be 60^°, so the crease angle is 60^° and the fold turns the edge through twice that.

In any such fold, ∠ RPQ and ∠ PQR both equal the crease angle at P.
An equilateral triangle has all three angles 60^°, so the crease angle is 60^°.
The folded edge PR is the original edge reflected in the crease: it turns through twice the crease angle.
Angle between the original edge and the folded edge = 2 × 60^° = 120^°.
Check on the straight edge at P: 60^° (crease to folded edge) + 60^° (crease to original edge) = 120^°.

answer60^°; 120^°

Common pitfalls

Answering 30^° for the crease, halving the 60^° as if the crease bisected the triangle's angle: the crease is a side of the triangle.
Giving 60^° for the turn of the edge; the fold doubles the crease angle.

Folded Strip: How Far the Direction Turns

propertyA fold turns the strip through twice the crease angle

A long strip of paper is folded once along a crease that makes an angle of 65^° with the strip's edge. Through what angle does the strip's direction turn? What is the acute angle between the edge before the fold and the edge after it?

The crease makes 65° with the edge; the fold copies it beyond the crease.

The crease makes 65^° with the edge on one side; the fold copies 65^° on the other side.

step 1 of 4

The edge after the fold is the mirror image of the edge before it, in the crease. Each side of the crease carries the same 65^°, so the turn is their sum; the acute angle between the two lines is what is left of 180^°.

The crease makes 65^° with the edge on one side; the fold copies 65^° on the other side.
The turn from the old direction to the new is 65^° + 65^° = 130^°.
The two edges cross at the fold point; the other angle between them is 180^° − 130^° = 50^°.
So the direction turns through 130^°, and the acute angle between the two edges is 50^°.

answer130^°; 50^°

Common pitfalls

Answering 65^° for the turn: the crease angle is half of it.
Measuring the turn from the crease instead of from the original edge.

Two Corner Folds on One Edge

propertyEach flap takes twice its crease angle off the straight edge; the wedge left between them is the rest

ABCD is a rectangle. Corner A is folded along the crease PQ (P on AB, Q on AD) and lands at A'; corner B is folded along the crease RS (R on AB, S on BC) and lands at B'. The flaps do not overlap. ∠ APQ = 55^° and ∠ BRS = 70^°. Find ∠ A'PR and ∠ B'RP.

At P the flap covers 55° + 55° = 110°.

At P: the fold makes ∠ A'PQ = ∠ APQ = 55^°, so the flap covers 2 × 55^° = 110^°.

step 1 of 4

Each fold copies its crease angle, so each flap covers twice that angle at its fold point on the edge AB. What each flap leaves on the straight edge is the answer at that point.

At P: the fold makes ∠ A'PQ = ∠ APQ = 55^°, so the flap covers 2 × 55^° = 110^°.
APB is a straight line: ∠ A'PR = 180^° − 110^° = 70^°.
At R: the fold makes ∠ B'RS = ∠ BRS = 70^°, so that flap covers 2 × 70^° = 140^°.
ARB is a straight line: ∠ B'RP = 180^° − 140^° = 40^°.

answer∠ A'PR = 70^°; ∠ B'RP = 40^°

Common pitfalls

Subtracting the crease angle once: 180^° − 55^° = 125^° leaves the flap's own copy inside the answer.
Reading the two flaps together as one 360^° turn: they sit at two different points of the edge.

Two Flaps That Overlap

propertyTwo flaps folded from one point cover twice their crease angles; what exceeds 180° is covered twice

ABCD is a rectangle and P is a point on AB. Corner A is folded along the crease PQ (Q on AD) and lands at A'; corner B is folded along the crease PS (S on BC) and lands at B'. ∠ APQ = 55^° and ∠ BPS = 50^°. Find ∠ QPS, the angle between the two creases, and ∠ A'PB', the angle of the part covered by both flaps.

Between the two creases: 180° − 55° − 50° = 75°.

APB is a straight line: ∠ QPS = 180^° − 55^° − 50^° = 75^°.

step 1 of 4

Both flaps hinge at the same point P on the straight edge. Each covers twice its crease angle; together that is more than 180^°, and the excess is the doubly covered wedge.

APB is a straight line: ∠ QPS = 180^° − 55^° − 50^° = 75^°.
The first flap covers 2 × 55^° = 110^° from PA; the second covers 2 × 50^° = 100^° from PB.
Together: 110^° + 100^° = 210^°, but the edge only offers 180^°.
The overlap, covered by both: 210^° − 180^° = 30^°, so ∠ A'PB' = 30^°.

answer∠ QPS = 75^°; ∠ A'PB' = 30^°

Common pitfalls

Giving the overlap as 180^° − 110^° − 100^° and getting a negative number: the sign is the overlap, not an error.
Taking ∠ A'PB' as the gap between the creases, 75^°.

Folded Corner: Areas of One and Two Layers

propertyThe area hidden by the fold equals the flap; sheet minus what shows is the flap

A rectangular sheet of paper is 12 cm by 8 cm. One corner is folded over along a crease so that the flap is a right-angled triangle with one leg the full 8 cm width of the sheet. After the fold, the area still showing is 80 cm². Find the area of the flap and the length of its other leg.

The whole sheet: 12 × 8 = 96 cm².

Area of the sheet = 12 × 8 = 96 cm².

step 1 of 5

The flap lands on the sheet and hides a region of its own area. So the area that disappears from view is the flap's area, and a right-angled triangle with one known leg gives the other.

Area of the sheet = 12 × 8 = 96 cm².
The flap covers part of the sheet, hiding an area equal to itself: area hidden = 96 − 80 = 16 cm².
So the flap's area is 16 cm².
The flap is a right-angled triangle with a leg of 8 cm: 12 × 8 × x = 16.
x = 16 × 2 ÷ 8 = 4 cm.

answer16 cm²; 4 cm

Common pitfalls

Halving 96 − 80 because the paper is doubled there: the hidden area is one flap's worth, not two.
Using 12 cm as the flap's leg because it is the sheet's length; the flap runs the full width, 8 cm.