Quadrilaterals · word problems

Quadrilateral Heuristics

10 question types · PSLE Paper 1 and 2 · one figure that follows the deduction

01

Trapezium: Co-Interior Angles

propertyBetween the parallel sides, each pair of angles on one leg adds to 180°

ABCD is a trapezium with AB parallel to DC. ∠ DAB = 112° and ∠ BCD = 67°. Find ∠ ADC and ∠ ABC.

ABCD112°67°
AB and DC are parallel; the legs AD and BC cross both.
AB ∥ DC, and AD crosses both: ∠ DAB and ∠ ADC are co-interior angles.
step 1 of 5

Each leg of the trapezium is a transversal across the two parallel sides, so the two angles on one leg are co-interior and add to 180°.

  1. AB ∥ DC, and AD crosses both: ∠ DAB and ∠ ADC are co-interior angles.
  2. ∠ ADC = 180° − 112° = 68°.
  3. BC crosses both too: ∠ ABC and ∠ BCD are co-interior angles.
  4. ∠ ABC = 180° − 67° = 113°.
  5. Check: the four angles of any quadrilateral add to 360°: 112 + 113 + 67 + 68 = 360.

answer∠ ADC = 68°; ∠ ABC = 113°

Common pitfalls

  • Pairing ∠ DAB with ∠ BCD as if they were opposite angles of a parallelogram: a trapezium has one pair of parallel sides, so only the angles on one leg are linked.
  • Making ∠ ABC = ∠ DAB: the two top angles are equal only in an isosceles trapezium, which nothing here says.
02

Isosceles Trapezium and a Diagonal

propertyEqual legs give equal base angles; the diagonal makes a Z with the parallel sides

PQRS is a trapezium with PQ parallel to SR and PS = QR. ∠ PSR = 72° and the diagonal PR makes ∠ PRS = 41°. Find ∠ SPR and ∠ QRP.

PQRS72°41°72°
Equal legs: the base angles at S and R are both 72°.
PS = QR, so the trapezium is isosceles and its base angles are equal: ∠ QRS = ∠ PSR = 72°.
step 1 of 5

The equal legs make the two base angles equal, so the angle at R is known and the diagonal splits it. In triangle PSR the third angle follows.

  1. PS = QR, so the trapezium is isosceles and its base angles are equal: ∠ QRS = ∠ PSR = 72°.
  2. In triangle PSR: ∠ SPR = 180° − 72° − 41° = 67°.
  3. The diagonal PR splits ∠ QRS: ∠ QRP = 72° − 41° = 31°.
  4. Check with the Z shape: PQ ∥ SR, so ∠ QPR = ∠ PRS = 41°.
  5. Then ∠ SPQ = 67° + 41° = 108°, and 108° + 72° = 180° on the leg PS.

answer∠ SPR = 67°; ∠ QRP = 31°

Common pitfalls

  • Assuming the diagonal bisects the angle at R and answering 36°: a trapezium's diagonal does not bisect its angles.
  • Using ∠ QRS = 108° by confusing the base angle with the top angle of the trapezium.
03

Right-Angled Trapezium: Rectangle plus Triangle

propertyDrop a perpendicular from the short parallel side: a rectangle and a right-angled triangle

ABCD is a trapezium with AB parallel to DC and right angles at A and D. AB = 8 cm, DC = 14 cm and AD = 8 cm. Find the area and the perimeter of ABCD.

ABCD8 cm14 cm8 cmE
BE, perpendicular to DC, cuts off a square ABED of side 8 cm.
Draw BE perpendicular to DC, with E on DC. ABED is a rectangle with AB = DE = 8 cm and AD = BE = 8 cm, so it is a square.
step 1 of 6

A perpendicular from B to DC cuts the trapezium into a square and a right-angled triangle whose legs are the overhang and the height. The slanted side is that triangle's hypotenuse.

  1. Draw BE perpendicular to DC, with E on DC. ABED is a rectangle with AB = DE = 8 cm and AD = BE = 8 cm, so it is a square.
  2. EC = DC − DE = 14 − 8 = 6 cm.
  3. Triangle BEC has legs 6 cm and 8 cm, so BC = 10 cm (6-8-10 right-angled triangle).
  4. Area = 8 × 8 + 12 × 6 × 8 = 64 + 24 = 88 cm2.
  5. Check with the trapezium rule: 12 × (8 + 14) × 8 = 88 cm2.
  6. Perimeter = 8 + 8 + 14 + 10 = 40 cm.

answer88 cm2; 40 cm

Common pitfalls

  • Using DC = 14 cm as the length of the slanted side BC in the perimeter.
  • Taking the area as 14 × 8 = 112 cm2: that is the full rectangle the trapezium sits in, and the corner triangle is not part of it.
04

Parallelogram: Adjacent and Opposite Angles

propertyAdjacent angles add to 180°; opposite angles are equal

ABCD is a parallelogram. ∠ DAB is 40° more than ∠ ABC. Find ∠ ABC and ∠ BCD.

ABCD??
AD and BC are parallel, so the angles at A and B add to 180°.
AD ∥ BC, so ∠ DAB + ∠ ABC = 180°.
step 1 of 5

Two angles on one side of a parallelogram are co-interior and add to 180°; with their difference given, that is a sum-and-difference share. The opposite angle is then a copy.

  1. AD ∥ BC, so ∠ DAB + ∠ ABC = 180°.
  2. ∠ DAB is 40° more than ∠ ABC: take the 40° off, 180° − 40° = 140° is two equal shares.
  3. ∠ ABC = 140° ÷ 2 = 70° and ∠ DAB = 70° + 40° = 110°.
  4. Opposite angles of a parallelogram are equal: ∠ BCD = ∠ DAB = 110°.
  5. Check: 110 + 70 + 110 + 70 = 360.

answer∠ ABC = 70°; ∠ BCD = 110°

Common pitfalls

  • Halving 180° and then adding 40° to one angle: 90° and 130° add to 220°, not 180°.
  • Giving ∠ BCD = ∠ ABC: C is opposite A, not B.
05

Rhombus: Diagonals Bisect the Angles

propertyA rhombus's diagonals bisect its angles and cross at right angles

ABCD is a rhombus with ∠ ABC = 116°. The diagonals AC and BD meet at O. Find ∠ ABO and ∠ BAO.

ABCDO116°
All four sides equal. The diagonals of a rhombus bisect its angles.
The diagonal BD bisects ∠ ABC: ∠ ABO = 116° ÷ 2 = 58°.
step 1 of 4

The diagonal BD halves the angle at B, and the diagonal AC halves the angle at A. The angle at A comes from the parallel sides first.

  1. The diagonal BD bisects ∠ ABC: ∠ ABO = 116° ÷ 2 = 58°.
  2. AD ∥ BC, so ∠ DAB = 180° − 116° = 64°.
  3. The diagonal AC bisects ∠ DAB: ∠ BAO = 64° ÷ 2 = 32°.
  4. Check in triangle ABO: 58° + 32° + ∠ AOB = 180° gives ∠ AOB = 90°, the right angle the diagonals of a rhombus always make.

answer∠ ABO = 58°; ∠ BAO = 32°

Common pitfalls

  • Bisecting 116° for both answers: the diagonal through A halves the angle at A, which is 64°.
  • Using this on a parallelogram that is not a rhombus: only equal sides make the diagonals bisect the angles.
06

Kite: One Line of Symmetry

propertyThe long diagonal is the mirror line: it bisects the two angles it passes through, and the other two angles are equal

PQRS is a kite with PQ = PS and RQ = RS. ∠ QPS = 84° and ∠ QRS = 32°. The diagonals meet at X. Find ∠ PQR and ∠ QPX.

PQRSX84°32°
Round the kite: ∠ PQR + ∠ PSR = 360° − 84° − 32° = 244°.
The four angles add to 360°: ∠ PQR + ∠ PSR = 360° − 84° − 32° = 244°.
step 1 of 5

Folding along PR puts Q onto S, so the angles at Q and S are equal and PR halves the angles at P and R.

  1. The four angles add to 360°: ∠ PQR + ∠ PSR = 360° − 84° − 32° = 244°.
  2. PR is the kite's line of symmetry, so ∠ PQR = ∠ PSR.
  3. ∠ PQR = 244° ÷ 2 = 122°.
  4. The same symmetry halves the angle at P: ∠ QPX = 84° ÷ 2 = 42°.
  5. Check in triangle PQX: the diagonals of a kite cross at right angles, so ∠ PQX = 180° − 90° − 42° = 48°, and 48° + ∠ XQR = 122° gives ∠ XQR = 74°, half of what is left of 180° after 32°.

answer∠ PQR = 122°; ∠ QPX = 42°

Common pitfalls

  • Halving the angles at Q and S too: the short diagonal QS does not bisect them.
  • Answering ∠ PQR = 90° because the diagonals cross at right angles: that right angle is at X, not at Q.
07

Parallelogram Cut by a Line from a Vertex

propertyA line across the parallelogram makes a triangle on one side and a Z with the other pair of parallels

ABCD is a parallelogram with ∠ ABC = 108°. E is a point on BC and ∠ BAE = 25°. Find ∠ AEC and ∠ EAD.

ABCDE108°25°47°
Triangle ABE: ∠ AEB = 180° − 108° − 25° = 47°.
In triangle ABE: ∠ AEB = 180° − 108° − 25° = 47°.
step 1 of 5

Triangle ABE has two known angles, which gives the angle at E on the straight line BC. The same angle appears again at A as an alternate angle with the parallel side AD.

  1. In triangle ABE: ∠ AEB = 180° − 108° − 25° = 47°.
  2. BEC is a straight line, so ∠ AEC = 180° − 47° = 133°.
  3. AD ∥ BC and AE crosses both: ∠ EAD and ∠ AEB are alternate angles.
  4. ∠ EAD = 47°.
  5. Check at A: ∠ BAD = 25° + 47° = 72°, and 72° + 108° = 180° as adjacent angles of a parallelogram must.

answer∠ AEC = 133°; ∠ EAD = 47°

Common pitfalls

  • Stopping at ∠ AEB = 47° when ∠ AEC, on the other side of E, is asked.
  • Making ∠ EAD equal to ∠ BAE: the line AE is not a diagonal and bisects nothing.
08

Isosceles Triangle Inside a Trapezium

propertyThe triangle's base angles carry across to the parallel side as alternate angles

ABCD is a trapezium with AB parallel to DC. E is a point on DC with EA = EB and ∠ AEB = 44°. ∠ ADE = 75°. Find ∠ AED and ∠ DAE.

ABCDE44°75°68°68°
EA = EB with 44° at the apex: base angles (180° − 44°) ÷ 2 = 68°.
EA = EB, so triangle ABE is isosceles with apex E: ∠ EAB = ∠ EBA = (180° − 44°) ÷ 2 = 68°.
step 1 of 4

The isosceles triangle gives its base angles at A and B; each of those is alternate to an angle at E on the parallel side DC. Then triangle ADE has two known angles.

  1. EA = EB, so triangle ABE is isosceles with apex E: ∠ EAB = ∠ EBA = (180° − 44°) ÷ 2 = 68°.
  2. AB ∥ DC and AE crosses both: ∠ AED and ∠ EAB are alternate angles, so ∠ AED = 68°.
  3. In triangle ADE: ∠ DAE = 180° − 75° − 68° = 37°.
  4. Check on the straight line DEC: ∠ AED + ∠ AEB + ∠ BEC = 68° + 44° + 68° = 180°, with ∠ BEC alternate to ∠ EBA.

answer∠ AED = 68°; ∠ DAE = 37°

Common pitfalls

  • Taking ∠ AED = 44°: the 44° is at the apex between EA and EB, and ∠ AED lies on the other side of EA.
  • Using ∠ DAE = ∠ ADE as if triangle ADE were isosceles; nothing makes DA = DE.
09

Rhombus and Parallelogram on One Side

propertyEach shape fixes its own angle at the shared vertex; the difference is the angle between them

ABCD is a rhombus and ABEF is a parallelogram, both drawn on the same side of AB. ∠ DAB = 70° and ∠ ABE = 130°. Find ∠ BAF and ∠ FAD.

ABCDEF70°130°
AF ∥ BE in the parallelogram, so ∠ BAF + ∠ ABE = 180°.
In parallelogram ABEF, AF ∥ BE, so ∠ BAF + ∠ ABE = 180°.
step 1 of 4

At A the rhombus contributes 70° and the parallelogram its own angle, found from the co-interior pair with ∠ ABE. The gap between the two figures is the difference.

  1. In parallelogram ABEF, AF ∥ BE, so ∠ BAF + ∠ ABE = 180°.
  2. ∠ BAF = 180° − 130° = 50°.
  3. AF lies inside ∠ DAB, since 50° < 70°: ∠ FAD = ∠ DAB − ∠ BAF = 70° − 50° = 20°.
  4. Check at B: the rhombus has ∠ ABC = 180° − 70° = 110°, so ∠ CBE = 130° − 110° = 20°; BC and AF sit at the same 20° gap, one on each side.

answer∠ BAF = 50°; ∠ FAD = 20°

Common pitfalls

  • Adding instead of subtracting at A: AF is between AB and AD, so the angles nest.
  • Using the rhombus's equal sides to make AF = AD: the parallelogram's sides have nothing to do with the rhombus's.
10

Shapes Joined Edge to Edge

propertyAt a shared vertex, add the angles each shape brings, and take the rest from 360°

ABCD is a square. Triangle ABE is isosceles with EA = EB, drawn outside the square on AB, with ∠ AEB = 40°. Triangle BCF is equilateral, drawn outside the square on BC. Find ∠ EAD and ∠ EBF.

ABCDEF40°70°70°
Triangle ABE, apex E: base angles (180° − 40°) ÷ 2 = 70°.
In triangle ABE: ∠ EAB = ∠ EBA = (180° − 40°) ÷ 2 = 70°.
step 1 of 5

Each shape brings a known angle to the vertex it shares. At A two angles sit side by side; at B three do, and the fourth is what is left of the full turn.

  1. In triangle ABE: ∠ EAB = ∠ EBA = (180° − 40°) ÷ 2 = 70°.
  2. At A: ∠ EAD = ∠ EAB + ∠ BAD = 70° + 90° = 160°.
  3. Triangle BCF is equilateral: ∠ CBF = 60°.
  4. At B, the angles round the point: ∠ EBA + ∠ ABC + ∠ CBF + ∠ EBF = 360°.
  5. ∠ EBF = 360° − 70° − 90° − 60° = 140°.

answer∠ EAD = 160°; ∠ EBF = 140°

Common pitfalls

  • Answering ∠ EBF = 180° − 70° − 60° = 50°: the angles at B go all the way round, and the square's 90° is one of them.
  • Giving the triangle's base angle as 40°: the 40° is at the apex E.
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