Quadrilaterals · applications

Applications: Quadrilaterals

11 problems

PSLE · O-Level · SAT · GCSE Higher

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°.

All steps
  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°

techniqueAngles in Parallel Lines · Special Quadrilaterals

examsO-Level · SAT · GCSE Higher

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.

All steps
  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°

techniqueSpecial Quadrilaterals · Angles in Parallel Lines

examsPSLE · O-Level · GCSE Higher

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.

All steps
  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

techniqueSpecial Quadrilaterals · Angles in a Triangle

examsPSLE · O-Level · GCSE Higher

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.

All steps
  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°

techniqueSpecial Quadrilaterals · Angles in Parallel Lines

examsPSLE · O-Level · GCSE Higher

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.

All steps
  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°

techniqueRhombus and Kite

examsPSLE · O-Level · SAT · GCSE Higher

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.

All steps
  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°

techniqueRhombus and Kite · Symmetry

examsPSLE · O-Level · SAT · GCSE Higher

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.

All steps
  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°

techniqueAngles in Parallel Lines · Angles in a Triangle

examsO-Level · SAT · GCSE Higher

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.

All steps
  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°

techniqueAngles in a Triangle · Angles in Parallel Lines

examsPSLE · O-Level · SAT · GCSE Higher

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.

All steps
  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°

techniqueRhombus and Kite · Special Quadrilaterals

examsPSLE · O-Level · SAT · GCSE Higher

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.

All steps
  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°

techniqueAngles at a Point · Angles in a Polygon

examsPSLE · O-Level · SAT · GCSE Higher

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.
11

A Square and a Rhombus Along One Diagonal

propertyA square's diagonal halves its right angles; carried on as a straight line, it fixes the rhombus's angle at the shared corner

ABCD is a square and CDEF is a rhombus. B, D and E lie on a straight line. (a) Find ∠ BDC. (b) Find ∠ DEF. (c) G is a point on AB such that ∠ AGD = 62°. Find ∠ GDB.

ABCDEF45 deg
(a) The diagonal halves the right angle at D: ∠ BDC = 45°.
The diagonal BD cuts the square into two identical right-angled triangles, so it halves the right angle at D. (a) ∠ BDC = 90° ÷ 2 = 45°.
step 1 of 5

The diagonal BD does two jobs. Inside the square it halves the right angles at B and D; carried on past D as the straight line BDE, it sets the angle of the rhombus at D. The rhombus's parallel sides then pass that angle to E.

All steps
  1. The diagonal BD cuts the square into two identical right-angled triangles, so it halves the right angle at D. (a) ∠ BDC = 90° ÷ 2 = 45°.
  2. BDE is a straight line, so the angles at D on the lower side of it add to 180°: ∠ CDE = 180° − 45° = 135°.
  3. In the rhombus CDEF, DC is parallel to EF, and DE runs between them, so ∠ CDE and ∠ DEF are co-interior angles that add to 180°. (b) ∠ DEF = 180° − 135° = 45°.
  4. Triangle ADG has a right angle at A, the corner of the square, so ∠ ADG = 180° − 90° − 62° = 28°.
  5. The diagonal also halves the right angle between DA and DC on the other side: ∠ ADB = 45°, and DG lies inside it. (c) ∠ GDB = 45° − 28° = 17°.

answer(a) ∠ BDC = 45° (b) ∠ DEF = 45° (c) ∠ GDB = 17°

techniqueSpecial Quadrilaterals · Rhombus and Kite · Angles on a Line

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Answering (b) with 135°, the same as ∠ CDE: in a rhombus the equal angles are the opposite ones, and ∠ CDE and ∠ DEF are neighbors on the side DE, so they add to 180°.
  • Answering (c) with 28°: that is ∠ ADG, measured from DA. ∠ GDB is measured from DG to the diagonal, so the 28° comes off the diagonal's 45°.