Hard GCSE Quadrilateral properties Questions

Challenging, exam-style GCSE Quadrilateral properties questions with worked solutions. Stretch yourself on the hardest parallelogram, angle bisector, alternate angles, rhombus problems.

parallelogramangle bisectoralternate anglesrhombusdiagonalsbisecting an angle
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
The vertices of a quadrilateral ABCDABCD are A(1,1)A(1, 1), B(6,1)B(6, 1), C(8,5)C(8, 5) and D(3,5)D(3, 5). What is the most precise name for the quadrilateral ABCDABCD?
Show worked solution

Worked solution

  1. Work out the vector along each side

    AB⃗=(50),BC⃗=(24),CD⃗=(−50),DA⃗=(−2−4)\vec{AB} = \binom{5}{0}, \quad \vec{BC} = \binom{2}{4}, \quad \vec{CD} = \binom{-5}{0}, \quad \vec{DA} = \binom{-2}{-4}

    Subtract the coordinates at the start of each side from the coordinates at its end.

  2. Compare the vectors of the opposite sides

    AB⃗=−CD⃗,BC⃗=−DA⃗\vec{AB} = -\vec{CD}, \quad \vec{BC} = -\vec{DA}

    AB and CD have the same numbers with opposite signs, so they are parallel and the same length. The same is true of BC and DA.

  3. Deduce that the shape is a parallelogram

    AB∥DC,AD∥BCAB \parallel DC, \quad AD \parallel BC

    Both pairs of opposite sides are parallel, which is exactly what makes a quadrilateral a parallelogram.

  4. Square the length of each side

    AB2=52+02=25,BC2=22+42=20AB^2 = 5^2 + 0^2 = 25, \quad BC^2 = 2^2 + 4^2 = 20

    Pythagoras on the two vectors gives squared lengths of 25 and 20.

  5. Rule out the rhombus

    25≠2025 \neq 20

    The two pairs of sides have different lengths, so the four sides are not all equal and the shape is not a rhombus. It is not a square either, for the same reason.

  6. Test the corner at B for a right angle

    BA⃗⋅BC⃗=(−5)(2)+(0)(4)=−10\vec{BA} \cdot \vec{BC} = (-5)(2) + (0)(4) = -10

    The scalar product of the two sides meeting at B is -10, not 0, so the angle at B is not a right angle.

  7. Rule out the rectangle

    ∠ABC≠90∘\angle ABC \neq 90^\circ

    A rectangle needs all four angles to be right angles, and the angle at B is not one.

  8. Rule out the kite

    AB=DC=5,the equal sides are opposite, not adjacentAB = DC = 5, \quad \text{the equal sides are opposite, not adjacent}

    A kite needs its equal sides to be next to each other. Here the equal sides face each other across the shape, which is the parallelogram pattern, not the kite pattern.

  9. Rule out the trapezium

    2 pairs of parallel sides, not 1\text{2 pairs of parallel sides, not 1}

    A trapezium has exactly one pair of parallel sides. This shape has two, so parallelogram is the more precise name.

  10. Work out the diagonals

    AC⃗=(74),BD⃗=(−34)\vec{AC} = \binom{7}{4}, \quad \vec{BD} = \binom{-3}{4}

    AC runs from (1, 1) to (8, 5) and BD runs from (6, 1) to (3, 5).

  11. Check that the diagonals bisect each other

    mid-point of AC=(4.5,3)=mid-point of BD\text{mid-point of } AC = (4.5, 3) = \text{mid-point of } BD

    Both diagonals have the same mid-point, so they cut each other exactly in half — another property that only the parallelogram family has.

  12. Check the diagonals are not equal

    AC2=72+42=65,BD2=32+42=25AC^2 = 7^2 + 4^2 = 65, \quad BD^2 = 3^2 + 4^2 = 25

    The diagonals have different lengths, which agrees with the shape not being a rectangle.

  13. Check the diagonals are not perpendicular

    7×(−3)+4×4=−21+16=−57 \times (-3) + 4 \times 4 = -21 + 16 = -5

    The scalar product is not zero, so the diagonals do not cross at right angles, which agrees with the shape not being a rhombus.

  14. Collect the evidence

    2 pairs of parallel sides, no right angles, sides not all equal\text{2 pairs of parallel sides, no right angles, sides not all equal}

    The shape is a parallelogram, and none of the extra conditions that would make it a rectangle, a rhombus or a square is satisfied.

  15. State the name of the shape

    Parallelogram\text{Parallelogram}

    Both pairs of opposite sides are parallel and equal, but the sides are not all equal and there is no right angle, so the most precise name is parallelogram.

Answer
Parallelogram\text{Parallelogram}
Question 2
6 markschallenging
The vertices of a quadrilateral ABCDABCD are A(0,0)A(0, 0), B(3,−2)B(3, -2), C(8,0)C(8, 0) and D(3,2)D(3, 2). What is the most precise name for the quadrilateral ABCDABCD?
Show worked solution

Worked solution

  1. Work out the vector along each side

    AB⃗=(3−2),BC⃗=(52),CD⃗=(−52),DA⃗=(−3−2)\vec{AB} = \binom{3}{-2}, \quad \vec{BC} = \binom{5}{2}, \quad \vec{CD} = \binom{-5}{2}, \quad \vec{DA} = \binom{-3}{-2}

    Subtract the coordinates at the start of each side from the coordinates at its end. The four vectors carry both the length and the direction of the sides.

  2. Square the length of each side

    AB2=32+22=13,BC2=52+22=29AB^2 = 3^2 + 2^2 = 13, \quad BC^2 = 5^2 + 2^2 = 29

    Using Pythagoras on each vector gives the squared lengths. Working with squares keeps everything as whole numbers, and two sides are equal exactly when their squares are.

  3. Square the lengths of the other two sides

    CD2=52+22=29,DA2=32+22=13CD^2 = 5^2 + 2^2 = 29, \quad DA^2 = 3^2 + 2^2 = 13

    The other two sides give 29 and 13 as well.

  4. Pair up the equal sides

    AB=DA=13,BC=CD=29AB = DA = \sqrt{13}, \quad BC = CD = \sqrt{29}

    There are two pairs of equal sides. Crucially, the equal sides are NEXT TO each other: AB and DA meet at A, and BC and CD meet at C.

  5. Check whether all four sides are equal

    13≠2913 \neq 29

    The two pairs have different lengths, so the four sides are not all equal. That rules out the rhombus and the square straight away.

  6. Test the opposite sides for being parallel

    AB⃗=(3−2),DC⃗=(5−2)\vec{AB} = \binom{3}{-2}, \quad \vec{DC} = \binom{5}{-2}

    For AB and DC to be parallel, one vector would have to be a multiple of the other. Here 3 x (-2) - (-2) x 5=−6+10=45 = -6 + 10 = 4, which is not zero, so they are not parallel.

  7. Test the other pair of opposite sides

    AD⃗=(32),BC⃗=(52)\vec{AD} = \binom{3}{2}, \quad \vec{BC} = \binom{5}{2}

    Again 3 x 2 - 2 x 5=6−10=−45 = 6 - 10 = -4, which is not zero, so AD and BC are not parallel either. The shape has no parallel sides at all, so it is neither a parallelogram nor a trapezium.

  8. Work out the two diagonals

    AC⃗=(80),BD⃗=(04)\vec{AC} = \binom{8}{0}, \quad \vec{BD} = \binom{0}{4}

    AC runs 8 across and 0 up; BD runs 0 across and 4 up.

  9. Check whether the diagonals cross at right angles

    8×0+0×4=08 \times 0 + 0 \times 4 = 0

    The scalar product of the two diagonal vectors is zero, so AC and BD are perpendicular — exactly what happens in a kite.

  10. Check whether the diagonals bisect each other

    mid-point of AC=(4,0),mid-point of BD=(3,0)\text{mid-point of } AC = (4, 0), \quad \text{mid-point of } BD = (3, 0)

    The mid-points are different, so the diagonals do NOT bisect each other. That confirms the shape is not a parallelogram of any kind.

  11. Check which diagonal is cut in half

    mid-point of BD=(3,0) lies on AC\text{mid-point of } BD = (3, 0) \text{ lies on } AC

    The point (3, 0) is on the line AC, so the axis AC does cut BD in half, even though BD does not return the favour. This one-way bisection is a kite property.

  12. Rule out the rhombus and the square

    AB2=13≠29=BC2AB^2 = 13 \neq 29 = BC^2

    Both of those shapes need all four sides equal, and these are not.

  13. Rule out the parallelogram

    AB⃗≠DC⃗\vec{AB} \neq \vec{DC}

    A parallelogram needs both pairs of opposite sides parallel, and neither pair is.

  14. Rule out the trapezium

    no pair of sides is parallel\text{no pair of sides is parallel}

    A trapezium needs at least one pair of parallel sides, and this quadrilateral has none.

  15. State the name of the shape

    Kite\text{Kite}

    Two pairs of equal adjacent sides, no parallel sides and diagonals that cross at right angles: the quadrilateral is a kite.

Answer
Kite\text{Kite}
Question 3
6 markschallenging
Which one of these statements about the diagonals of a quadrilateral is always true?
Show worked solution

Worked solution

  1. Use the fact that the diagonals of a parallelogram bisect each other

    OA=OC,OB=ODOA = OC, \quad OB = OD

    In any parallelogram the diagonals cut each other in half at the crossing point O.

  2. Add the right angle at the crossing point

    ∠AOB=∠BOC=90∘\angle AOB = \angle BOC = 90^\circ

    The extra condition in the question says the diagonals meet at right angles, so every angle at O is a right angle.

  3. Compare triangle AOB with triangle COB

    OA=OC,OB=OB,∠AOB=∠COB=90∘OA = OC, \quad OB = OB, \quad \angle AOB = \angle COB = 90^\circ

    The two triangles share the side OB, have equal sides OA and OC, and have equal angles between them, so they are congruent.

  4. Read off the equal sides

    AB=CBAB = CB

    Matching sides of the congruent triangles are equal, so two sides of the parallelogram that are next to each other are equal.

  5. Use the parallelogram property

    AB=CD,BC=ADAB = CD, \quad BC = AD

    Opposite sides of a parallelogram are already equal, and now two adjacent sides are equal as well.

  6. Put the two facts together

    AB=BC=CD=DAAB = BC = CD = DA

    All four sides are the same length, and a quadrilateral with four equal sides is a rhombus.

  7. Rule out the first of the other statements

    counter-example 1\text{counter-example 1}

    "A quadrilateral whose diagonals cross at right angles must be a rhombus" is false. The diagonals of a kite cross at right angles, and a kite is not a rhombus.

  8. Rule out the second of the other statements

    counter-example 2\text{counter-example 2}

    "The diagonals of a rhombus are equal in length" is false. In a rhombus with angles of 60° and 120° the two diagonals have clearly different lengths; they are only equal when the rhombus is a square.

  9. Rule out the third of the other statements

    counter-example 3\text{counter-example 3}

    "A square is the only quadrilateral whose diagonals cross at right angles" is false. The diagonals of a rhombus and of a kite also cross at right angles.

  10. Rule out the fourth of the other statements

    counter-example 4\text{counter-example 4}

    "All four sides of a parallelogram are equal in length" is false. That is a rhombus. A general parallelogram only has its opposite sides equal.

  11. Check the converse is also true

    rhombus⇒AC⊥BD\text{rhombus} \Rightarrow AC \perp BD

    The diagonals of every rhombus do cross at right angles, so for a parallelogram the two conditions are exactly equivalent.

  12. Compare with the rectangle result

    equal diagonals⇒rectangle\text{equal diagonals} \Rightarrow \text{rectangle}

    Equal diagonals in a parallelogram force a rectangle; perpendicular diagonals force a rhombus. A shape with both is a square.

  13. Be careful about dropping the word parallelogram

    kite:AC⊥BD\text{kite}: AC \perp BD

    The diagonals of a kite also cross at right angles, and a kite is usually not a rhombus. It is only because the shape is already known to be a parallelogram that the conclusion follows.

  14. Check that only one statement survives

    4 false, 1 true\text{4 false, 1 true}

    Each of the other four statements has a counter-example, so exactly one statement is left standing.

  15. State the true statement

    true\text{true}

    A parallelogram whose diagonals cross at right angles must be a rhombus.

Answer
A parallelogram whose diagonals cross at right angles must be a rhombus\text{A parallelogram whose diagonals cross at right angles must be a rhombus}
Question 4
5 markschallenging
Which one of these statements about squares, rhombuses and rectangles is true?
Show worked solution

Worked solution

  1. Write down what a rhombus needs

    4 equal sides\text{4 equal sides}

    A quadrilateral is a rhombus exactly when all four of its sides are the same length. Nothing is said about its angles.

  2. Write down what a square needs

    4 equal sides,4 right angles\text{4 equal sides},\quad \text{4 right angles}

    A square needs four equal sides AND four right angles. So a square has everything a rhombus has, and more.

  3. Deduce that every square is a rhombus

    square⊂rhombus\text{square} \subset \text{rhombus}

    Every square has four equal sides, so every square is a rhombus. That already rules out the statements that say a rhombus can never be a square.

  4. Find a rhombus that is not a square

    60∘, 120∘, 60∘, 120∘60^\circ, \ 120^\circ, \ 60^\circ, \ 120^\circ

    A rhombus with angles of 60° and 120° has four equal sides but no right angles, so it is a rhombus that is not a square.

  5. Find a rhombus that is a square

    90∘, 90∘, 90∘, 90∘90^\circ, \ 90^\circ, \ 90^\circ, \ 90^\circ

    A rhombus whose angles happen to be right angles is a square. So some rhombuses are squares and some are not.

  6. Choose the right word

    sometimes\text{sometimes}

    Because there are examples of both kinds, the honest word is "sometimes": a rhombus is sometimes a square, never always and never never.

  7. Rule out the first of the other statements

    counter-example 1\text{counter-example 1}

    "A rhombus is always a square" is false. A rhombus with angles of 60° and 120° is not a square.

  8. Rule out the second of the other statements

    counter-example 2\text{counter-example 2}

    "A rhombus is never a square" is false. A square has four equal sides, so a square is a rhombus — a rhombus can be a square.

  9. Rule out the third of the other statements

    counter-example 3\text{counter-example 3}

    "A square is never a rhombus" is false. A square has four equal sides, which is exactly what makes a shape a rhombus.

  10. Rule out the fourth of the other statements

    counter-example 4\text{counter-example 4}

    "A rectangle is always a square" is false. A 6 cm by 3 cm rectangle is not a square.

  11. Say the same thing about rectangles

    square⊂rectangle\text{square} \subset \text{rectangle}

    A square is a rectangle whose sides happen to be equal, so a rectangle is also sometimes a square — but a 6 cm by 3 cm rectangle is not one.

  12. Place the square in the family

    square=rhombus∩rectangle\text{square} = \text{rhombus} \cap \text{rectangle}

    A square is exactly a shape that is both a rhombus and a rectangle: four equal sides and four right angles.

  13. Watch the direction of the statement

    square⇒rhombus,rhombus⇏square\text{square} \Rightarrow \text{rhombus}, \quad \text{rhombus} \not\Rightarrow \text{square}

    Statements like these are one-way streets. Every square is a rhombus, but that does not let you turn the sentence round.

  14. Check that only one statement survives

    4 false, 1 true\text{4 false, 1 true}

    Each of the other four statements has a counter-example, so exactly one statement is left standing.

  15. State the true statement

    true\text{true}

    A rhombus is sometimes a square.

Answer
A rhombus is sometimes a square\text{A rhombus is sometimes a square}
Question 5
5 markschallenging
The diagonals of a quadrilateral bisect each other. Which one of these statements about the quadrilateral is always true?
Show worked solution

Worked solution

  1. Set up the two triangles the diagonals create

    OA=OC,OB=ODOA = OC, \quad OB = OD

    Call the crossing point O. The diagonals bisect each other, so O is the mid-point of AC and also the mid-point of BD.

  2. Compare triangle AOB with triangle COD

    ∠AOB=∠COD\angle AOB = \angle COD

    Angle AOB and angle COD are vertically opposite angles where the diagonals cross, so they are equal.

  3. Match the two triangles

    OA=OC,OB=OD,∠AOB=∠CODOA = OC, \quad OB = OD, \quad \angle AOB = \angle COD

    Two sides and the angle between them match, so triangle AOB and triangle COD are congruent copies of each other.

  4. Read off what that gives about the sides

    AB=CDAB = CD

    Matching sides of the two congruent triangles are equal, so AB and CD are the same length.

  5. Read off what that gives about the direction of the sides

    ∠BAO=∠DCO⇒AB∥DC\angle BAO = \angle DCO \Rightarrow AB \parallel DC

    The matching angles are equal, and they are alternate angles at the line AC, so AB and DC are parallel.

  6. Do the same with the other pair of triangles

    BC∥AD,BC=ADBC \parallel AD, \quad BC = AD

    Triangles BOC and DOA are congruent in exactly the same way, so the other pair of opposite sides is equal and parallel too. Both pairs of opposite sides are parallel, which is the definition of a parallelogram.

  7. Rule out the first of the other statements

    counter-example 1\text{counter-example 1}

    "A quadrilateral whose diagonals bisect each other must be a rhombus" is false. The diagonals of a 6 cm by 3 cm rectangle bisect each other, and a rectangle is not a rhombus.

  8. Rule out the second of the other statements

    counter-example 2\text{counter-example 2}

    "A quadrilateral whose diagonals bisect each other must be a rectangle" is false. The diagonals of a rhombus with angles of 60° and 120° bisect each other, and it has no right angles.

  9. Rule out the third of the other statements

    counter-example 3\text{counter-example 3}

    "A quadrilateral whose diagonals bisect each other must be a kite" is false. The diagonals of a sloping parallelogram bisect each other, and it has no pair of equal adjacent sides.

  10. Rule out the fourth of the other statements

    counter-example 4\text{counter-example 4}

    "The diagonals of a trapezium always bisect each other" is false. Diagonals that bisect each other force a parallelogram, and a trapezium has only one pair of parallel sides.

  11. Test the claim on a rectangle

    rectangle: diagonals bisect each other\text{rectangle: diagonals bisect each other}

    A rectangle has diagonals that bisect each other, and a rectangle is a parallelogram, so it fits the true statement.

  12. Test the claim on a kite

    kite: only one diagonal is bisected\text{kite: only one diagonal is bisected}

    In a kite the axis of symmetry cuts the other diagonal in half, but it is not cut in half itself, so a kite does not satisfy the condition at all.

  13. Say what the statement does not claim

    parallelogram⇏rectangle\text{parallelogram} \not\Rightarrow \text{rectangle}

    Bisecting diagonals force a parallelogram and nothing more. To force a rectangle the diagonals would also have to be equal, and to force a rhombus they would also have to cross at right angles.

  14. Check that only one statement survives

    4 false, 1 true\text{4 false, 1 true}

    Each of the other four statements has a counter-example, so exactly one statement is left standing.

  15. State the true statement

    true\text{true}

    A quadrilateral whose diagonals bisect each other must be a parallelogram.

Answer
A quadrilateral whose diagonals bisect each other must be a parallelogram\text{A quadrilateral whose diagonals bisect each other must be a parallelogram}

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