Hard GCSE Notation and drawing Questions

Challenging, exam-style GCSE Notation and drawing questions with worked solutions. Stretch yourself on the hardest coordinates, side lengths, gradients, classifying quadrilaterals problems.

coordinatesside lengthsgradientsclassifying quadrilateralsright anglesdiagonal properties
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A quadrilateral has two pairs of parallel sides, all four sides equal in length, and diagonals that are equal in length. Give its name.
Show worked solution

Worked solution

  1. Write down the three clues

    2 pairs parallel;  4 equal sides;  equal diagonals2 \text{ pairs parallel};\; 4 \text{ equal sides};\; \text{equal diagonals}

    All three must hold at once.

  2. Use the first clue

    2 pairs parallel⇒parallelogram2 \text{ pairs parallel} \Rightarrow \text{parallelogram}

    Two pairs of parallel sides is the definition of a parallelogram.

  3. Eliminate the kite

    kite: 0 pairs parallel\text{kite: 0 pairs parallel}

    A kite that is not a rhombus has no parallel sides at all, so it fails the first clue.

  4. Eliminate the one-pair trapezium

    1≠21 \ne 2

    A trapezium with exactly one pair of parallel sides fails the first clue.

  5. Eliminate the isosceles trapezium

    1≠21 \ne 2

    An isosceles trapezium also has only one pair of parallel sides, so it fails too.

  6. Use the second clue

    4 equal sides⇒rhombus4 \text{ equal sides} \Rightarrow \text{rhombus}

    A parallelogram with all four sides equal is a rhombus.

  7. Eliminate the rectangle that is not a square

    6≠36 \ne 3

    A rectangle like (0, 0), (6, 0), (6, 3), (0, 3) has sides 6, 3, 6, 3, which are not all equal.

  8. Eliminate the parallelogram with no right angles

    AB2=25,  BC2=13AB^2 = 25, \; BC^2 = 13

    The parallelogram (0, 0), (5, 0), (7, 3), (2, 3) has unequal adjacent sides, so its four sides are not all equal.

  9. Take stock

    it must be a rhombus\text{it must be a rhombus}

    The first two clues force the shape to be a rhombus. The third clue must now separate the square from the other rhombuses.

  10. Test a rhombus that is not a square

    (0,0),(5,0),(8,4),(3,4)(0,0), (5,0), (8,4), (3,4)

    This rhombus has diagonals with squared lengths 80 and 20.

  11. Eliminate it

    80≠2080 \ne 20

    Its diagonals are NOT equal, so it fails the third clue. A rhombus that is not a square is ruled out.

  12. Explain why equal diagonals force right angles

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

    In a parallelogram the diagonals are equal exactly when all four angles are right angles, which makes it a rectangle.

  13. Combine the two names

    rhombus+rectangle=square\text{rhombus} + \text{rectangle} = \text{square}

    The shape is a rhombus (four equal sides) AND a rectangle (four right angles). A shape that is both is a square.

  14. Check with a real square

    (0,0),(4,0),(4,4),(0,4)(0,0), (4,0), (4,4), (0,4)

    Its diagonals are (4, 4) and (-4, 4), both with squared length 32, so they ARE equal. All four sides are 4. Every clue is satisfied.

  15. State the answer

    Square\text{Square}

    The only quadrilateral satisfying all three clues is the square.

Answer
Square\text{Square}
Question 2
6 markschallenging
Ollie says: "If a quadrilateral has two pairs of equal sides then it must be a parallelogram." Give a counterexample and explain why Ollie is wrong.
Show worked solution

Worked solution

  1. Read the claim carefully

    2 pairs of equal sides⇒parallelogram?2 \text{ pairs of equal sides} \Rightarrow \text{parallelogram?}

    Ollie has not said WHERE the equal sides are. That is the weakness in the claim.

  2. Say what a parallelogram needs

    equal sides must be OPPOSITE\text{equal sides must be OPPOSITE}

    In a parallelogram the two pairs of equal sides are opposite each other.

  3. Spot the other possibility

    equal sides could be ADJACENT\text{equal sides could be ADJACENT}

    Two pairs of equal sides could instead sit next to each other. That is a kite.

  4. Choose a counterexample

    A(0,4),B(3,0),C(0,−6),D(−3,0)A(0,4), B(3,0), C(0,-6), D(-3,0)

    Take this quadrilateral, with the vertices in order.

  5. Find AB

    AB2=32+(−4)2=25⇒AB=5AB^2 = 3^2 + (-4)^2 = 25 \Rightarrow AB = 5

    A to B is 3 across and 4 down.

  6. Find DA

    DA2=32+42=25⇒DA=5DA^2 = 3^2 + 4^2 = 25 \Rightarrow DA = 5

    D to A is 3 across and 4 up.

  7. Find BC

    BC2=(−3)2+(−6)2=45BC^2 = (-3)^2 + (-6)^2 = 45

    B to C is 3 left and 6 down.

  8. Find CD

    CD2=(−3)2+62=45CD^2 = (-3)^2 + 6^2 = 45

    C to D is 3 left and 6 up.

  9. Confirm two pairs of equal sides

    AB=DA=5,BC=CD=45AB = DA = 5, \quad BC = CD = \sqrt{45}

    There really are two pairs of equal sides, so the shape satisfies Ollie's condition.

  10. Notice where the equal sides are

    AB,DA meet at AAB, DA \text{ meet at } A

    The two 5s meet at A and the two root-45s meet at C. The equal sides are adjacent, not opposite.

  11. Test the first pair of opposite sides

    mAB=−43,mDC=−2m_{AB} = -\frac{4}{3}, \quad m_{DC} = -2

    AB and DC have different gradients, so they are not parallel.

  12. Test the second pair of opposite sides

    mBC=2,mAD=43m_{BC} = 2, \quad m_{AD} = \frac{4}{3}

    BC and AD have different gradients too, so they are not parallel.

  13. Conclude it is not a parallelogram

    0 pairs of parallel sides\text{0 pairs of parallel sides}

    No pair of opposite sides is parallel, so the shape is definitely not a parallelogram.

  14. Name the counterexample

    kite\text{kite}

    Two pairs of ADJACENT equal sides makes this a kite, and a kite is not a parallelogram.

  15. State what Ollie should have said

    equal OPPOSITE sides⇒parallelogram\text{equal OPPOSITE sides} \Rightarrow \text{parallelogram}

    Ollie is wrong: the equal sides must be OPPOSITE each other. If they are adjacent you get a kite instead.

Answer
Wrong: a kite is a counterexample\text{Wrong: a kite is a counterexample}
Question 3
6 markschallenging
A quadrilateral has vertices A(0, 0), B(4, 2), C(8, 0) and D(4, -6), joined in that order. How many lines of symmetry does it have?
Show worked solution

Worked solution

  1. Find the length of AB

    AB2=42+22=20AB^2 = 4^2 + 2^2 = 20

    A(0, 0) to B(4, 2) is 4 across and 2 up.

  2. Find the length of BC

    BC2=42+(−2)2=20BC^2 = 4^2 + (-2)^2 = 20

    B(4, 2) to C(8, 0) is 4 across and 2 down.

  3. Find the length of CD

    CD2=(−4)2+(−6)2=52CD^2 = (-4)^2 + (-6)^2 = 52

    C(8, 0) to D(4, -6) is 4 left and 6 down.

  4. Find the length of DA

    DA2=(−4)2+62=52DA^2 = (-4)^2 + 6^2 = 52

    D(4, -6) to A(0, 0) is 4 left and 6 up.

  5. Compare the sides

    AB=BC=20,CD=DA=52AB = BC = \sqrt{20}, \quad CD = DA = \sqrt{52}

    The equal sides come in adjacent pairs (both root-20 sides meet at B, both root-52 sides meet at D).

  6. Rule out the rhombus

    20≠5220 \ne 52

    The sides are not all equal, so this is a kite, not a rhombus.

  7. Check for parallel sides

    mAB=12,mDC=64=32m_{AB} = \frac{1}{2}, \quad m_{DC} = \frac{6}{4} = \frac{3}{2}

    The gradients differ, so AB is not parallel to DC. The other pair differs too, so a kite has no parallel sides.

  8. Find the diagonal AC

    A(0,0)→C(8,0)A(0,0) \rightarrow C(8,0)

    AC is the horizontal segment along y=0y = 0 from (0, 0) to (8, 0).

  9. Find the diagonal BD

    B(4,2)→D(4,−6)B(4,2) \rightarrow D(4,-6)

    BD is the vertical segment along x=4x = 4 from (4, 2) to (4, -6).

  10. Test BD as a mirror line

    x=4:  A(0,0)↔C(8,0)x = 4: \; A(0,0) \leftrightarrow C(8,0)

    Reflecting in x=4x = 4 sends A(0, 0) to (8, 0), which is C, and sends C to A. B and D lie on the line and stay put. So BD is a line of symmetry.

  11. Test AC as a mirror line

    y=0:  B(4,2)↦(4,−2)≠D(4,−6)y = 0: \; B(4,2) \mapsto (4, -2) \ne D(4,-6)

    Reflecting in y=0y = 0 sends B(4, 2) to (4, -2), but D is at (4, -6). The shape does not map onto itself, so AC is NOT a mirror line.

  12. Explain the failure

    20≠52\sqrt{20} \ne \sqrt{52}

    A fold along AC would have to swap B and D, which needs the sides at B to match the sides at D. They are root 20 and root 52, so they do not.

  13. Rule out any other mirror line

    a mirror line must pass through the centre\text{a mirror line must pass through the centre}

    Any mirror line of a quadrilateral must either be a diagonal or join the midpoints of opposite sides. Neither mid-line works here, since opposite sides have different lengths.

  14. Check the rotational symmetry

    order 1\text{order } 1

    A half turn would need opposite sides to be equal, and AB is root 20 while CD is root 52, so the order of rotational symmetry is only 1.

  15. Count the lines of symmetry

    11

    The kite has exactly 1 line of symmetry, the diagonal BD.

Answer
11
Question 4
6 markschallenging
A quadrilateral has vertices A(-4, 3), B(0, 6), C(4, 3) and D(0, -6), joined in that order. Name the quadrilateral, and say which of its diagonals is a line of symmetry.
Show worked solution

Worked solution

  1. Find the length of AB

    AB2=42+32=25⇒AB=5AB^2 = 4^2 + 3^2 = 25 \Rightarrow AB = 5

    A(-4, 3) to B(0, 6) is 4 across and 3 up.

  2. Find the length of BC

    BC2=42+(−3)2=25⇒BC=5BC^2 = 4^2 + (-3)^2 = 25 \Rightarrow BC = 5

    B(0, 6) to C(4, 3) is 4 across and 3 down.

  3. Find the length of CD

    CD2=(−4)2+(−9)2=97CD^2 = (-4)^2 + (-9)^2 = 97

    C(4, 3) to D(0, -6) is 4 left and 9 down.

  4. Find the length of DA

    DA2=(−4)2+92=97DA^2 = (-4)^2 + 9^2 = 97

    D(0, -6) to A(-4, 3) is 4 left and 9 up.

  5. Compare the sides

    AB=BC=5,CD=DA=97AB = BC = 5, \quad CD = DA = \sqrt{97}

    The equal sides are ADJACENT: AB and BC both meet at B, and CD and DA both meet at D.

  6. Rule out the rhombus

    25≠9725 \ne 97

    The four sides are not all equal, so it is not a rhombus. Two pairs of adjacent equal sides makes it a kite.

  7. Check there are no parallel sides

    mAB=34,mDC=94m_{AB} = \frac{3}{4}, \quad m_{DC} = \frac{9}{4}

    AB and DC have different gradients, so they are not parallel; the same is true of the other pair. So it is not a parallelogram or a trapezium.

  8. Find the diagonal AC

    A(−4,3)→C(4,3)A(-4,3) \rightarrow C(4,3)

    AC is the horizontal segment from (-4, 3) to (4, 3).

  9. Find the diagonal BD

    B(0,6)→D(0,−6)B(0,6) \rightarrow D(0,-6)

    BD is the vertical segment from (0, 6) to (0, -6), which lies on the line x=0x = 0.

  10. Test BD as a mirror line

    x=0:  A(−4,3)↔C(4,3)x = 0: \; A(-4,3) \leftrightarrow C(4,3)

    Reflecting in the line x=0x = 0 sends A(-4, 3) to (4, 3), which is C, and sends C to A. B and D lie ON the line, so they stay put. The kite maps exactly onto itself.

  11. Test AC as a mirror line

    y=3:  B(0,6)↦(0,0)≠Dy = 3: \; B(0,6) \mapsto (0, 0) \ne D

    Reflecting in the line y=3y = 3 sends B(0, 6) to (0, 0), but D is at (0, -6). The shape does not map onto itself, so AC is NOT a line of symmetry.

  12. Explain which diagonal works

    the axis joins B and D\text{the axis joins } B \text{ and } D

    The mirror line of a kite joins the two vertices where the EQUAL sides meet: here that is B (where the two 5s meet) and D (where the two root-97s meet).

  13. Note the equal angles

    ∠A=∠C\angle A = \angle C

    Reflection in BD swaps A and C, so the angles at A and C are equal. These are the angles between the sides of DIFFERENT lengths, and they are the kite's single pair of equal opposite angles.

  14. Check the diagonals are perpendicular

    AC⃗⋅BD⃗=(8)(0)+(0)(−12)=0\vec{AC} \cdot \vec{BD} = (8)(0) + (0)(-12) = 0

    AC is horizontal and BD is vertical, so they cross at right angles, as in every kite.

  15. State the answer

    kite; axis BD\text{kite; axis } BD

    ABCD is a kite that is not a rhombus, and its line of symmetry is the diagonal BD.

Answer
Kite (not a rhombus); line of symmetry BD\text{Kite (not a rhombus); line of symmetry } BD
Question 5
5 markschallenging
How many diagonals does a dodecagon have?
Show worked solution

Worked solution

  1. Name the polygon

    dodeca-=12\text{dodeca-} = 12

    The prefix "dodeca" means twelve, so a dodecagon has 12 sides.

  2. Count the vertices

    n=12n = 12

    A 12-sided polygon has 12 vertices.

  3. Recall what a diagonal is

    joins non-adjacent vertices\text{joins non-adjacent vertices}

    A diagonal joins two vertices that are not next to each other.

  4. Count the vertices you cannot join to

    itself+2 neighbours=3\text{itself} + 2 \text{ neighbours} = 3

    From any one vertex, three vertices are unavailable: the vertex itself and the two either side of it.

  5. Count the diagonals from one vertex

    12−3=912 - 3 = 9

    That leaves 9 vertices you CAN join to with a diagonal.

  6. Count over all the vertices

    12×9=10812 \times 9 = 108

    Each of the 12 vertices has 9 diagonals leaving it, giving 108.

  7. Spot the double counting

    AC and CA are the same diagonalAC \text{ and } CA \text{ are the same diagonal}

    The diagonal from A to C is exactly the same line as the diagonal from C to A, but it has been counted at both ends.

  8. Halve the total

    1082=54\frac{108}{2} = 54

    Dividing by 2 removes every duplicate.

  9. Write the general formula

    n(n−3)2\frac{n(n-3)}{2}

    This argument works for any polygon, giving the standard formula.

  10. Check the formula on a quadrilateral

    4×12=2\frac{4 \times 1}{2} = 2

    A quadrilateral should have 2 diagonals, and the formula agrees.

  11. Check the formula on a pentagon

    5×22=5\frac{5 \times 2}{2} = 5

    A pentagon should have 5 diagonals, and the formula agrees.

  12. Check the formula on a hexagon

    6×32=9\frac{6 \times 3}{2} = 9

    A hexagon should have 9 diagonals, and the formula agrees.

  13. Apply the formula to n=12n = 12

    12×92=1082\frac{12 \times 9}{2} = \frac{108}{2}

    Substituting n=12n = 12 gives 12 times 9, divided by 2.

  14. Work out the value

    1082=54\frac{108}{2} = 54

    This comes to 54.

  15. State the answer

    5454

    A dodecagon has 54 diagonals.

Answer
5454

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