Hard GCSE Notation and drawing Questions

Write down the three clues

All three must hold at once.

Use the first clue

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

Eliminate the kite

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

Eliminate the one-pair trapezium

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

Eliminate the isosceles trapezium

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

Use the second clue

A parallelogram with all four sides equal is a rhombus.

Eliminate the rectangle that is not a square

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

Eliminate the parallelogram with no right angles

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

Take stock

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

Test a rhombus that is not a square

This rhombus has diagonals with squared lengths $80$ and $20$.

Eliminate it

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

Explain why equal diagonals force right angles

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

Combine the two names

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

Check with a real square

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.

State the answer

The only quadrilateral satisfying all three clues is the square.

Read the claim carefully

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

Say what a parallelogram needs

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

Spot the other possibility

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

Choose a counterexample

Take this quadrilateral, with the vertices in order.

Find $AB$

A to $B$ is $3$ across and $4$ down.

Find $DA$

$D$ to A is $3$ across and $4$ up.

Find $BC$

$B$ to $C$ is $3$ left and $6$ down.

Find $CD$

$C$ to $D$ is $3$ left and $6$ up.

Confirm two pairs of equal sides

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

Notice where the equal sides are

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

Test the first pair of opposite sides

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

Test the second pair of opposite sides

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

Conclude it is not a parallelogram

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

Name the counterexample

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

State what Ollie should have said

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

Find the length of $AB$

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

Find the length of $BC$

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

Find the length of $CD$

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

Find the length of $DA$

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

Compare the sides

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

Rule out the rhombus

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

Check for parallel sides

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

Find the diagonal $AC$

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

Find the diagonal $BD$

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

Test $BD$ as a mirror line

Reflecting in $x = 4$ 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.

Test $AC$ as a mirror line

Reflecting in $y = 0$ 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.

Explain the failure

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.

Rule out any other mirror line

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.

Check the rotational symmetry

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 $1$.

Count the lines of symmetry

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

Find the length of $AB$

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

Find the length of $BC$

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

Find the length of $CD$

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

Find the length of $DA$

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

Compare the sides

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

Rule out the rhombus

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

Check there are no parallel sides

$AB$ 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.

Find the diagonal $AC$

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

Find the diagonal $BD$

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

Test $BD$ as a mirror line

Reflecting in the line $x = 0$ 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.

Test $AC$ as a mirror line

Reflecting in the line $y = 3$ 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.

Explain which diagonal works

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).

Note the equal angles

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.

Check the diagonals are perpendicular

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

State the answer

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

Name the polygon

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

Count the vertices

A $12$-sided polygon has $12$ vertices.

Recall what a diagonal is

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

Count the vertices you cannot join to

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

Count the diagonals from one vertex

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

Count over all the vertices

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

Spot the double counting

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.

Halve the total

Dividing by $2$ removes every duplicate.

Write the general formula

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

Check the formula on a quadrilateral

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

Check the formula on a pentagon

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

Check the formula on a hexagon

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

Apply the formula to $n = 12$

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

Work out the value

This comes to $54$.

State the answer

A dodecagon has $54$ diagonals.