4MA1 Polygons

A quadrilateral has diagonals that are equal in length, bisect each other and cross at right angles.

Write down the mathematical name of the quadrilateral. [1]

Why this scores

Mark scheme:

• B1 for square

Why this scores full marks: Equal diagonals that bisect each other rule in a rectangle or a square; diagonals crossing at right angles rule in a rhombus, a square or a kite. Only a square has all three properties.

A regular polygon has $40$ lines of symmetry.

Work out the size of each interior angle of the polygon. [2]

Why this scores

Mark scheme:

• M1 for $360 \div 40$ (= $9$) or $(40 - 2) \times 180 \div 40$ oe
• A1 for $171$ cao

Why this scores full marks: The candidate turns the symmetry fact into the number of sides first — a regular polygon with $n$ sides has $n$ lines of symmetry and rotational symmetry of order $n$ — and then uses the exterior angle. Giving $9^{\circ}$ (the exterior angle) as the answer is the usual way to lose the A1.

Write down two properties of the diagonals of a rhombus that are not always properties of the diagonals of a parallelogram. [2]

Why this scores

Mark scheme:

• B1 for one correct property, e.g. the diagonals are perpendicular
• B1 for a second correct property, e.g. the diagonals bisect the angles of the rhombus (oe, such as "each diagonal is a line of symmetry")

Why this scores full marks: Both properties are true of every rhombus but not of every parallelogram. "The diagonals bisect each other" scores nothing, because every parallelogram has that property too; "the diagonals are equal" is wrong, because that is a property of a rectangle, not a rhombus.

The angles $DAB$, $ABC$, $BCD$ and $CDA$ of the quadrilateral $ABCD$ are in the ratio 1 : 2 : 3 : 4

Show that $AB$ is parallel to $DC$. [3]

Why this scores

Mark scheme:

• M1 for $360 \div 10$ (= $36$)
• A1 for angles $DAB = 36$ and $CDA = 144$ (or $ABC = 72$ and $BCD = 108$)
• A1 for $36 + 144 = 180$ (or $72 + 108 = 180$) and a conclusion that co-interior angles add up to $180^{\circ}$, so $AB$ is parallel to $DC$

Why this scores full marks: The candidate picks the pair of angles that lie between $AB$ and $DC$ — the two angles at the ends of side $AD$ — shows that they add up to exactly $180^{\circ}$, and ends with a statement. Adding the angles at $A$ and $B$ ($36 + 72 = 108$) tests the wrong pair of lines. So $ABCD$ is a trapezium.

Each exterior angle of a regular polygon is $(2x - 30)^{\circ}$.

Each interior angle of the same polygon is $(7x + 30)^{\circ}$.

Work out the number of sides of the polygon. [4]

Why this scores

Mark scheme:

• M1 for $2x - 30 + 7x + 30 = 180$ oe
• A1 for $x = 20$
• M1 for $360 \div (2 \times 20 - 30)$ or $360 \div$ their exterior angle
• A1 for $36$ cao

Why this scores full marks: The candidate substitutes $x$ back into the exterior angle before dividing into $360$, and checks that the interior angle, $7(20) + 30 = 170^{\circ}$, pairs with $10^{\circ}$ to make $180^{\circ}$. Dividing $360$ by $x = 20$ itself gives $18$, a common wrong answer.

The sum of the interior angles of a polygon is $6$ times the sum of its exterior angles.

(a) Work out the number of sides of the polygon. [2]

(b) The polygon is regular. Work out the size of each interior angle. Give your answer correct to $1$ decimal place. [2]

Why this scores

Mark scheme:

• M1 for $(n - 2) \times 180 = 6 \times 360$ oe (a)
• A1 for $14$ cao (a)
• M1 for $2160 \div 14$ or $180 - 360 \div 14$ (b)
• A1 for awrt $154.3$ (b)

Why this scores full marks: The candidate uses the fact that the exterior angles always total $360^{\circ}$, writes down the unrounded value $154.285\ldots$ and then rounds once, to the accuracy asked for.

$ABCDEF$ is a hexagon. The diagonal $AD$ is a line of symmetry of the hexagon.

Angle $FAB = 96^{\circ}$ and angle $CDE = 132^{\circ}$.

Angle $ABC$ is $42^{\circ}$ larger than angle $BCD$.

Work out the size of angle $ABC$. Give a reason for each stage of your working. [5]

Why this scores

Mark scheme:

• B1 for $720$ as the angle sum of a hexagon
• M1 for using the symmetry: angle $AFE$ = angle $ABC$ and angle $DEF$ = angle $BCD$
• M1 for $96 + 132 + 2(x + 42) + 2x = 720$ oe
• A1 for $144$ cao
• B1 for reasons including the angle sum of a polygon is $(n - 2) \times 180^{\circ}$ and angles that are reflections of each other in a line of symmetry are equal

Why this scores full marks: The candidate uses the line of symmetry to pair up the four unknown angles, so there are only two different ones. Check: $96 + 144 + 102 + 132 + 102 + 144 = 720$. Scripts that count each unknown angle once write $228 + (x + 42) + x = 720$ and get $x = 225$; listing all six angles then gives a total of $1212^{\circ}$, not $720^{\circ}$, so a quick check exposes the error.

$ABCDE$ is a regular pentagon. The diagonal $BE$ is drawn.

Show that $BE$ is parallel to $CD$. Give a reason for each stage of your working. [5]

Why this scores

Mark scheme:

• B1 for $108$ as the interior angle
• M1 for $(180 - 108) \div 2$ (= $36$)
• M1 for $108 - 36$ (= $72$)
• A1 for $72 + 108 = 180$ and the conclusion that co-interior angles add up to $180^{\circ}$, so $BE$ is parallel to $CD$
• B1 for reasons including base angles of an isosceles triangle are equal and angles in a triangle add up to $180^{\circ}$

Why this scores full marks: The candidate works forwards from facts about the pentagon and never assumes the lines are parallel. The final line names the pair of co-interior angles, shows they total exactly $180^{\circ}$ and states the conclusion — in a "show that" the A1 is for that complete argument, not for the number $72$.

The interior angle of a regular polygon with $n$ sides is $(10n + 30)^{\circ}$.

Find the two possible values of $n$. Show clear algebraic working. [5]

Why this scores

Mark scheme:

• M1 for $180 - \dfrac{360}{n}$ or $\dfrac{(n - 2) \times 180}{n}$
• M1 for equating to $10n + 30$ and clearing the fraction
• A1 for $n^{2} - 15n + 36 = 0$ oe
• M1 for $(n - 3)(n - 12)$ or a correct substitution into the quadratic formula
• A1 for $3$ and $12$

Why this scores full marks: Both roots are positive whole numbers of at least $3$, so both are valid and both are given, with a check of each. Trying values of $n$ one at a time is not the algebraic working the question requires.

$ABCD$ is an isosceles trapezium with $AB$ parallel to $DC$ and $AD = BC$. The diagonals $AC$ and $BD$ are drawn.

Angle $ADC = 116^{\circ}$ and angle $BDC = 28^{\circ}$.

(a) Give a reason why angle $BCD = 116^{\circ}$. [1]

(b) Work out the size of angle $DAC$. Give a reason for each stage of your working. [4]

Why this scores

Mark scheme:

• B1 for a correct reason in (a), e.g. the base angles of an isosceles trapezium are equal, or the trapezium is symmetrical (a)
• M1 for angle $ACD = 28$ (b)
• M1 for $180 - 116 - 28$, or angle $DAB = 64$ (co-interior angles) and $64 - 28$ with angle $BAC = 28$ (b)
• A1 for $36$ cao (b)
• B1 for reasons including congruent triangles (or symmetry of the isosceles trapezium) for angle $ACD$ = angle $BDC$, and angles in a triangle add up to $180^{\circ}$ (b)

Why this scores full marks: The key step is angle $ACD = 28^{\circ}$, and the candidate proves it instead of assuming it: the diagonals of an isosceles trapezium are mirror images of each other in its line of symmetry. The common error is to use angle $ADC$ and angle $BDC$ in the same triangle, or to take angle $ACD$ to be angle $BDC$ "because they are alternate" — they are not alternate angles.