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Either paper 3 marks
Here is a quadrilateral.
Work out the size of the angle marked \(x\).
Show answer
\(110^\circ\)The interior angles of a quadrilateral add to \(360^\circ\).
\(90 + 70 + 90 = 250\)
\(x = 360 - 250 = 110\)
\(x = 110^\circ\)
- M1angle sum of a quadrilateral \(360^\circ\)
- M1sum of the given angles \(250\)
- A1\(x = 110^\circ\)
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Either paper 4 marks
The diagram shows a hexagon.
(a)Find the sum of the interior angles of a hexagon.
(b)Find the size of the angle marked \(x\).
Show answer
(a) \(720^\circ\)
(b) \(135^\circ\)A hexagon splits into \(6 - 2 = 4\) triangles, and the angles of each triangle add to \(180^\circ\).
\(4 \times 180 = 720\)
The interior angles add to \(720^\circ\).
Add the five angles that are given:
\(130 + 115 + 110 + 110 + 120 = 585\)
Take that from the total for a hexagon:
\(x = 720 - 585 = 135\)
\(x = 135^\circ\)
- (a) M1\(4\) triangles, \(4 \times 180\)
- (a) A1\(720^\circ\)
- (b) M1sum of the five given angles \(585\)
- (b) A1\(x = 720 - 585 = 135^\circ\)
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Either paper 3 marks
Here is a pentagon.
Find the size of the angle marked \(x\).
Show answer
\(93^\circ\)A pentagon splits into \(5 - 2 = 3\) triangles, so its interior angles add to \(3 \times 180 = 540^\circ\).
\(98 + 108 + 133 + 108 = 447\)
\(x = 540 - 447 = 93\)
\(x = 93^\circ\)
- M1angle sum of a pentagon \(540^\circ\)
- M1sum of the given angles \(447\)
- A1\(x = 93^\circ\)
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Either paper 3 marks
A regular polygon has \(40\) sides.
(a)Work out the size of each exterior angle.
(b)Work out the size of each interior angle.
Show answer
(a) \(9^\circ\)
(b) \(171^\circ\)The exterior angles of any polygon add to \(360^\circ\), and in a regular polygon they are all equal.
\(360 \div 40 = 9\)
Each exterior angle is \(9^\circ\).
An interior angle and the exterior angle at the same vertex lie on a straight line.
\(180 - 9 = 171\)
Each interior angle is \(171^\circ\).
- (b) M1exterior angle \(9\) and \(180 - 9\)
- (b) A1interior angle \(171^\circ\)
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Either paper 4 marks
Here is a regular dodecagon.
(a)Calculate the size of one exterior angle of the dodecagon.
(b)Calculate the size of the angle marked \(x\).
Show answer
(a) \(30^\circ\)
(b) \(150^\circ\)Walking once round the outside of any polygon turns you through \(360^\circ\) in total, and in a regular polygon each turn is the same.
A dodecagon has \(12\) sides, so
\(360 \div 12 = 30\)
Each exterior angle is \(30^\circ\).
\(x\) is an interior angle. An interior angle and the exterior angle at the same corner lie on a straight line, so they add to \(180^\circ\).
\(x = 180 - 30 = 150\)
\(x = 150^\circ\)
- (a) M1\(360 \div 12\)
- (a) A1exterior angle \(30^\circ\)
- (b) M1interior angle = \(180 - 30\)
- (b) A1\(x = 150^\circ\)
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Either paper 2 marks
The sum of the angles in any quadrilateral is \(360^\circ\).
A quadrilateral can be split into 2 triangles, so \(2 \times 180^\circ = 360^\circ\).
Idris writes \(3 \times 180^\circ = 540^\circ\), so the sum of the angles in any pentagon must be \(540^\circ\).
Is Idris correct?
Tick a box. Yes / No
Show working to support your answer.
Show answer
Yes. The angles in a pentagon add up to \(540^\circ\), and Idris claimed \(540^\circ\).A pentagon has 5 sides, so it can be split into \(5 - 2 = 3\) triangles.
Each triangle has angles adding to \(180^\circ\), so the sum of the angles in a pentagon is
\(3 \times 180^\circ = 540^\circ\)
Compare this with Idris's figure of \(540^\circ\).
Idris's figure is correct, so the answer is Yes.
- M1finds the number of triangles from the 5 sides, \(5 - 2 = 3\) (or \((5 - 2) \times 180\)), giving \(3 \times 180 = 540\) for a pentagon; repeating the pupil's own product without deriving the 3 earns nothing
- A1Yes, comparing 540 with Idris's 540
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Either paper 3 marks
Part of a regular polygon is shown. The jagged edge is where the rest of it has been torn off.
The diagram is not drawn accurately.
(a)Assume that the polygon is a nonagon.
Calculate the size of angle \(x\).
(b)In fact, the polygon has fewer sides than a nonagon.
What does this mean about the size of angle \(x\)? Tick one box.
A. It is the same as the answer to part (a)
B. It is less than the answer to part (a)
C. It is more than the answer to part (a)
D. It could be any of the above
Show answer
(a) \(40^\circ\)
(b) C. It is more than the answer to part (a)The exterior angles of a regular polygon are all equal and add to \(360^\circ\).
\(360 \div 9 = 40\)
The exterior angle is \(40^\circ\) and the interior angle is \(180 - 40 = 140^\circ\).
Angle \(x\) is the exterior angle, so \(x = 40^\circ\).
An exterior angle is \(360^\circ\) divided by the number of sides, so the more sides there are, the smaller each exterior angle is. An interior angle is \(180^\circ\) minus the exterior angle, so it moves the opposite way.
Angle \(x\) is the exterior angle, and the polygon has fewer sides than before, so angle \(x\) is larger than the answer to part (a).
The correct box is: It is more than the answer to part (a).
- (a) M1exterior angle \(360 \div 9 = 40\)
- (a) A1angle \(x = 40^\circ\)
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Either paper 3 marks
Each interior angle of a regular polygon is \(144^\circ\).
Calculate the number of sides the polygon has.
Show answer
\(10\ \text{sides}\)An interior angle and its exterior angle make a straight line, so start with the exterior angle.
\(180 - 144 = 36\)
The exterior angles add to \(360^\circ\) and are all equal, so
\(10 = 360 \div 36 = 10\)
The polygon has \(10\) sides.
- M1exterior angle \(180 - 144 = 36\)
- M1\(360 \div 36\)
- A1\(10\) sides
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Either paper 4 marks
The interior angle of a regular polygon with \(5\) sides is \(108^\circ\).
Here is the regular polygon.
(a)At each vertex of the polygon there is a reflex angle outside the polygon.
Calculate the sum of the reflex angles at \(4\) of the vertices.
(b)Harry is asked how many lines of symmetry a heptagon has. He assumes it is regular.
State the number of lines of symmetry of a regular heptagon.
(c)Omar then says, "What if the heptagon is not regular?"
For a heptagon that is not regular, which statement is true? Write down the letter of the true statement.
A. There could be 0 or 1
B. There could be 0, 1 or 2
C. There must be 0
D. There could be any number up to 7
Show answer
(a) \(1008^\circ\)
(b) \(7\ \text{lines}\)
(c) A. There could be 0 or 1At each vertex the interior angle and the reflex angle make a full turn.
\(360 - 108 = 252\)
There are \(4\) of these vertices, so
\(4 \times 252 = 1008\)
The sum of the reflex angles is \(1008^\circ\).
A regular polygon has one line of symmetry for each side.
A regular heptagon has \(7\) lines of symmetry.
Only the regular heptagon has \(7\) lines of symmetry, and a polygon that is not regular can always have no lines at all or just one.
Any other number of lines of symmetry has to divide exactly into \(7\) and be smaller than \(7\).
So the true statement is: There could be 0 or 1.
- (a) M1reflex angle \(360 - 108 = 252\) and multiplying by \(4\)
- (a) A1sum \(1008^\circ\)
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Either paper 5 marks
In the diagram, \(CD\) and \(FG\) are parallel lines. \(FC\) and \(GD\) are straight lines.
The lines \(FC\) and \(GD\) are extended to meet at the point \(A\).
Show that triangle \(ACD\) is isosceles.
Give a reason for each stage of your working.
Show answer
\(\angle ADC = 180 - 133 = 47^\circ\) (angles on a straight line add up to \(180^\circ\)); \(\angle ACD = 47^\circ\) (corresponding angles, parallel lines); two equal angles, so the triangle is isoscelesFind one base angle of the triangle from the angle on its straight line.
\(\angle ADC + \angle CDG = 180^\circ\) (angles on the straight line \(ADG\) add up to \(180^\circ\))
\(\angle ADC = 180 - 133 = 47^\circ\)
Find the other base angle from the parallel lines.
\(\angle ACD = \angle GFC = 47^\circ\) (corresponding angles, because \(CD\) is parallel to \(FG\))
\(\angle ADC = \angle ACD = 47^\circ\), so triangle \(ACD\) has two equal angles and is isosceles.
- B1\(\angle ADC = 180 - 133 = 47^\circ\)
- C1reason: angles on the straight line ADG add up to 180°
- B1\(\angle ACD = 47^\circ\)
- C1reason: corresponding angles, because CD is parallel to FG
- C1two equal angles, so the triangle is isosceles