Cambridge Essentials Mathematics Core 9 GM1.1 Answers
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Cambridge Essentials Mathematics Core 9 GM1.1 Answers
GM1.1 Answers
1 a
b
2
Shape Name Regular Irregular Convex Concave A Decagon B Octagon C Pentagon D Quadrilateral E Heptagon F Hexagon G Quadrilateral H Triangle I Triangle J Hexagon
Original Material © Cambridge University Press 2010 1 Cambridge Essentials Mathematics Core 9 GM1.1 Answers
3 Check that students have drawn a regular, an irregular convex and a concave pentagon.
Here are some examples:
4 a Sides must all be of equal length and the interior angles must be equal.
b A regular quadrilateral is a square.
c
5 a The sum of the interior angles increases by 180° each time.
b
Number of sides 3 4 5 6 7 8 Sum of interior angles 180 360 540 720 900 1080
c i 160° ii 115° iii 155° iv 120° v 141° iv 97°
6 a A regular pentagon has five equal angles. The sum of these angles is 540°. Divide this
number
by 5. This gives 108° for each interior angle.
b The interior angle of a regular octagon = 135°
c The interior angle of a dodecagon = 150°
Original Material © Cambridge University Press 2010 2 Cambridge Essentials Mathematics Core 9 GM1.1 Answers
7 The missing angle = 900° – 748° = 152°
8 a (n – 2) × 180° works out the sum of the interior angles of a polygon with n sides.
b i (5 – 2) × 180° = 540°; x = 540° – 395 =145° and y = 180° – 145 = 35°
ii (8 – 2) × 180° = 1080°; x = 1080° – 937° = 143° and y = 180° – 143 = 37°
iii (4 – 2) × 180° = 360°; x = 360° – 275° = 85° and y = 180° – 85° = 95°
9 The formula says that the sum of the interior angles of a quadrilateral add to 360°.
Alternatively, divide the shape into two triangles.
As you can see, adding up the angles in each triangle is the same as adding up the four
interior angles of the quadrilateral. Therefore, the sum of the interior angles of the concave
quadrilateral is 2 × 180° = 360°. This is the same as the answer that the formula gave. This
indicates that the formula works for convex and concave polygons.
10 a = 268°, b = 139°, c = 253°
11 a 144° b 157.5°
12 a (n – 2) × 180° = (7 – 2) × 180° = 5 × 180° = 900°
b Each interior angle of a regular heptagon is 900° ÷ 7 =128.6°
c Each side should be 4 cm and each angle 128.6.
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13 a
b Exterior angles add up to 360°, which is a full turn.
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14 i a, b and c
ii Exterior angles total 360° for each polygon.
15 a 40°, 140° b 30°, 150° c 72°, 108°
16 Use the formula (n – 2) × 180° to find the sum of the interior angles of a hexagon:
(n – 2) × 180° = (6 – 2) × 180° = 4 × 180° = 720°
Then divide this answer by 6 to find each interior angle:
Each interior angle = 720° ÷ 6 = 120°
The exterior angle is 180° – 120° = 60°
Alternatively, use the fact that the sum of exterior angles = 360° and divide this answer by
6 to get exterior angle of regular hexagon:
Each exterior angle = 360° ÷ 6 = 60°
Interior angle = 180° – 60° = 120°
17 a Equal sides, equal angles b The interior angle increases. c ii has the greatest number of sides d i 8 ii 20 iii 15
18 a 24 b 30 c 40
19 a Angle x is 45° because angles around a point total 360°.
b Angle y is 67.5° because an isosceles triangle has two equal angles:
2y = 180 – x = 180 – 45 = 135; y = 67.5
c z = 135° because the angle z = 2 × y.
20 a True; five equal angles round a point
Original Material © Cambridge University Press 2010 5 Cambridge Essentials Mathematics Core 9 GM1.1 Answers
b x = 72° because angles round a point total 360°
c The angles in an equilateral triangle are all 60° and x = 72
1 d An isosceles triangle has two equal angles: y = (180 – 72) = 54 2
e Angle z = 2 × y so z = 108°
21 a a = 40° (corresponding angles are equal), b = 41° (angles in a triangle),
c = 41° (corresponding angles are equal)
b x = 140° (corresponding angles are equal), y = 140° (alternate angles or vertically
opposite angles), z = 20° (angles in an isosceles triangle)
c a = 55° (vertically opposite angles), c = 55° (vertically opposite angles),
b = 80° (angles in a triangle)
22 a 360° b 120 c 60° d Yes
e Corresponding angles
23 a v = 40° b w = 70° c x = 40° d u = 70°
e z = 40° f y = 70° g They are both isosceles triangles.
Original Material © Cambridge University Press 2010 6