A regular hexagon has all sides and angles equal. Write down the size of one interior angle of a regular hexagon.
(1)
2
Find the exterior angle of a regular pentagon.
(1)
3
Work out the sum of the interior angles for each polygon.
(a)Triangle(1)
(b)Quadrilateral(1)
(c)Heptagon(1)
4
A regular octagon has interior angles all equal. Calculate the size of one interior angle.
(2)
5
Calculate the exterior angle of a regular decagon.
(2)
6
An irregular pentagon has interior angles 90 degrees, 110 degrees, 95 degrees and 85 degrees. Work out the missing interior angle.
(2)
7
A regular polygon has each interior angle equal to 150 degrees. Find the number of sides of the polygon.
(3)
8
The interior angles of a quadrilateral add to 360 degrees. Two of the angles are equal and the other two are 95 degrees and 85 degrees. Find the size of each of the equal angles.
(2)
9
Each exterior angle of a regular polygon is 24 degrees. How many sides does the polygon have?
(2)
10
A regular polygon has 14 sides. Calculate both the sum of its interior angles and the measure of one interior angle.
(3)
11
A regular triangle is equilateral. A student says one exterior angle is 100 degrees. Explain why this statement is incorrect and give the correct exterior angle size.
(2)
12
A regular polygon has interior angle 162 degrees. Work out n, the number of sides.
(2)
13
The angles in a quadrilateral are in the ratio 3:4:5:6. Find the size, in degrees, of the largest angle.
(3)
14
A regular polygon and a regular star-shaped polygon both have the same exterior angle of 40 degrees at a certain vertex. For the regular polygon, find the number of sides.
(2)
15
A teacher draws a regular polygon on the board. The sum of its interior angles is 2340 degrees. How many sides does the polygon have?
(3)
16
In a regular polygon, three exterior angles are removed and their total is 108 degrees. If the remaining exterior angles are all equal and there are 12 exterior angles in total, find the measure of each of the remaining exterior angles.
(3)
17
Stretch question. The interior angles of a convex polygon are in arithmetic progression. The smallest angle is 100 degrees and the largest is 170 degrees. There are 8 sides. Find the common difference of the progression and write down all eight interior angles.
(3)
18
Stretch question. A regular polygon is joined to its centre to form isosceles triangles. If each triangle has a vertex angle of 20 degrees at the centre, how many sides does the polygon have and what is each base angle of the isosceles triangle?
(3)
Mark scheme · KS3.M-G10 Angles in Polygons
Question 1
B1 120 degrees cao
Answer: 120 degrees
Question 2
B1 72 degrees cao
Answer: 72 degrees
Question 3
(a) B1 180 degrees cao
(a) Answer: 180 degrees
(b) B1 360 degrees cao
(b) Answer: 360 degrees
(c) B1 900 degrees cao
(c) Answer: 900 degrees
Question 4
M1 use sum formula: (8-2)*180 = 1080 or divide by 8
A1 135 degrees cao
Answer: 135 degrees
Question 5
M1 use 360/n = 360/10 or compute interior then 180-interior
A1 36 degrees cao
Answer: 36 degrees
Question 6
M1 use sum for pentagon: (5-2)*180 = 540 and subtract given angles
A1 160 degrees cao
Answer: 160 degrees
Question 7
M1 use relation interior = 180 - 360/n leading to 150 = 180 - 360/n or 360/n = 30
M1 rearrange to n = 360/30
A1 n = 12 cao
Answer: 12
Question 8
M1 compute remaining sum: 360 - (95 + 85) = 180
A1 each equal angle = 90 degrees cao
Answer: 90 degrees
Question 9
M1 use n = 360/exterior = 360/24
A1 n = 15 cao
Answer: 15
Question 10
M1 sum = (14-2)*180 = 12*180 = 2160
M1 interior = sum/14 = 2160/14 or use 180 - 360/14
A1 2160 degrees and 154.285... -> 154 2/7 degrees (or 154.2857 degrees) cao
Answer: 2160 degrees; 154.285714 degrees
Question 11
M1 explain that interior of equilateral is 60 so exterior = 180 - 60 = 120, not 100: identification of error using exterior = 180 - interior
A1 correct exterior angle is 120 degrees cao (award for clear correction)
Answer: Exterior angle = 120 degrees
Question 12
M1 use exterior = 180 - interior = 18 then n = 360/18
A1 n = 20 cao
Answer: 20
Question 13
M1 let common multiplier = k, sum = (3+4+5+6)k = 18k and set = 360
M1 find k = 360/18 = 20
A1 largest angle = 6k = 120 degrees cao
Answer: 120 degrees
Question 14
M1 use n = 360/exterior = 360/40
A1 n = 9 (since 360/40 = 9) cao
Answer: 9
Question 15
M1 use sum formula (n-2)*180 = 2340
M1 rearrange: n-2 = 2340/180 = 13
A1 n = 15 cao
Answer: 15
Question 16
M1 total of all exterior angles = 360 so remaining sum = 360 - 108 = 252
M1 number of remaining angles = 12 - 3 = 9 so each = 252/9
A1 each remaining exterior angle = 28 degrees cao
Answer: 28 degrees
Question 17
M1 recognise arithmetic progression with first a = 100, last a + 7d = 170 so 7d = 70