Edexcel GCSE Maths (1MA1), Higher tier · Geometry and measures
Practise Angles, parallel lines and polygons. 5 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
(a) Explain why a regular polygon cannot have an interior angle of 100°.[2]
(b) A regular polygon has interior angles of 156°. How many sides does it have?[2]
Show the answer and mark scheme
(a)Answer: The exterior angle would be 80°, and \(360 \div 80 = 4.5\), which is not a whole number of sides.
M1 for exterior angle \(180 - 100 = 80\)
C1 for \(360 \div 80 = 4.5\), which is not a whole number, so there is no such polygon
Worked solution: Exterior angle \(= 180 - 100 = 80^\circ\). The exterior angles of a polygon add up to 360°, so \(n = 360 \div 80 = 4.5\). A polygon must have a whole number of sides.
(b)Answer: 15
M1 for \(360 \div (180 - 156)\)
A1 for 15
Worked solution: Exterior angle \(= 24^\circ\), so \(n = 360 \div 24 = 15\).
Question 3Hard4 marks
Regular polygon P has \(n\) sides. Regular polygon Q has \(2n\) sides. Each interior angle of Q is 15° larger than each interior angle of P.
Find the value of \(n\).[4]
Show the answer and mark scheme
Answer: 12
P1 for the exterior angles \(\frac{360}{n}\) and \(\frac{360}{2n}\)
P1 for recognising that the exterior angles also differ by 15° (or interior angles \(180 - \frac{360}{n}\) and \(180 - \frac{180}{n}\))
P1 for \(\frac{360}{n} - \frac{180}{n} = 15\) oe
A1 for 12
Worked solution: Exterior angle of P \(= \frac{360}{n}\); of Q \(= \frac{360}{2n} = \frac{180}{n}\). A larger interior angle means a smaller exterior angle by the same amount: \(\frac{360}{n} - \frac{180}{n} = 15\), so \(\frac{180}{n} = 15\) and \(n = 12\). (Check: 150° and 165°.)