Skip over navigation
Cambridge University Faculty of Mathematics NRich logo
menu search
  • Teachers expand_more
    • Early years
    • Primary
    • Secondary
    • Post-16
    • Events
    • Professional development
  • Students expand_more
    • Primary
    • Secondary
    • Post-16
  • Parents expand_more
    • Early Years
    • Primary
    • Secondary
    • Post-16
  • Problem-Solving Schools
  • About NRICH expand_more
    • About us
    • Impact stories
    • Support us
    • Our funders
    • Contact us
  • search

Or search by topic

Number and algebra

  • The Number System and Place Value
  • Calculations and Numerical Methods
  • Fractions, Decimals, Percentages, Ratio and Proportion
  • Properties of Numbers
  • Patterns, Sequences and Structure
  • Algebraic expressions, equations and formulae
  • Coordinates, Functions and Graphs

Geometry and measure

  • Angles, Polygons, and Geometrical Proof
  • 3D Geometry, Shape and Space
  • Measuring and calculating with units
  • Transformations and constructions
  • Pythagoras and Trigonometry
  • Vectors and Matrices

Probability and statistics

  • Handling, Processing and Representing Data
  • Probability

Working mathematically

  • Thinking mathematically
  • Developing positive attitudes
  • Cross-curricular contexts

Advanced mathematics

  • Decision Mathematics and Combinatorics
  • Advanced Probability and Statistics
  • Mechanics
  • Calculus

For younger learners

  • Early Years Foundation Stage

Funky Hexagon

Age 11 to 14
ShortChallenge Level Yellow starYellow star
Secondary curriculum
  • Problem
  • Solutions

Using the formula for internal angles of a polygon

The internal angles of an $n$ sided polygon sum to $(n-2) \times 180^\text{o}$, so for a hexagon the internal angles sum to $180(6 - 2)=720$ degrees.

The interior angles not shown are  $360^\text{o}-75^\text{o}=285^\text{o}$ and $ (360 - x)^\text{o}$. So $$\begin{align}26+27+24+23+285+(360-x)&=720\\
\Rightarrow 385+360-x&=720\\
\Rightarrow 745-x&=720\\
\Rightarrow x&=25\end{align}$$


Using triangles inside the hexagon

Drawing lines from the vertex with the $x^\text{o}$ angle to the other vertices makes four triangles, A, B, C and D.



The sum of the angles in each of the triangles is $180^\text{o}$, so the sum of the angles in all four triangles is $180^\text{o}\times4=720^\text{o}$. So
$26+27+24+23+$ the red angle $+$ the blue angle $=720$
The red angle makes a full circle with the $75^\text{o}$ angle, so it must be $360^\text{o}-75^\text{o}=285^\text{o}$.
The blue angle makes a full circle with the $x^\text{o}$ angle, so it must be $(360-x)^\text{o}$. This gives $$\begin{align}26+27+24+23+285+(360-x)&=720\\
\Rightarrow 385+360-x&=720\\
\Rightarrow 745-x&=720\\
\Rightarrow x&=25\end{align}$$


Using external angles on a diagram

Imagine starting at the $23^\text{o}$ angle and travelling to the left, and walking along the perimeter of the hexagon.
You would arrive at the $26^\text{o}$ angle and then need to rotate through the green angle shown to be facing in the correct direction to walk along the next side. The green angle is $154^\text{o}$ because $154 + 26 = 180.$
The full journey is shown on the diagram below. The two angles that are shown in red are red because the turn is the other way (anticlockwise, not clockwise).

  

By the time you arrive back at the $23^\text{o}$ angle and rotate $153^\text{o}$, you are facing in the same direction you started in, so you have turned through a total of $360^\text{o}$.

You have also turned forwards through the green angles and backwards through the red angles, so the sum of the green angles subtract the sum of the red angles must be $360^\text{o}$. $$\begin{align}154 - 105 + 153 + 156 + -(180 - x ) + 157 &= 360\\
\Rightarrow 515-180+x&=360\\
\Rightarrow 515+x&=540\\
\Rightarrow x&=25\end{align}$$


You can find more short problems, arranged by curriculum topic, in our short problems collection.

You may also like

Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

Calendar Capers

Choose any three by three square of dates on a calendar page...

Days and Dates

Investigate how you can work out what day of the week your birthday will be on next year, and the year after...

  • Tech help
  • Accessibility Statement
  • Sign up to our newsletter
  • Twitter X logo

The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

University of Cambridge logo NRICH logo