Skip to main content
Links to the University of Cambridge website
Links to the NRICH website Home page
Maths at Home
Summer Challenges 2024
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
Site search
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
Mathematical mindsets
For younger learners
Early Years Foundation Stage
Advanced mathematics
Decision Mathematics and Combinatorics
Advanced Probability and Statistics
Mechanics
Calculus
Symmetric Angles
Age
14 to 16
Short
Challenge Level
Secondary curriculum
Problem
Solutions
As the figure has rotational symmetry of order $4$, $ABEF$ is a square.
Area $ABEF=4\times$area$\triangle BDA=4\times \frac{1}{2}BD \times DA=2(DB)^2=24$cm$^2$, so $BD=\sqrt{12}$cm$=2\sqrt{3}$cm.
As $ABEF$ is a square, $\angle ABD=45^{\circ}$ so $\angle CBD=45^{\circ} -15^{\circ} =30^{\circ}$.
Since $\tan {30^{\circ}}= \frac{CD}{BD}=\frac{CD}{2\sqrt{3}}$, we have $CD=2\sqrt{3}\tan{30^{\circ}}$cm.
Now consider the following equilateral triangle with side lengths $2$:
The vertical line is perpendicular to the base-line and so bisects both the angle at the top vertex and the base-line.
Consider the left right-angled triangle. Pythagoras' theorem gives $a=\sqrt{3}$ and then we have $\tan{30^{\circ}}=\frac{1}{a}=\frac{1}{\sqrt{3}}$
Therefore $CD=2\sqrt{3}\tan{30^{\circ}}$cm$=2$cm.
This problem is taken from the
UKMT Mathematical Challenges
.
You can find more short problems, arranged by curriculum topic, in our
short problems collection
.