This problem is in two parts. The first part provides two building blocks which will help you to solve the final challenge. These can be attempted in either order. Of course, you are welcome to go straight to the Final Challenge!
Click on a question below to get started.

Question A
This question is about triangles with their bases on the same line and a shared side, like the ones below:

The numbers in the triangles represent their areas.

The diagrams are not drawn to scale - can you create diagrams with the correct areas drawn to scale?

Convince yourself that there are many different possibilities

Make a note of the base lengths of your triangles.

Notice anything interesting? Convince yourself it always happens.

Question B
The red line is half the length of the blue line. Numbers and letters inside a triangle represent its area. What can you say about $A+B$?

For each triangle with base lengths as shown, write C in terms of D.

FINAL CHALLENGE
Look at the diagram below (which is not drawn to scale).

The areas of three of the triangles are shown.

What is the area of the quadrilateral $APOQ$?

*Click below to reveal a method for finding the area. The statements have been muddled up, however. Can you rearrange them into the correct order?*

A set of printable cards is also avilable.

A. Substituting this back into equation (1) gives $E = 12$.

B. Triangles $QAO$ and $OAB$ have the same height above the line $BQ$, so $\frac{F}{OQ}=\frac{E+8}{BO}$.

C. Triangles $PAO$ and $OAC$ have the same height above the line $PC$, so $\frac{E}{PO}=\frac{F+5}{CO}$.

D. Combining these two equations gives $\frac{E}{8} = \frac{F+5}{10}$.

E. The area of quadrilateral $AQOP$ is $E+F = 10+12=22$.

F. Triangles $PBO$ and $OBC$ also have the same height above $PC$, so $\frac{8}{PO}=\frac{10}{CO}$.

G. Solving for $F$ gives $F=10$.

H. Doubling equation (2) and rearranging gives $10E = 20F - 80$.

I. Draw in the line $AO$. Label the area of triangle $AOP$ as $E$, and that of $AOQ$ as $F$.

J. Clearing denominators gives $10E = 8F+40$. Call this equation (1).

K. Clearing denominators gives $10F = 5E+40$. Call this equation (2).

L. Combining these two equations gives $\frac{F}{5} = \frac{E+8}{10}$.

M. Triangles $QCO$ and $OBC$ also have the same height above $BQ$, so $\frac{5}{OQ}=\frac{10}{BO}$.

N. Substituting this into equation (1) gives $20F - 80 = 8F+40$.