Cut-Up Square
Weekly Problem 15 - 2015
In the diagram, two lines have been drawn in a square. What is the ratio of the areas marked?
In the diagram, two lines have been drawn in a square. What is the ratio of the areas marked?
Problem
Image
The diagram shows a square, a diagonal and a line joining a vertex to the midpoint of a side. What is the ratio of area P to area Q?
If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.
Student Solutions
Image
First, we label the corners of the square, the midpoint, and the point at which the two lines intersect $A$, $B$, $C$, $D$, $E$ and$F$ respectively. We also label the four triangles created within the square as 1, 2, 3 and 4.
Call the area of triangle a.
Angle $AFD$ is equal to angle $EFC$ as they are vertically opposite, and angle $FAD$ is equal to angle $FCD$ as they are alternate. So, triangles 1 and 3 are similar. As $E$ marks the midpoint of a side of the square, length $CE$ is half the value of length $AD$.
This means that the area of triangle 3 is equal to 4a, as its base and height are double that of triangle 1.
It also means that length $AF$ is equal to twice the value of length $FC$. If we let $AF$ be the base of triangle 3 and $FC$ be the base of triangle 2, then these triangles have the same height (equal to the perpendicular distance from $D$ to the line $AF$) so the area of triangle 2 is equal to half that of triangle 3 - that is, 2a.
As line $AC$ is a diagonal, the combined areas of triangles 2 and 3 is equal to half the area of the square. 4a + 2a = 6a, so the square has a total area of 12a. The combined area of triangle 1 and triangle 4 is then also 6a. As we have defined the area of triangle 1 as a, we then know that triangle 4 has an area of 5a.
As the ratio of $P:Q$ is simply the ratio area of triangle 2:area of triangle 4, we then know that this ratio is 2a:5a=2:5.