You may also like

problem icon

Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle

problem icon

Triangle Midpoints

You are only given the three midpoints of the sides of a triangle. How can you construct the original triangle?

problem icon

Fermat's Poser

Find the point whose sum of distances from the vertices (corners) of a given triangle is a minimum.

Octagonal Ratio

Age 14 to 16 Short Challenge Level:
Dividing the octagon into rectangles and triangles

By drawing some extra lines you get this picture:
 
The red and green triangles are right-angled isosceles triangles with a hypotenuse equal to the side length of the octagon.

The blue and yellow triangles are also right-angled isoseles triangles, and as the central square's length is the same as the side lenght of the octagon, these triangles are congruent to the red and green triangles.

There is one rectangle and two triangles shaded in each of the four colours, so the four shaded areas are equal. Therefore the ratio of the shaded to unshaded parts of the original octagon is $1:3$.

Using the centre of the octagon

AE divides the octagon in half.
The green triangle is $\frac{1}{8}$ of the total area of the octagon.
Triangle ABE has the same base and twice the height, so twice the area, so is $\frac{1}{4}$ of the total area of the octagon.
So Trapezium BCDE is $\frac{1}{4}$ of the total area.

So the ratio is $1:3$. 
 


Finding areas of shapes
We can label the side length as $1$ since we are looking for ratios between areas, so it doesn't matter how big the octagon is. Three helpful shapes are drawn in. Finding their areas will enable us to find the areas of the shaded and unshaded parts of the octagon.


 The area of the red triangle is $\frac{1}{2}\times x\times x=\frac{1}{2}x^2.$ Applying Pythagoras' Theorem to the red triangle gives $x^2+x^2=1,$ so $x^2$ must be $\frac{1}{2},$ so $\frac{1}{2}x^2$ must be $\frac{1}{4}.$ So the area of the red triangle is $\frac{1}{4}.$



The area of the yellow rectangle is $1\times x=x.$ Since $x^2=\frac{1}{2},$ $x=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt2}.$ So the area of the yellow rectangle is $\frac{1}{\sqrt2}.$


  


 The area of the blue square is $1$.





 So the area of the shaded part of the octagon is $\frac{1}{4}+\frac{1}{\sqrt2}+\frac{1}{4}=\frac{1}{2}+\frac{1}{\sqrt2}$



 
And the area of the unshaded part of the octagon is $\frac{1}{\sqrt2}+1+\frac{1}{\sqrt2}+\frac{1}{4}+\frac{1}{\sqrt2}+\frac{1}{4}=1\frac{1}{2}+\frac{1}{\sqrt2},$ which is the same as $3\times\left(\frac{1}{2}+\frac{1}{\sqrt2}\right),$ which is $3$ times the shaded area.

So the ratio of the shaded area to the unshaded area is $1:3.$
You can find more short problems, arranged by curriculum topic, in our short problems collection.