Edexcel GCSE Maths Exam 2023 - Solving the Final Octagon Question
The dreaded Edexcel GCSE Maths 2023 final question
Edexcel GCSE Maths Paper 2: The Final Question
The final question on this year's GCSE maths paper from Edexcel has confused a lot of people and even been branded 'sadistic' by one furious pupil. It's also been pointed out that a very similar question has been used recently on an A-level practice paper for older students.
Once you break it down, however, the maths involved are not too complicated. As with all grade nine questions (remember this is the tricky question at the end of the paper), the skill being tested is not whether you can apply particular maths knowledge to a problem. It's whether you can first work out which maths knowledge you need to apply and then apply it correctly.
Once you break it down, however, the maths involved are not too complicated. As with all grade nine questions (remember this is the tricky question at the end of the paper), the skill being tested is not whether you can apply particular maths knowledge to a problem. It's whether you can first work out which maths knowledge you need to apply and then apply it correctly.
Eight octagons joined together to form a star shape in the middle.
The question
Look at the diagram above. We have eight regular octagons joined together in a circle to form a shaded star shape in the centre. Each octagon has side length a.
We need to find the area of the shaded star in the form P(2+√2)a^2.
We need to find the area of the shaded star in the form P(2+√2)a^2.
A close look at the top of the star
How to solve it
In the image above we have zoomed in to the top of the star. The octagons are all regular octagons; hence every side is of length a, as can be seen on the diagram.
The length of x
We can now add a straight line joining the two horizontal lines on top. This is the red line in the diagram. This forms a triangle at the top of the shape. We want to work out the length of this line, x.
We can fill in the angles of the triangle by noting that the base angles of the triangle are the exterior angles of an octagon. The exterior angles are 360° ÷ 8 = 45°. The top angle is simply two exterior angles together, hence a right angle of 90°.
We now have a right-angled triangle with a hypotenuse of length x and two shorter sides of length a. Pythagoras' theorem, therefore, gives us:
x = √(a^2 + a^2) = √(2a^2) = a^2 √2
We can fill in the angles of the triangle by noting that the base angles of the triangle are the exterior angles of an octagon. The exterior angles are 360° ÷ 8 = 45°. The top angle is simply two exterior angles together, hence a right angle of 90°.
We now have a right-angled triangle with a hypotenuse of length x and two shorter sides of length a. Pythagoras' theorem, therefore, gives us:
x = √(a^2 + a^2) = √(2a^2) = a^2 √2
A square and triangles
Zooming back out, we have now added a red square to the inside of the star shape, splitting it up into a square and four equal triangles. If we can find the area of the square and triangles, we can solve the problem.
It can be seen easily from the diagram that the square has a side length of 2a + a√2. Therefore the area of the square is (2a + a√2)^2 = (2 + √2)2a^2 = (6 + 4√2)a^2.
This last step comes from expanding the brackets (2 + √2)(2 + √2).
The triangles are each right-angled triangles with perpendicular height and width of a. Therefore the area of each triangle is 1/2 × a × a = 1/2 × a^2.
It can be seen easily from the diagram that the square has a side length of 2a + a√2. Therefore the area of the square is (2a + a√2)^2 = (2 + √2)2a^2 = (6 + 4√2)a^2.
This last step comes from expanding the brackets (2 + √2)(2 + √2).
The triangles are each right-angled triangles with perpendicular height and width of a. Therefore the area of each triangle is 1/2 × a × a = 1/2 × a^2.
Bringing it all together
The total area is the area of the square plus the four triangles, hence:
area = (6 + 4√2)a^2 + 4 × 1/2 × a^2
= (6 + 4√2)a^2 + 2a^2
= (8 + 4√2)a^2
= 4(2 + √2)a^2
which is in the form as required.
area = (6 + 4√2)a^2 + 4 × 1/2 × a^2
= (6 + 4√2)a^2 + 2a^2
= (8 + 4√2)a^2
= 4(2 + √2)a^2
which is in the form as required.
Comments
What do you think of this question? Is it the trickiest GCSE exam paper question ever?
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