# Geometry

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#1
How would I go about calculating the area of the octagon in terms of x,I’ve tried using trig but it doesn’t seem to be the right approach
Last edited by The A.G; 1 month ago
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1 month ago
#2
Get the side length in terms of x, then either
* Subtract two small squares (4 corners) from the large enclosing square or
* Split the octagon up into 8 isosceles triangles (connect centre to vertices)
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
Get the side length in terms of x, then either
* Subtract two small squares (4 corners) from the large enclosing square or
* Split the octagon up into 8 isosceles triangles (connect centre to vertices)
My initial attempt was trying to make isosceles triangles but the angles are awkward so I wasn’t able to use sine or cos rules. I resorted to the four corners but it took quite a lot of work considering it being only 6 marks
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1 month ago
#4
(Original post by The A.G)
My initial attempt was trying to make isosceles triangles but the angles are awkward so I wasn’t able to use sine or cos rules. I resorted to the four corners but it took quite a lot of work considering it being only 6 marks
It shouldnt be a problem. The two equal sides are pythagoras on x/2 and half the side and the angle is 360/8 = ... so the area (sin rule) is easy?
Similarly the 4 corners are "sqrt(2)", so you can can almost write them down.
Last edited by mqb2766; 1 month ago
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#5
Considering one isosceles triangle you’ll have the height being x/2, wouldn’t the equal sides(hypotenuse) be longer than x/2 ?
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1 month ago
#6
(Original post by The A.G)
Considering one isosceles triangle you’ll have the height being x/2, wouldn’t the equal sides(hypotenuse) be longer than x/2 ?
Yes, I edited that typo. As in the originnal post, find the side length and use pythogaras on side/2 and x/2.
The two approaches are not that much different, but the square on is proabably a bit quicker.
Last edited by mqb2766; 1 month ago
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1 month ago
#7
(Original post by The A.G)
Considering one isosceles triangle you’ll have the height being x/2, wouldn’t the equal sides(hypotenuse) be longer than x/2 ?
Note, there is a slightly quicker way of computing (the area of the triangle and hence) the area of the octagon.

If you imagine a rectangle of of width x and height z (the side length) which is x/(1+sqrt(2)) stretching across the octagon, then the area of the rectangle is 4 of the isosceles triangles, so simply double the area of the rectangle to get the area of the octagon.

Its a useful trick for regular polygons with an even number of sides and was on the smc a couple of years ago for a decagon (q9)
https://www.ukmt.org.uk/sites/defaul...l_extended.pdf
Last edited by mqb2766; 1 month ago
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