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Geometry

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 approachE330C9E8-5E3D-4DED-8FC3-C945E1392D37.jpeg
(edited 2 years ago)
Reply 1
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mqb2766
21
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)
(edited 2 years ago)
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
Reply 3
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mqb2766
21
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.
(edited 2 years ago)
Considering one isosceles triangle you’ll have the height being x/2, wouldn’t the equal sides(hypotenuse) be longer than x/2 ?
Reply 5
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mqb2766
21
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.
(edited 2 years ago)
Reply 6
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mqb2766
21
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/default/files/ukmt/senior-mathematical-challenge/SMC2018_sol_extended.pdf
(edited 2 years ago)

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