Mathematical Investigation
Introduction – I have been set the task of finding equable shapes and
eventually find a possible formula for them. I will intend to do this
by first using the rule of an equable shape, where the perimeter and
the area have the same numerical value and extending it to regular
polygons. E.g.
Taking the sides as X
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Area = Perimeter
X² = 4X
X² - 4X = 0
X(X-4) = 0
X = 0 X = 4
I intend to set out the results, eventually in a table and possibly in
a graph depending on the type of results I get. I hope to see a
pattern in the results so I can predict the results in more
complicated shapes. I intend to extend the task to surface area and
volume and find a formula for them. I am only going to do regular
polygons because irregular polygons could get very complicated and I
doubt a general formula could be found for them.
The first shape I am going to take is an equilateral triangle because
it is a regular polygon and it has the least amount of sides a shape
can have. The area of a triangle is ½ base * height and I am going to
use L for the sides E.g. 4L means there’s 4 sides. To find out the
area of a triangle trigonometry is going to have to be used.
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Area = Perimeter
½ base * height = 3 L
½ * L * H = 3L
Line Callout 2: To find the length of H trigonometry is going to have to be used. Since I am trying to find H Tan (opposite over adjacent) is going to be used. Text Box: Tan 30º = ½ L
H
H = ½ L
Tan30
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½ L * ½ L