P=2L+6
A=3L
3L=2L+6
3L-2L=6
L=6
This shows that the length must be six for a rectangle with a fixed width of three to be equable.
Now I will look at the general rectangle width Length L and Width W.
Area = LW
Perimeter = 2(L+W)
LW = 2(L+W)
LW = 2L + 2W
LW-2L = 2W
L(W-2) = 2W
L = This is the formula to find L (the length) given W (the width)
Test the formula
Suppose the width of a rectangle is 42
L =
L =
L = 2.1
Area = 42 x 2.1 = 88.2
Perimeter = 2(42 + 2.1) = 88.2
This shows that this rectangle is equable as the area and the perimeter are the same. This can work for any rectangle with a width greater than 2, as long as you know the width (W). Now I will make W (width) the subject.
LW = 2L+2W
LW – 2W = 2L
W (L-2) = 2L
W = For values of L2
Using this formula, we can take the length (L) of any rectangle (assuming that the length is greater than two) and calculate the width needed to make the rectangle equable.
Section 4
Triangles
All Regular polygons have some traits that are the same. All the sides have the same width and all interior angles are the same. A regular triangle is otherwise known as an equatorial triangle. We also have to remember to get equality the area and perimeter has to be equal.
First I will look at a triangle
In algebraic form, the perimeter of this equilateral triangle is 3X
We do not know the area of the equilateral triangle so we will split it in half and find the formula for a right angle triangle.
Area of one right-angle triangle = (equation 1)
But sin 60 =
Rearrange this to get H = X sin 60 (equation 2)
Now I will substitute formula two into formula one to get:
Area =
This equals = 1/2 X²sin60
Therefore for an equilateral triangle to be equable (area = perimeter) then:
3X = ½ X²sin60
3 = ½X sin 60
6 = X sin 60
= X
Therefore X = 6.92820323…
=6.93 (2dp)
Test
If the perimeter = 3X = 20.79 (2dp)
If the Area = ½ X² Sin 60 = 20.79 (2dp)
The above formulas show that the equilateral triangle with sides of 6.93 (to 2 decimal places) is an equable shape. This also proves that my formula is correct, and is fully functional.
Section five
Regular polygons
I have finished looking at “common” shapes like square, rectangle and triangle, so now I will move onto other regular polygons. The first regular polygon I will be looking at is a pentagon.
In order to find the perimeter I need to use the formula 5X. X being the length of one side. To make the pentagon equable then 5X (the perimeter) will have to be equal to the area. In order to find the formula of the pentagon I will separate it into five congruent isosceles triangles, because there is no easy formula for a pentagon. The area of the pentagon would be 5the area of one isosceles triangles.
An equable triangle: 5X = 5the area of one isosceles triangles (H being the height)
The area of one triangle = (equation 1)
To express H in terms of X using tangent.
Tan 36= O/A = Opposite/ Adjacent
Tan 36=
H Tan 36 = (H)
2HTan36 = X (2)
H = (÷2Tan36) (equation 2)
Substitute (2) into (1) to get area of one triangle
=
=
Area of pentagon =
For the pentagon to be equable,
Area = Perimeter
= 5X
= 1 (÷5X)
X = 4 Tan 36 (×4 Tan 36)
This shows us that for a regular pentagon to be equable the sides (X) must be equal to 4 Tan 36.
Test
Perimeter = 5X = 14.53
Area = 5 = 5X = 14.53
The area and the perimeter are numerically equal and therefore show that my formula is correct. This also shows that 4Tan36 does give the length to make a regular pentagon equable.
Section five – part two
Regular hexagons
After finding the formula for regular pentagons I have decided to look at regular hexagons. This is the second regular polygon I will be looking at.
For the shape to be equable the area has to be equal to the perimeter. Again I am going to split the hexagon into six equal segments, because it has six equal sides. I am doing this because there is no easy formula for the hexagon so I can find the formula for one isosceles triangle and multiply it by five.
Perimeter = 6X
Area of one isosceles triangle = (Equation 1)
Tan 30=
This shows that Tan 30 (this is the angle) is equal to half the length of the isosceles triangle divided by the height
H Tan 30° = X/2 (×H)
2 Tan 30° = X (×2)
H = (÷ 2 Tan 30º) (equation 2)
Substitute (equation 1) into (equation 2) to get:
Area of one isosceles triangle =
=
So therefore:
Area of regular hexagon =
Perimeter = 6X
And so for the hexagon to be equable:
6X =
1 = (÷6X)
4 Tan 30° = X (÷4 Tan 30°)
X = 4 Tan 30°
This shows us that the numerical value of 4 Tan 30° equals the length of side required to make this regular hexagon equable.
Section six
Three dimensional shapes
For a three dimensional shape to be equable we need to change the conditions for equability. The volume of the shape now needs to be the same as the surface area, instead of the area being equal to the perimeter. I started off with a cube.
I am going to let the width of a cube be A.
This would make the volume of the cube = A³
This would make the Surface area of the cube = 6A²
If then:
Volume = surface area
A³ = 6A² (A²)
A = 6
This shows that the cubes length, width and weight to be 6 for the cube to be equable.
Now I will find equability with a sphere.
(R = radius)
Volume of sphere = R³
Surface Area of sphere = 4R²
R³ = 4R² (4R²)
R = 1 (x3)
R = 3
This shows that for a sphere to be equable the radius has to be 3.
Now I will be doing a closed cylinder. A closed cylinder is a cylinder with a base AND a lid.
The volume of a closed cylinder = R²H
The surface area of a closed cylinder = 2R² + 2RH
R²H = 2R² + 2RH (R)
RH = 2R + 2H (Make R the subject)
RH – 2R = 2H (Factorise)
R(H-2) = 2H
R =
H2
This shows that for a closed cylinder to be equable the height has to be greater than two.