Isoperimetric Quotients Essay Example

Isoperimetric Quotients Essay Isoperimetric Quotients of plane shapes are calculated using the formula: I.Q. = 4? x Area of shape (Perimeter of shape)à ¯Ã‚ ¿Ã‚ ½ I am going to investigate isoperimetric quotients of plane shapes and interpret my findings. We will write a custom essay sample on Isoperimetric Quotients specifically for you for only $16.38 $13.9/page Order now We will write a custom essay sample on Isoperimetric Quotients specifically for you FOR ONLY $16.38 $13.9/page Hire Writer We will write a custom essay sample on Isoperimetric Quotients specifically for you FOR ONLY $16.38 $13.9/page Hire Writer Firstly, I am going to look at flat shapes. Using the formula, I will calculate the isoperimetric quotients of the shapes. Starting with the smallest 2D shape- a triangle- I will calculate the I.Q s of right-angled triangles. I will also do this with isosceles and equilateral triangles. I will move on to quadrilaterals and look at the I.Q s. Maybe there will be something about the results that will help me with further plane shapes; pentagon, hexagon, heptagon, nonagon, decagon and possibly a circle. With comparison, the results might show something about the shapes, such as a pattern. Triangles I am now going to study right-angled triangles. Right-angled triangles I will first look at the 3, 4, 5 right-angled triangles and then enlargements of it. 1. Perimeter= 3+4+5= 12 cm Area= 1/2 x 4 x 3= 6cmà ¯Ã‚ ¿Ã‚ ½ I.Q. = 4 x ? x 6 12à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.5236 I will look at similar enlarged right-angled triangles, based on the 3, 4, 5 triangle. 2. Perimeter= 6+8+10= 24 cm Area= 1/2 x 6 x 8= 24 cm à ¯Ã‚ ¿Ã‚ ½ I.Q= 4 x ? x 24 24à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.5236 3. Perimeter= 16+20+12=48 cm Area= 1/2 x 16 x 12= 96 cmà ¯Ã‚ ¿Ã‚ ½ I.Q. = 4 x ? x 96 48à ¯Ã‚ ¿Ã‚ ½ =0.5236 Similar 345 right-angled triangles have the same I.Q. The answers are all the same. I predict that similar enlarged shapes make the same I.Q. answer. I am now going to look at other right-angled triangles that are not versions of the 3, 4, 5. Other right angled triangles Firstly, in these triangles, I will need to find the hypotenuse length so I will use Pythagoras theorem. 4. Hypotenuse = 4à ¯Ã‚ ¿Ã‚ ½ + 6à ¯Ã‚ ¿Ã‚ ½ = 52 = V52 = 7. 2111 cm Perimeter = 4+6+7.2111 = 17.2111 cm Area= 1/2 x 4 x 6 = 12 cm à ¯Ã‚ ¿Ã‚ ½ I.Q. = 4 x ? x 12 17.2111à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.5090 5. H = 8à ¯Ã‚ ¿Ã‚ ½ + 12à ¯Ã‚ ¿Ã‚ ½ = 208 = V 208 = 14.422 cm P = 12+8+14.422 = 34.422 A = 1/2 x 8 x 12 = 48 I.Q. = 4 x ? x 48 34.422à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.5090 6. H = 3à ¯Ã‚ ¿Ã‚ ½ + 2à ¯Ã‚ ¿Ã‚ ½ = 13 = V13 = 3.6 P = 3+2+3.6 = 8.6 A= 1/2 x 2 x 3 = 3 I.Q. = 4 x ? x 3 8.6à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.5097 7. H = 6à ¯Ã‚ ¿Ã‚ ½+4à ¯Ã‚ ¿Ã‚ ½ = 52 =V52 = 7.211 P = 6+4+7.211 = 17.211 A= 1/2 x 4 x 6 = 12 I.Q. = 4 x ? x 12 17.211à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.5091 8. H = 7à ¯Ã‚ ¿Ã‚ ½+9à ¯Ã‚ ¿Ã‚ ½ = 130 = V130 = 11.4018 P = 7+9+11.4018 = 27.4018 A = 1/2 x 7 x 9 = 31.5 I.Q. = 4 x ? x 31.5 27.4018à ¯Ã‚ ¿Ã‚ ½ I.Q.= 0.5271 9. H = 18à ¯Ã‚ ¿Ã‚ ½+14à ¯Ã‚ ¿Ã‚ ½= 520 = V520 = 22.8035 cm P = 18+14+22.8035 = 54.8035 cm A= 1/2 x 14 x 18 = 126 cmà ¯Ã‚ ¿Ã‚ ½ I.Q. = 4 x ? x 126 54.8035à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.5271 Different right-angled triangles have different I.Q. s but enlarged versions have the same I.Q. s as smaller triangles. Isosceles Triangles 1. P = 6+6+4 = 16 cm H = cà ¯Ã‚ ¿Ã‚ ½=aà ¯Ã‚ ¿Ã‚ ½+b à ¯Ã‚ ¿Ã‚ ½ 6à ¯Ã‚ ¿Ã‚ ½ = aà ¯Ã‚ ¿Ã‚ ½ + 2à ¯Ã‚ ¿Ã‚ ½ 6à ¯Ã‚ ¿Ã‚ ½ 2à ¯Ã‚ ¿Ã‚ ½ = aà ¯Ã‚ ¿Ã‚ ½ 32=aà ¯Ã‚ ¿Ã‚ ½ V32 = a 5.66 = a A of isosceles= 1/2 x 4 x 5.66 = 11.32 cmà ¯Ã‚ ¿Ã‚ ½ I.Q. = 4 x ? x 11.32 16à ¯Ã‚ ¿Ã‚ ½ I.Q.= 0.556 I predict that an enlarged version of the isosceles triangle will have the same I.Q. 2. P = 8+12+12 = 32 H = cà ¯Ã‚ ¿Ã‚ ½=aà ¯Ã‚ ¿Ã‚ ½+b à ¯Ã‚ ¿Ã‚ ½ 12à ¯Ã‚ ¿Ã‚ ½ = aà ¯Ã‚ ¿Ã‚ ½ + 4à ¯Ã‚ ¿Ã‚ ½ 12à ¯Ã‚ ¿Ã‚ ½ 4à ¯Ã‚ ¿Ã‚ ½ = aà ¯Ã‚ ¿Ã‚ ½ 128= aà ¯Ã‚ ¿Ã‚ ½ V128 = a 11.314 cm = a A of isosceles= 1/2 x 8 x 11.314 = 45.256 I.Q. = 4 x ? x 45.256 32à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.55537 The answers were the same my prediction was correct. (Do not round for more accurate answers.) I will now look at a different isosceles and an enlarged version. 3. P= 8+8+5 = 21 H = cà ¯Ã‚ ¿Ã‚ ½=aà ¯Ã‚ ¿Ã‚ ½+b à ¯Ã‚ ¿Ã‚ ½ 8à ¯Ã‚ ¿Ã‚ ½ = aà ¯Ã‚ ¿Ã‚ ½ + 2.5à ¯Ã‚ ¿Ã‚ ½ 8à ¯Ã‚ ¿Ã‚ ½ 2.5à ¯Ã‚ ¿Ã‚ ½ = aà ¯Ã‚ ¿Ã‚ ½ 57.75= aà ¯Ã‚ ¿Ã‚ ½ V57.75 = a 7.599342077 cm = a A= 1/2 x 5 x 7.599342077 = 18.99835519 I.Q. = 4 x ? x 18.99835519 21à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.541361388 I am now looking at an enlarged version of the above triangle; I predict that the I.Q. will be 0.54136188. 4. P= 16+16+10 = 42cmà ¯Ã‚ ¿Ã‚ ½ H = cà ¯Ã‚ ¿Ã‚ ½=aà ¯Ã‚ ¿Ã‚ ½+b à ¯Ã‚ ¿Ã‚ ½ 16à ¯Ã‚ ¿Ã‚ ½ = aà ¯Ã‚ ¿Ã‚ ½ + 5à ¯Ã‚ ¿Ã‚ ½ 16à ¯Ã‚ ¿Ã‚ ½ 5à ¯Ã‚ ¿Ã‚ ½ = aà ¯Ã‚ ¿Ã‚ ½ 231= aà ¯Ã‚ ¿Ã‚ ½ V231 = a 15.19868415 cm = a A= 1/2 x 10 x 15.19868415 = 75.99342075 cm à ¯Ã‚ ¿Ã‚ ½ I.Q. = 4 x ? x 75.99342075 42à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.541361388 The I.Q. s of the isosceles enlarged triangle is equal to the smaller triangle, therefore my prediction was correct. The I.Q. s of enlarged isosceles triangles are equal. However, when they are not similar or enlarged isosceles triangles they have different I.Q. s Equilateral Triangles I think that all equilateral triangles will have the same I.Q. because each one is an enlargement of another; they are all similar. 1. I am going to us 1/2absinC to find the area because all the angles in an equilateral triangle are 60à ¯Ã‚ ¿Ã‚ ½. A= 1/2 x absinC = 1/2 x 3 x 3 x sin 60 = 3.897114317 P= 3+3+3 = 9 I.Q. = 4 x ? x 3.897114317 9à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.604599788 I will now look at other similar enlarged versions. 2. A= 1/2 x absinC = 1/2 x 5 x 5 x sin 60 = 10.82531755 P= 5+5+5 = 15 I.Q. = 4 x ? x 10.82531755 15à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.604599788 3. A= 1/2 x absinC = 1/2 x 11 x 11 x sin 60 = 52.39453693 P= 11+11+11 = 33 I.Q. = 4 x ? x 52.39453693 33à ¯Ã‚ ¿Ã‚ ½ I.Q. = 0.604599788 The I.Q. for each triangle was 0.604599788. They were all equal. I will now construct a table comparing all of the isoperimetric quotients I have found. I will use this to try and discover a pattern between the I.Q. s.