ENLARGING AREAS AND VOLUMES

Transcription

1 ENLARGING AREAS AND VOLUMES First of all I m going to investigate the relationship between the scale factor and the enlargement of the area of polygons: I will use my own examples. Scale factor: 2 A 1 : 8cm 2 A 2 : 32cm 2 What is the change of A 1 in A 2? A 1: 2cm x 4cm A 2: (2cm x 2) x (4cm x 2) Scale Factor: A 2/A 1 = 2cm x 2 x 4cm x 2 2cm x 4cm = 2cm x 4cm x 2 x 2 2cm x 4cm = 2 x 2 Sf = 2 2 Hypothesis: The enlarged area is the original area times the scale factor squared. E A(X) =X x Sf 2 Trying it out:

2 Calculations : 1 a) 3 x 4 / 2 = 6 b) 6 x 8 /2 = a) 3.14 x 1 x 1 = 3.14 b) 3.14 x 3 x 3 = a) 1.5 x x ( 3.14 x 0.75 x 0.75 ) = b) 6 x x ( 3.14 x 3 x 3 ) = a) ( ) x 0.5 / 2 = 1.25 b) ( ) x 2.5 / 2 = 62.5 Fig Original Area (cm 2 ) Enlarged Area(cm 2 ) Scale Factor Area 1 a) 6 b) a) 3.14 b) a) b) a) 1.25 b) Conclusion: The hypothesis is correct as it works for different polygons, regular and irregular, and also for different scale factors. This is because the sides are directly proportional to the area, and the sides are proportional to the corresponding ones in the enlarged figure: L 2 L 1 A 2 L 2 L 1 A 1 L 2 = L 1 x Sf And A 2 = L 2 x L 2 Substituting L 2 = L 1 x Sf A 2 = L 1 x Sf x L 1 x Sf A 2 =(L 1) 2 x Sf 2 A 2 = A 1 x Sf 2 Furthermore, we can say that since we are figuring out the area of a two dimensional polygon, to do so, we need two factors, length and width. Since we are multiplying each by the scale factor, the product will be the original area (length x width) multiplied by the scale factor twice, in other words, squared.

3 We can now solve Bettys original problem. If her friends garden is 6m long the scale factor is 6 / 5= 1.2. So the area will be 40 x 1.2 x 1.2 = 57.6 m 2 Now to investigate the relationship between the scale factor and the enlargement of the volume of 3D bodies: 3cm 3cm 6cm 6cm 3cm V 1 : 27cm 3 V 2 : 216cm 2 6cm What is the change of V 1 in V 2 V 1: 3cm x 3cm x 3cm V 2: (3cm x 2) x (3cm x 2) x (3cm x 2) A 1,2 = 3cm x 2 x 3cm x 2 x 3cm x 2 3cm x 3cm x 3cm = 3cm x 3cm x 3cm x 2 x 2 x 2 3cm x 3cm x 3cm = 2 x 2 x 2 = 2 3 Hypothesis: The enlarged area is the original area times the scale factor cubed. E v(x) =X x Sf 3 Trying it out: 1. Volume 1 a) 3 x 1 x 1 x 3.14 = 9.42 b) 6 x 2 x 2 x 3.14 = Enlargement / 9.42 = 8 times 2. Volume 2 a) 2 x 2 x 3 / 2 = 6 b) 6 x 6 x 9 / 2 = 162 Enlargement 162 / 6 = 27 times

4 3. Volume 3 a ) 1 x 1 x 2 / 3 = 2 / 3 b) 4 x 4 x 8 / 3 = 42 2 / 3 Enlargement 42 2/3 divided by 2/3 = Volume 4 a) 2 x 2 x 2 x 3.14 x 4 / 3 = b) 10 x 10 x 10 x 3.14 x 4 / 3 = Enlargement / = 125 Fig Original Volume (cm 3 ) Enlarged Volume (cm 3 ) Scale Factor Volume /3 42 2/ / / Conclusion: The hypothesis is correct as it works for different 3D shapes, with different scale factors. This is because the sides are directly proportional to the volume, and the sides of the original figure are proportional to the corresponding ones in the enlarged one: L 2 L 1 V 2 L 2 V 1 L 1 L 1 L 2 L 2 = L 1 x Sf V 2 = L 2 x L 2 x L 2 = L 1 x Sf x L 1 x Sf x L 1 x Sf V 2 = (L 1) 3 x (Sf) 3 = V 1 x (Sf) 3 Furthermore, we can say that since we are figuring out the area of a three dimensional polygon, to do so, we need three factors, length, width and height. Since we are multiplying each by the scale factor, the product will be the original area (length x width x height) multiplied by the scale factor thrice, in other words, cubed.

5 We can now solve Bettys problem. If the volume is a half of the original then we must have (Sf) 3 = 0.5. Taking the cube root we get Sf = So the diameter of the bottle would be 8 x = cm Application of Discoveries to solve problems: 1. The area of a rectangle is 56cm 2 if the length and width are doubled what will be the area of the new rectangle? Explain why. According to the rule I had discovered earlier, the enlarged area of a figure is equal to the original area times the scale factor squared. In this case, since the area of the rectangle is 56cm 2 and the scale factor is two, since both the length and width are doubled, the enlarged area will be: 56cm 2 x 2 2 = 224cm 2 [~>>Answer: 224cm 2 <<~] 2. The volume of a Ball of radius is 2cm is 200cm 3, what will the volume of a ball radius 6cm be? I must find out what the scale factor is by dividing the radius of the enlarged ball by the original. Then I must apply it to the rule I discovered, finding out, in this way, the volume of the enlarged ball: Sf = 6cm = 3 2cm E v(x) =X x Sf 3 E v(x) = 200cm 3 x 3 3 E v(x) = 5400 cm 3 [~>>Answer: 5400cm 3 <<~] 3. The big shape below is an enlargement of the small shape. What is the area of the big shape? Show your working out giving reasons. A:? cm 2 A: 30cm 2 As exercise 2, I must find the Sf to apply it to the rule. Sf = 6cm = 3 4cm E A(X) =X x Sf 2 E A(X) = 30cm 2 x E A(X) = 67.5 cm 2 [~>>Answer: 67.5cm 2 <<~] 5

6 4. The big cylinder is an enlargement of the small cylinder. Find the volume of the small cylinder. Show you working out explain how you did it. Volume =? Volume = 1000cm 3 Once again, the scale factor must be calculated in order to calculate the volume: Sf = 10cm = 2.5 4cm E v(x) =X x Sf cm 3 = X x X = 64 cm 3 [~>>Answer: 64cm 3 <<~] 5. The diameter of the Earth is 8000 miles. The diameter of the Sun is miles. How many times bigger in Volume is the Sun compared to the Earth? Going back to the rule discovered, the volume of the enlarged figure, in this case, the Sun s, is the volume of the original figure, the Earth s, multiplied by the scale factor cubed. In this case, the scale factor can be calculated by dividing the Sun s volume by the Earth s: Sf = m 8000m = 200 We do not even need to know the radius, nor calculate the volume of the bodies, since the volume of the Sun will be the Earth s times 200 cubed (200 3 = ). So, the Sun s volume will be times bigger than the Earth.

7 Reflection Reflect on the methods used in this project (How did you discover the rules) and the reliability of your discoveries. How might this rule be useful in real life? Can you think of any real life situations where things are enlargements of each other? I discovered the rule by finding out the difference of the original and enlarged areas and volumes by dividing the latter by the former. Then, I compared the answer with the scale factor and realized that the answer was the scale factor squared (or cubed). I tried the same with other shapes and as the rule happened in all the cases, I was able to generalize it as a reliable formula, which could be applied to all cases of enlargement of 2D and 3D polygons and bodies. In real life, many cases of things being enlargements of others exist. One very obvious and common case is that of the maps. In these, scales are used and the real distances and the like would be the enlargements of the symbols drawn. Another example is when people design something. Sometimes, the real object is drawn on a piece of paper or sometimes even made in a little model, also using a scale.