Ratios and Scale Factors

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Presentation transcript:

Ratios and Scale Factors Slideshow 33, Mathematics Mr. Richard Sasaki

Objectives Understand how to find and use the Ratio of Lengths and Ratio of Areas for given shapes Be able to find the value (weight) of a ratio Be able to find missing edges of similar triangles with the use of ratios Be aware that ratios can be formed for similar triangles in more than one way

Starting Exercise Find the unknowns in the equations with ratios. 72:64=9: 8 2:6=14: 42 4 :7=8:14 32: =12:24 64 6 :12=7:14 30:36=5: 6 :6=4:3 8 1:2= :4 2 4:2=20: 10 6: =12:18 9 63:2∙ =7∙2 9 2 +1:5=18:30 27=16:2∙ 12 6:3∙ −1=9:12 3 15:3=2∙ −1:1 3 13 −8:4=30:24 1 2 2 :14=1:2 14 6:6∙ =54:27 6:2= 2 −3∙ :18 (−6) (−6) 12:15= 2 :45 ±6 or 6:2= 2 −3∙ :18 9 9

Ratios of Different Properties Have a look at the two rectangles below. 𝑀 𝑁 𝐴 𝐵 1.5 𝑐𝑚 𝑋 3 𝑐𝑚 𝑌 𝐶 𝐷 4 𝑐𝑚 𝑃 𝑄 8 𝑐𝑚 In what way are their sizes different? We can either consider their areas or their dimensions (length and width). Note: These ratios are not scales! Ratio of Lengths: 𝑨𝑩 : 𝑴𝑵 = 𝟏 :𝟐 Ratio of Areas: 𝑿:𝒀= 𝟏 𝟐 : 𝟐 𝟐 =𝟏:𝟒 We can use these to calculate missing lengths. Also, Ratios of Volume exist for 3D shapes.

1 :2 1 :3 1 :4 1 :5 1 :10 1 :4 1 :9 1 :16 1 :25 1 :100 Ratio of Lengths - 1 :2 𝐴=32 𝑐 𝑚 2 Ratio of Areas - 1 :4 Ratio of Lengths - 2 :1 𝐵=20 𝑐 𝑚 2 Ratio of Areas - 4 :1 Ratio of Lengths - 2 :3 𝐶=45 𝑐 𝑚 2 Ratio of Areas - 4 :9 𝐷= 𝜋 4 𝑐 𝑚 2 𝐸=4𝜋 𝑐 𝑚 2 Ratio of Lengths - 1 :4 Ratio of Areas - 1 :16 𝑥= 1 2 𝑐𝑚, 𝑦=2 𝑐𝑚

Scale Factors and Ratio Values Here are two similar ideas with inverse principles. If we look at the Ratio of Lengths for the two squares, we get . 𝑎:𝑏 1 𝑏 𝑎 ∴ The value of the ratio is . 𝑎 𝑏 𝑎 𝑐𝑚 𝑏 𝑐𝑚 What is a Scale Factor? Scale Factors are an enlargement of one shape to another (a transformation). Scale factors consider lengths, not areas. Less than 1 The shape has shrunk. Exactly 1 The same size as the original. More than 1 The shape has grown.

Scale Factor: 1 2 Scale Factor: 3 Value of Ratio: 1 3 Value of Ratio: 2 𝑆𝑐𝑎𝑙𝑒 𝐹𝑎𝑐𝑡𝑜𝑟= 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑅𝑎𝑡𝑖𝑜 −1 Non-existent ∆𝐴𝐵𝐶~∆𝐸𝐹𝐷 𝐸𝐹 =6 𝑐𝑚 𝑎𝑑=𝑏𝑐

Finding Lengths of Shapes Once again, for two shapes to be similar, they must… Have the same angles as one another Have lengths in the same proportion to each other The lengths must be in corresponding places (in terms of the angles for this to work. We could only complete a question like this using the sine or cosine rule. 3 𝑐𝑚 45 𝑜 80 𝑜 45 𝑜 5 𝑐𝑚 10 𝑐𝑚 Let’s do questions where angles and lengths are in the same place! 80 𝑜 ? 𝑐𝑚

Finding lengths of Shapes Look at the two shapes below. 𝐵 𝑌 We can say . ∆𝐴𝐵𝐶~ ∆𝑌𝑍𝑋 𝑎 𝑐𝑚 𝑧 𝑐𝑚 Here 𝑎 corresponds to and 𝑏 corresponds to . 𝑦 𝐶 𝑧 𝐴 𝑏 𝑐𝑚 𝑋 𝑍 𝑦 𝑐𝑚 Using the Ratio of Lengths, the equations for similarity are… or (or vice-versa). 𝑎:𝑏=𝑦:𝑧 𝑎:𝑦=𝑏:𝑧 𝑎 𝑏 = 𝑦 𝑧 Using the first, we can say which implies that . 𝑎𝑧=𝑏𝑦

Finding lengths of Shapes Example Look at the two similar rectangles below. 3 𝑐𝑚 5 𝑐𝑚 7 𝑐𝑚 𝑥 𝑐𝑚 Write an equation with ratios for the rectangles given and hence, solve for 𝑥. Equation: 3:5=7:𝑥 (or one of the other three) 3 5 = 7 𝑥 ⇒ 𝑥= 35 3 3𝑥=7∙5⇒

Answers 𝑏= 35 6 𝑎= 10 3 𝑐= 240 17 𝑑= 24 5 𝑥= 73 16 𝑥=4 Scale Factor = 11 9 Length = 275 9 or 30. 5 𝑐𝑚