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Geometric Similarities Math 416

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Geometric Similarities Time Frame 1) Similarity Correspondence 2) Proportionality (SSS) (Side-side-side) 3) Proportionality (SAS) (side-angle-side) 4) Similarity Postulates 5) Deductions 6) Dimensions 7) Three Dimensions

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Similarity Correspondence Similarity – Two shapes are said to be similar if they have the same angles and their sides are proportional Note – we see shape by angles & we see size with side length

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Consider A C D B Z Y W X Similar & Why?

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Proportionality (SSS) We say the two shapes are similar because their angles are the same and their sides are proportional We can note corresponding points A X D W B Y C Z

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Angles We note corresponding angles < ADC = < XWZ (95°) < DCB = < WZY (85°) < CBA = < ZYX (80°) < BAD = < YXW (100°)

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Notes Hence we would say ADCB XWZY Hence we note corresponding angles < ADC = < XWZ < DCB = < WZY < CBA = < ZYX < BAD = < YXW

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Proportionality Next is proportionality which we will state as a fraction AD=8 DC=16 CB=32 BA=24 XW 5 WZ 10 ZY 20 YX 15 What is the proportion (not in a fraction)? 8/5 which is reduced to 1.6

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Question #1 Identify the similar figures and state the similarity relationship, side proportion and angle equality A C Z T C B BIG SMALL BIG MED SMALL

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Notes for Solution By observing you need to establish the relationship. Look at angles or side lengths Important: An important trick when comparing angles and sides is that the biggest angles is always across the biggest side, the smallest from the smallest and medium from the medium.

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Solution #1 Triangle ABC ˜ TCZ AB = BC = CA TC CZ ZT < ABC = < TCZ < BCA =

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Important Note Make sure the middle angle letters are all different because the middle letter is the actual angle that you are looking at. AC = CA < ACB = < BCA Both the above are the same

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Question #2 K R L T X Q MED SMALL MED With isosceles (or equilateral triangles) you may get two (or three) different answers). However, you are only required to provide one.

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Solution a for #2 The question is to identify similar figures and state the similarity relationship, side proportion and graph equality. QK = KT = QT RX XL RL < QKT = < RXL < KTQ = < XLR < TQK = < LRX QKT ˜ RXL

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Solution b for #2 You can also have another solution Triangle QKT is still congruent to RLX QK = KT = QT RL LX RX < QKT = < RLX < KTQ = < LXR < TQK = XRL

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More Notes There are other ways of establishing similarity in triangles At this point we will abandon reality for simple effective but not accurate drawings of triangles… (it is not to scale). Please complete #1 a – o For Question #3, again, state similarity relationship, side proportion and angle equality.

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Question #3 T Q R C B A If the three sides are proportional to the corresponding three sides in the other triangle, the two will be similar.

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Solution Notes You need to check… SMALL with SMALL MEDIUM with MEDIUM BIG with BIG

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Solution #3 ABC ˜ QRT Med Small Big Small Med Big 18 = 21 = = 0.6 = 0.6; YES SIMILAR

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Proportionality SAS We can also show similarity in triangles if we can find two set corresponding sides proportional and the contained angles equal; we can determine similarity 14°18 A B C Z Y X 42 14° 15 35

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Question #4 Show if the triangle is similar Solution… since

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Triangle Similarity Postulates There are three main postulates we use to state similarity SSS all corresponding sides proportional SAS two sets of corresponding sides and the contained angle are equal AA two angles (the third is automatically equal since in a triangle, the interior angle must add up to 180°) are equal

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Example #1 Why are the following statements true? QPT ˜ ZXA 42° 84° 54° AA PT Z A X Q

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Example #2 Why are the following statements true? KTR ˜ PMN 51° Solution: since 24/16 = 27/18 R KT P N M SAS 24

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Example # CB A P T K Since 16 = 24 = /3 = 8/3 = 8/3 S S S

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Parallel Lines Facts: If two tranversals intersect three parallel lines, the segments between the lines are proportional a bd Therefore, a = c b d c

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Notes Also note that… A C B BC = 1 AC 2

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Parallel and the Triangle If a parallel line to a side of a triangle intersects the other two sides it creates two similar triangles A E B C D Therefore, ABE ˜ ACD

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Question #1 3 4x 9 Find x 3 = 9 4 x 3x = 36 x = 12

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Question #2 C x B E A D We note BE // CD Thus, ABE ˜ ACD AB = BE = AE AC CD AD x = 50 = AE x AD

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Question #2 Soln Cont We only need x = 50 x x = 50(x+40) 150x = 50x x = 2000 x = 20

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Proportion Ratio Consider

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Dimensions SIDE SMALLBIGRATIO SIDE 1D1101:10 or 1/10 AREA 2D11001:100 OR 1/100

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Dimensions In general in 1D if a:b then in 2D a 2 : b 2 Ex. In 1D if 5:3 then in 2D? In 2D then 25:9 You can go backwards by using square root Ex. In 2D if 36:49 then in 1D 6:7

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3D or Volume Consider

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3D or Volume SmallBigRatio Side 1D151:5 or 1/5 Volume 3D11251:125 or 1/125

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3D or Volume In general in 1D if a:b then in 3D… Then in 3D a 3 : b 3 Ex. In 1D if 6:7 then in 3D In 3d 216:343 You can go backwards by using the cube root Ex. In 3D if 27:8 then in 1D In 1D 3:2

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Practice Complete the following 1 D Length 2 D Area 3 D Volume 2:94:81 8:729 2:114:121 8:1331 5:3 25:9125:27

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3D Question #1 Two spheres have a volume ratio of 64:125. If the radius of the large one is 11cm, what is the radius of the small one? Big Small r 11 x 3D Ratio 64:125 1D Ratio 4:5 Thus 4 = x x = 44 x = 8.8

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3D Question #2 V=200m 3 V = ? A Base = 100m 2 A base = 16 m 2

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Question #2 Solution Big Small Area of Base Volume 200 x 1 Ratio 10 / 4 2 Ratio 100/16 3 Ratio 1000/64 Thus 200 = 1000 x x = x = 12.8

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