Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Euler angles wikipedia , lookup
Rational trigonometry wikipedia , lookup
Renormalization group wikipedia , lookup
History of trigonometry wikipedia , lookup
Trigonometric functions wikipedia , lookup
Euclidean geometry wikipedia , lookup
Pythagorean theorem wikipedia , lookup
Similarity What is Similarity? Similarity means “like something but not exactly the same”. In geometry, similarity means the same in shape but not in size. Two polygons which have the same shape are called Similar Polygons. When objects are similar, one is an accurate enlargement of the other. The symbol for similarity is ~ Let’s see if the following pairs of shapes are similar or not Pair 1 – The lengths have doubled and the corresponding angles are the same. This makes these polygons similar W Pair 2 – S T X Z V Y U In the above pairs the lengths of the sides may be different but if you compare them, their ratio is the same. Pair 2 - ST = TU = UV = VS WX XY YZ ZW In the above pairs of polygons, the corresponding angles are also equal Pair 2 – S=W, T=X, Y = U, Z = V STUV~WXYZ This leads to the following conclusion – Two polygons can only be similar if - The ratios of pairs of corresponding sides must be equal - Corresponding angles must be equal. Sometimes it is possible that 2 polygons do not follow the above conditions. Then we say that the polygons are not similar Examples 18cm 6cm 27cm 75o 80o 21 cm Not similar because corresponding angles are different 9cm 120o 7cm 120o 32 cm 8cm not similar because ratio of corresponding sides is different Questions Are the following pairs of polygons similar or not? 100 cm 10 cm 50cm Yes, the above pair of polygons is similar because they obey the conditions 5 cm 102o 70o 6 cm 21cm 130o 90o 88o 100o 60o 16 cm 80o 4 cm The above shapes are not similar as the corresponding angles and the ratio of corresponding sides is not the same Which of the following must be similar to each other? a) 2 equilateral triangles - similar b) 2 squares - similar c) 2 rectangles – must not be similar but could be d) 2 circles? – must be similar Now that we have established the two conditions required for polygons to be similar, we can easily use them to find the missing angle or a missing length of a side of a polygon. Examples Below are two similar polygons. Find angle T and length of side QR P 6cm 5 cm T 86o 12 cm 10cm Q x S 8 cm U W R 16 cm 14 cm V Angle T is 86 because corresponding angles of similar polygons are equal i.e. angle P = angle T Side QR can be found in the following way: We know that ratios of the corresponding sides are the same so PQ = QR TU UV o 5 = x 10 14 Cross multiply: 10x = 70 10 10 So x = 7 cm Some more questions…. 1. Find angle I C 3 cm 60o D G 9 cm H 80o 7cm x 21 cm 18 cm E 100o F 4cm E J 12 cm I Answer – Angle I = angle E Angle E = 360o-(100o + 60o + 80o) (all internal angles of a quadrilateral add up to 360o) Angle E = I = 120o From the above polygon, find the length of side DE DE (x) = CD/GH = x/HI 3/9 = x/18 Cross multiply, 54 cm = 9x 6 cm = x 2. In the given pair of similar polygons, the ratio of the corresponding sides is 2:3. a) Find angle R b) Find all the sides of polygon QRST R 6 cm N Q Cross M 44o multiplication can also be used; 5 cm 9 cm your answer will be the same. P O T S 7 cm Answer – angle N =angle R Angle N = 44o then angle R = 44o b) For this you have to multiply each of the lengths of the sides of MNOP by 3/2. We use 3/2 (reciprocal of the scale factor) because we are increasing the value of the sides of MNOP by this ratio. Your answers will be QR = 9cm, RS = 13.5 cm, ST = 10.5cm, TQ =7.5cm Similar Triangles We have seen that for polygons to be similar all corresponding angles must be equal and the ratio of the corresponding sides must be the same. In triangles, this is not necessary. Let’s see how: Look at the following triangles, 35o 55o To find C , you have to add all three angles in the triangle, A +B+C =180 90+55+C = 180 C = 180 – 145 C= 35 If you look at the second triangle, angle E is 35 We can now say that C = E and we already know that A = D These triangles can be called similar because two angles of triangle 1 are equal to two angles of triangle 2. AA similarity Postulate – If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. If this happens, then by default the third angle is also equal Also the ratio of the corresponding sides is equal BA= AC = CB FD DE EF Are the following triangles similar? If Yes, show how? 1) 5 2 4 2.5 3 6 The above triangles are similar because all three angles are equal, even though only two are indicated; the third one is equal by default. All sides are in the same proportion i.e. (2.5/5) = (2/4) = (3/6) All these reduce to ½ = 1:2 2) B D 4 21 16 9 E A 7 F C 6 The above triangles are not equal because the corresponding angles are not equal and the corresponding sides are not in the same ratio Areas and perimeters of similar triangles Perimeter - Sum of All sides Of a Polygon Relationship between the corresponding sides of similar triangles and their scale factor 6 J G 2 9 L 3 I 12 4 K H All the corresponding angles are equal in the above diagram 6 = 9 = 12 2 3 4 All the above ratios simplify to 3:1 3:1 is also known as the scale factor of the similar triangles The reduced ratio of corresponding sides of similar triangles is called the scale factor Relationship between the scale factor and the perimeter of similar triangles Using the above example; Perimeter of Triangle 1 = 27 cm Perimeter of triangle 2 = 9cm 27 cm = 3 = 3:1 which is the scale factor of the similar triangle 9 cm 1 Ratio of the perimeter is equal to the reduced ratio of corresponding sides This leads us to the following theorem: Theorem 60: if the scale factor of two similar triangles is a:b, then the ratio of their perimeters will also be a:b Answers a) scale factor = reduced ratio of corresponding sides 5/20 = 7/28 = 9/36 All these reduce to ¼ Scale factor = 1:4 Questions…. 1) For the following pair of similar triangles a) find the scale factor of the corresponding sides b) find the ratio of their perimeters X 5cm U 7cm Y 28cm 20cm Z 9cm V W 36cm b) Application of theorem 60. Perimeter of XYZ = 21cm and of UVW = 84cm. Ratio = 21/84 = ¼ = 1:4 2) The scale factor of 2 similar triangles is 3:4. The perimeter of triangle 1 is 24 cm. what is the perimeter of triangle 2? Answer: scale factor = ratio of perimeters (theorem 60) 3:4 = 21: x Cross multiply = 4(21) = 3x 84/3 = 3x/3 28cm = x Perimeter of triangle 2 = 28cm Answer – a) Perimeter of triangle FGH (z) = Scale factor = 16/4 = 4/1 = 4:1 Triangle CDE has a perimeter of 76cm 4:1 = 76cm: x (cross multiply) z = 19cm 3) For the following pair of similar triangles, a) Find the perimeter of triangle FGH b) Find the remaining sides of triangle FGH C F 16cm 4cm y 28cm G D H E x b) Use scale factor to find remaining sides e.g for x, you can do 4/1 = 32/x and cross multiply x = 8cm y = 7cm 32cm 4) If triangle ABC has a perimeter of 27cm and triangle EFG has a perimeter of 13.5 cm, what is the ratio of the perimeter? The sides of triangle EFG are 6 cm, 5cm and 2.5 cm. find the sides of triangle ABC? -Answer Ratio of perimeter = perimeter of ABC/perimeter of EFG = 27cm/13.5 cm = 2/1 = 2:1 Sides of triangle ABC = multiplication of sides by the scale factor 2/1 x 6 = 12cm 2/1 x 5 = 10cm 2/1 x 2.5 = 5cm Area Look at the below pair of similar triangles; 3cm 24cm 28cm 14 cm Scale factor = 1:2 (see before on how to get scale factor) Area of triangle 1 = 42 cm2 Area of triangle 2 = 168 cm2 What is the ratio of the areas? 42 = 1 = 1:4 168 4 Relationship between scale factor and area ratio? Scale factor = 1:2 1= 1 2 4 area ratio: 1:4 (1/2) = (1/2)2 As you can see, the ratio of the areas is the square of the ratio of the scale factor proving that: Theorem 61: if two similar triangles have a scale factor a:b, then the ratios of their areas is a2:b2 Questions…. 1) Two similar triangles have an area of 18cm2 and 32 cm2 respectively; if the base of the smaller triangle is 6cm, find the base of the larger triangle? Answer – Scale factor = (ratio of areas) 2 – theorem 61 32 = (a/b) 2 18 Simplify the fraction = 16/9 Now reduce it by finding the square root. Your answer is 4/3 and that is your scale factor. So to find the base, all you have to do is cross multiply; 4:3 = x: 6 x = 8cm Answer – x/4 = (20/5)2 x/4 = (4/1)2 find square root of (4/1)2 x/4 = 2/1 x=4x2 x = 8cm 2) Below are two similar triangles. Find x 20cm2 5cm2 4cm x 3) The perimeters of two similar triangles is in the ratio 5:6. The sum of their areas is 122 cm2 Find the area of each triangle. Answer – Let 5x be a side in triangle 1 and 6x be the corresponding side in triangle 2 Perimeter ratio = scale factor Scale factor = (ratio of areas)2 Therefore the ratio of perimeter = (ratio of areas)2 5/6 = (5x/6x)2 = 25x2/36x2 Area of each triangle = 25x2 + 36x2 = 122 cm2 61x2 = 122cm2 x2 = 2 cm Area of Area of 1 = 25x2 = 25 x 2 = 50cm2 2 = 36x2 = 36 x 2 = 72 cm 2 Try This! Two similar rectangles have been drawn below. Find A (Area) using the concept of theorem 61 A 8cm 18cm2 6cm 16 cm 3cm Answer – the process is the same. Find the scale factor which in this case is 8:3 Then square it, you’ll get 64:9. Then cross multiply 64:9 = A: 18 A = 128cm2 Similarity in 3-D Shapes We already know that if two shapes are similar their corresponding sides are in the same ratio, and their corresponding angles are equal.3D shapes can also be similar provided they follow the same rules Examples of similar 3D shapes: a) Below are two cuboids 6cm 3cm 1cm 2cm 2cm 4cm The length of cuboid 1 is twice the length of cuboid 2 The width of cuboid 1 is twice the width of cuboid 2 The height of cuboid 1 twice the height of cuboid 2 Therefore the scale factor is 2:1, all angles are same and so the cuboids are similar. b) Two cones can also be similar. 3x y slant height x 3y radius – 3z radius - z The radius of cone 2 is thrice that of cone 1 The height of cone 2 is thrice that of cone1 The slanted height of cone 2 is thrice that of cone 1 All angles are also equal Therefore, the cones are also similar. Are the following shapes similar or not? 8cm 24cm Radius – 6 cm Radius – 30cm Below are 2 cubes – are they similar? a aa a a b b b No, these shapes are not similar because the ratio of the corresponding sides is not equal 6/30 is not equal to 8/24. b All cubes are similar without any exception. This is because the length, width and height of a cube is the same. So the ratio of corresponding sides will also be the same i.e Length = width = height a= a = a b b b Scale factor = a:b When dealing with 3D shapes, we only talk about 2 operations namely a) Volume b) Surface Area Volume of similar 3D figures What is the ratio of their volumes? Cube1 has a volume of a3 Cube 2 has a volume of b3 The ratio of their volumes is a3:b3 What is the ratio of the corresponding sides (scale factor)? a: b This means that for a cube of scale factor a: b the ratio of their volume is a3:b3 This leads to the following conclusion: for any similar 3D shapes with a scale factor of a: b, the ratio of the volumes will be a3:b3 Example: Look at the shapes below 60cm3 480cm3 5 cm 10cm The ratio of corresponding sides is 5:10 which when reduced is 1:2 The ratio of the volume is 60:480 which when reduced you get 1:8 Therefore: 1 = 1 which is the same as ½ = (1/2)3 2 8 This proves our conclusion. The theorem can also be expressed as follows: Volume 1 = (a/b)3. Volume 2 Questions 1) Two similar jugs have heights of 4cm and 6cm respectively. If the capacity of the smaller jug is 50cm3, find the capacity of the larger jug. Answer – Scale factor = 4/6 = 2/3 = 2:3 Volume 1 = a 3 Volume 2 b 50 cm3 = (2/3)3 x 50cm3/x = 8/27 Cross multiply; x = 50 x 27 = 1350 x= 1350/8 x= 168.75 cm3 2) Below are 2 cones. All the dimensions in each are given. Show that they are similar. (Find volume of each as well) 50cm 25cm 48cm 24cm Radius = 7cm radius = 14cm Answer - the above cones are similar, Scale factor of the cones – 7 = 24 = 25 = 1:2 (all sides are corresponding to each other) 14 48 50 Volume of cone 1 = 1/3 1/3 x x 73 1/3 x 343 r3 Volume of cone 2 = 1/3 1/3 x x 143 1/3 x 2744 r3 V1 = 1x 343 x3 = 343 V2 3 2744 2744 Answer is 1/8 which is the same as (1/2)3, proving that the cones are similar 2) By referring to the similar cylinders below, find V 5 cm 20 cm3 15cm V Answer – Scale factor – 5/15 = 1:3 In volume, scale factor is cubed = (1/3)3 = 1/27 1/27 = 20/V V = 20 x 27 V = 540cm3 3) Find the missing length Answer – Volume 1 = (a/b)3 Volume 2 3 3cm = (2/x)3 24cm3 (3/24) = 8/x3, cross multiply 24 x 8 = 3x3 192/3 = 3x3/3 64 = x3 4 cm = x 24cm3 3c 3 3cm 2cm x 4) The area of the bases of two similar glasses is in the ratio 4:25. Find the ratio of their volumes Answer Your first step is to find the reduced ratio (scale factor) 4:25 is the area ratio that means that it is in the form of (a:b)2 Reduce the ratio to get the scale factor. Reduce it by finding the square root Your answer will be 2/5 Now use the normal process to find volume ratio 2/5 = (2/5)3 Final ans – 8/125 = 8:125 (ratio of volumes) Surface Area This is the same as the process which you use when you look for areas of 2D shapes Let’s look at an example: Below are 2 similar cones 50cm 25cm 48cm 24cm Radius = 7cm radius = 14cm The scale factor/dimensions are in the ratio 1:2. Let’s find the surface area of the cones S.A of cone = rl+ r2 Cone 1 = 224 Cone 2 = 896 Cone 2/cone 1 = 224 /896 - Pi = 1:4 (Scale factor) 2 = area ratio (1/2)2 = 1/4 So if scale factor is 1:2, then the area ratio is (1/2)2 – Theorem 61 Questions…. 1) Two similar pyramids have volumes of 64cm3 and 343cm3. What is the ratio of their surface areas? Answer First step is to find the reduced ratio (scale factor) 64/343 is the volume ratio and so it is in the form of (a: b)3 Reduce it by finding the cube root of the numbers. Your answer will be 4/7 Now use the normal process to find area a:b = (a:b)2 4:7 = (4:7)2 Final answer = 16:49 2) Two similar cylinders have a scale factor of 3:5. The surface area of the smaller one is 9cm2. Find the surface area of the larger one? Answer: Use theorem 61 (3:5) = (3:5)2 Answer = 9/25 or 9:25 So the surface area of the larger one will be 25cm2 3) Two similar cones have surface areas in the ratio 4:9. Find the ratio of; a) Their lengths b) Their volumes Answer – Same process as used before. Find square root of 4:9 as it is an area ratio Your answer is 2:3…. (a) Now cube this ratio because you have to find the ratio of the volumes Your answer is 8:27….. (b) 4) The ratio of the surface area of two spheres is 4:1. If the surface area of the larger sphere is 5542cm2. Find the surface area of the smaller one and hence find radius of each sphere. 5542cm2 xcm2 Answer Find ratio of smaller sphere; Cross multiplication required 5542/x = 4/1 x = 5542/4 x = 1385.5 cm2 Find radius of each sphere; Here you will have to use the formula, Sphere 1 = 4 r2 5542 cm2 = 4 r2 4 4 r2 = 441 r = 21 cm Sphere 2 = 4 r2 1385.5 cm2 = 4 r2 4 4 110.25 = r2 r = 10.5 cm Surface Area of Sphere = 4 r2 5) Two similar open cylinders are painted together. The larger one has a surface area of 256cm2 and the scale factor of the two cylinders is 4:3. Find the surface area of the second cylinder. ii) If the radius of the second cylinder is 12cm, what would be the radius of the first? Answer – i) surface area ratio = (4:3)2 = 16:9 Cross multiply = 9 x 256 = 2304/16 S. a of second cylinder = 144cm2 ii) Scale factor = 4:3 Radius of small cylinder = 12cm Radius of large cylinder (x) =? Cross multiply = 12 cm x 4 = 48/3 x = 16cm Try This!! A few revision questions 1) Are triangles ABC and DBE similar? ABC and DBE are similar, because all of the corresponding angles are equal and And all corresponding angles are equal i.e. angle BDE = angle BAC and angle BED = angle BCA 2) The Easypac Soup Company produces small cylindrical tins of radius 3cm, and similar large tins of radius 5cm. If a small tin contains 243ml of soup, how much does a large tin contain? You need to work out the ratio of the volume; Volume ratio for a:b = (a:b)3 (3/5) = (3/5)3 = 27/125 Now you can cross multiply 27/125 = 243/x 125 x 243 = x 27 x = 1125 ml 3) A 33o 55o 40.3 cm This question also uses a concept from the circle theorem. The theorem states that angles formed by a chord in the same sector are equal. In this case it is angle A = angle D and angle B = angle C B 26.8 cm 92o X 20.1cm 55o 33o D a) Find X b) Calculate XD Answer – X = 920, vertically opposite angle XD = cross multiplication of corresponding sides AX/XD = BX/XC 40.3/XD = 26.8/20.1 XD = 30.2cm C 4) Two cones are similar, 8cm 4cm 11 cm l cm a) find l b) When full the larger cone contains 172cm3 of water. How much water does the smaller one contain? Answer – l= 8cm/4cm = 11cm/l cm L cm = 5.5 cm For b) let x be the amount of water in the smaller cone Find scale factor which is 2:1 (8/4 and 11/5.5) Now cube it to find volume ratio = (2:1)3 = 8:1 Cross multiply 8/1 = 172/x x = 21.5 cm3