Mathematical Similarity

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Mathematical Similarity
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Presentation transcript:

Mathematical Similarity Scales (Representative Fractions) Scale Drawings Working Out Scale Factor Similar Triangles 1 Similar Triangles 2 www.mathsrevision.com Similar Triangles 3 with Algebra Similar Figures Scale Factor in 2D (Area) Surface Area of similar Solids Scale Factor 3D (Volume)

Working out Scale Factor Give one distance from the map and the corresponding actual distance we can work out the scale of the map. Example : The map distance from Ben Nevis to Ben Doran is 2cm. The real-life distance is 50km. What is the scale of the map. Map Real Distance 2  50km 1  50 ÷ 2 = 25km Scale Factor 1 : 250 000

Working out Scale Factor Example : The actual length of a Olympic size swimming pool is 50m. On the architect’s plan it is 10cm. What is the scale of the plan. Plan Real Distance 10  50m 1  5 Scale Factor 1 : 500

Similar Triangles 1 Learning Intention Success Criteria To explain how the scale factor applies to similar triangles. 1. Understand how the scale factor applies to similar triangles. 2. Solve problems using scale factor.

Conditions for similarity Two shapes are similar only when: Corresponding sides are in proportion and Corresponding angles are equal All rectangles are not similar to one another since only condition 2 is true.

If two objects are similar then one is an enlargement of the other The rectangles below are similar: Find the scale factor of enlargement that maps A to B A B 8 cm 16 cm 5 cm 10 cm Not to scale! Scale factor = x2 Note that B to A would be x ½

If two objects are similar then one is an enlargement of the other The rectangles below are similar: Find the scale factor of enlargement that maps A to B A B 8 cm 12 cm 5 cm 7½ cm Not to scale! Scale factor = x1½ Note that B to A would be x 2/3

The 3 rectangles are similar. Find the unknown sides 8 cm If we are told that two objects are similar and we can find the scale factor of enlargement then we can calculate the value of an unknown side. A B C 3 cm Not to scale! 24 cm x cm y cm 13½ cm The 3 rectangles are similar. Find the unknown sides Comparing corresponding sides in A and B: 24/8 = 3 so x = 3 x 3 = 9 cm Comparing corresponding sides in A and C: 13½ /3 = 4½ so y = 4 ½ x 8 = 36 cm

The 3 rectangles are similar. Find the unknown sides 5.6 cm If we are told that two objects are similar and we can find the scale factor of enlargement then we can calculate the value of an unknown side. A B C 2.1 cm Not to scale! x cm 7.14 cm 26.88 cm y cm The 3 rectangles are similar. Find the unknown sides Comparing corresponding sides in A and B: 7.14/2.1 = 3.4 so x = 3.4 x 5.6 = 19.04 cm Comparing corresponding sides in A and C: 26.88/5.6 = 4.8 so y = 4.8 x 2.1 = 10.08 cm

Similar Triangles Similar triangles are important in mathematics and their application can be used to solve a wide variety of problems. The two conditions for similarity between shapes as we have seen earlier are: Corresponding sides are in proportion and Corresponding angles are equal Triangles are the exception to this rule. only the second condition is needed Two triangles are similar if their

Similar Triangles

70o 45o 65o These two triangles are similar since they are equiangular. 50o 55o 75o These two triangles are similar since they are equiangular. If 2 triangles have 2 angles the same then they must be equiangular = 180 – 125 = 55

Scale factors Enlargement Scale factor? 3 2 ESF = = 8cm 8 8 12 12cm 12 Reduction Scale factor? 2 3 5cm RSF = = 7.5cm Can you see the relationship between the two scale factors?

Scale factors a 9cm 27cm 5cm b 15cm Find a given ESF = 3 a 9 ESF = 3 = By finding the RSF Find the value of b. 5cm b 1 3 b 15 5 15 RSF = = = 15cm

Finding Unknown sides (1) 20 cm 18 cm 12 cm 6 cm b c Since the triangles are equiangular they are similar. So comparing corresponding sides.

Starter Questions Q1. Find the roots to 1 decimal place Q2. A freezer is reduced by 20% to £200 in a sale. What was the original price. Q3. Calculate

Similar Triangles 2 Learning Intention Success Criteria To explain how the scale factor applies to similar triangles with algebraic terms. 1. Understand how the scale factor applies to similar triangles with algebraic terms. 2. Solve problems using scale factor that contain algebraic terms.

Determining similarity B E D Triangles ABC and DEC are similar. Why? C Angle ABC = angle DEC (Alt angles) Angle BAC = angle EDC (Alt angles) Angle ACB = angle ECD (Vertically Opposite) Since ABC is similar to DEC we know that corresponding sides are in proportion ABDE BCEC ACDC The order of the lettering is important in order to show which pairs of sides correspond.

A B C D E If BC is parallel to DE, explain why triangles ABC and ADE are similar Angle BAC = angle DAE (common to both triangles) Angle ABC = angle ADE (corresponding angles between parallels) Angle ACB = angle AED (corresponding angles between parallels) A line drawn parallel to any side of a triangle produces 2 similar triangles. A D E B C B C Triangles EBC and EAD are similar Triangles DBC and DAE are similar

The two triangles below are similar: Find the distance y. C B 20 cm y A E D 5 cm 45 cm AD AE 50 5 20 y RSF = = = 50 5x20 y = = 2 cm

Alternate approach to the same problem The two triangles below are similar: Find the distance y. C Alternate approach to the same problem B 20 cm y A E D 5 cm 45 cm AE AD 5 50 y 20 ESF = = = 50 5x20 y = = 2 cm

A A E B D C C D So perimeter = 3 + 8 + 4 + 4.8 = 19.8 cm 7.5 cm 3 cm In the diagram below BE is parallel to CD and all measurements are as shown. Calculate the length CD Calculate the perimeter of the Trapezium EBCD 4.8 m 6 m 4 m 4.5 m 10 m A C D 7.5 cm 3 cm 8 cm So perimeter = 3 + 8 + 4 + 4.8 = 19.8 cm

Similar Triangles 3 Learning Intention Success Criteria To explain how the scale factor applies to similar triangles with harder algebraic terms. 1. Understand how the scale factor applies to similar triangles with harder algebraic terms. 2. Solve problems using scale factor that contain harder algebraic terms.

In a pair of similar triangles the ratio of the corresponding sides is constant, always producing the same enlargement or reduction. B A E D C 4 m 3 m 6 m Find the values of x given that the triangles are similar. Corresponding sides are in proportion AC AB DC EB = 3 + x 3 6 4 x 1.5 = 3x6 = 4(3 + x) 18 = 12 + 4x 4x = 6 x = 1.5

In a pair of similar triangles the ratio of the corresponding sides is constant, always producing the same enlargement or reduction. S T P Q R 3 18 Find the values of x given that the triangles are similar. 5 Corresponding sides are in proportion ST PQ PT PR = x 11.25 5 3 x 18 - x = 3x = 5(18 - x) 3x = 90 - 5x 8x = 90 x = 11.25

Similar Figures Learning Intention Success Criteria To explain how the scale factor applies to other similar figures. 1. Understand how the scale factor applies to other similar figures. 2. Solve problems using scale factor.

Scale Factor applies to ANY SHAPES that are mathematically similar. 1.4 cm 7 cm 2cm y z 3 cm 15 cm 6 cm Given the shapes are similar, find the values y and z ? 3 15 5 1 Scale factor = ESF = = 5 Scale factor = RSF = = 0.2 y is 2 x 5 = 10 z is 6 x 0.2 = 1.2

Scale Factor applies to ANY SHAPES that are mathematically similar. 8 cm 7.5 cm b 10 cm 4 cm 2cm a 3 cm Given the shapes are similar, find the values a and b ? 10 4 4 10 Scale factor = RSF = = 0.4 Scale factor = ESF = = 2.5 a is 8 x 0.4 = 3.2 b is 2 x 2.5 = 5

Area of Similar Shape Learning Intention Success Criteria To explain how the scale factor applies to area. 1. Understand how the scale factor applies to area. 2. Solve area problems using scale factor.

Area of Similar Shape www.mathsrevision.com Draw an area with sides 2 units long. Draw an area with sides 4 units long. 4 2 x y 2 x y Area = 2 x 2 = 4 Area = 4 x 4 = 16 www.mathsrevision.com It should be quite clear that second area is four times the first. The scaling factor in 2D (AREA) is (SF)2. For this example we have this case SF = 2 (2)2 = 4.

Another example of similar area ? Work out the area of each shape and try to link AREA and SCALE FACTOR Connection ? 4cm 2cm 6cm 12cm Small Area = 4 x 2 = 8cm2 Large Area = 12 x 6 = 72cm2 4 12 Scale factor = ESF = = 3 Large Area = (3)2 x 8 = 9 x 8 = 72cm2

The following two shapes are said to be similar. Example The following two shapes are said to be similar. If the smaller shape has an area of 42cm2. Calculate the area of the larger shape. 4cm 3cm Working ESF = So area S.F = Area of 2nd shape = X 42 = X 42 = 74.67cm2

Questions 1. 2.

3. 4.

Surface Area of Similar Solids Learning Intention Success Criteria To explain how the scale factor applies to surface area. 1. Understand how the scale factor applies to surface area. 2. Solve surface area problems using scale factor.

The same rule applies when dealing with Surface Area Example : Work out the surface area of the larger cuboid. 4cm 2cm 3cm 12cm Surface Area of small cuboid : 2(2x3) + 2(4x3) + 2(2x4) = 52cm2 4 12 Surface Area of large cuboid : ESF = = 3 (3)2 x 52 = 468 cm2

Volumes of Similar Solids Learning Intention Success Criteria To explain how the scale factor applies to volume. 1. Understand how the scale factor applies to 3D – volume. 2. Solve volume problems using scale factor.

Volumes of Similar Solids Draw a cube with sides 2 units long. Draw a cube with sides 4 units long. 2 x y z 2 4 2 Volume = 2 x 2 x 2 = 8 x y z Volume = 4 x 4 x 4 = 64 2 4 Scale factor = ESF = = 2 Using our knowledge from AREA section, (SF)2. For VOLUME the scale factor is (SF)3 = (2)3 = 8

Large Volume = (2)3 x 12 = 8 x 12 = 96cm3 Another example of similar volumes ? Work out the volume of each shape and try to link volume and scale factor 6cm 4cm 3cm 2cm Connection ? V = 3 x 2 x 2 = 12cm3 V = 6 x 4 x 4 = 96cm3 3 6 Scale factor = ESF = = 2 Large Volume = (2)3 x 12 = 8 x 12 = 96cm3

2cm 6cm Given that the two boxes are similar, calculate the volume of the large box if the small box has a volume of 15ml 2 6 ESF = = 3 So volume of large box = = = 405 ml

Example ESF = So volume of large jug = = = 2.7 litres

(a) SD : B = 4 : 3 SD : M = 8 : 5 (b) RSF = 900cm2 (c) RSF = 1562.5cm3

ESF = So volume of large jug = = = £1.35

Ratio of their surface area is 1 : 4.6 (to 1 d.p.) (SF )3 = SF = 2.1544 So surface area ratio = (SF)2 = = 4.64 Ratio of their surface area is 1 : 4.6 (to 1 d.p.)