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Prepared by: Alia. Ayub. Ayub. 11B SIMILAR POLYGONS What does the word similar mean? It means not exactly the he same but pretty close. Congruent polygons In order to be congruent, two figures must be the same size and same shape shape.. Congruent figures are exact replicas of each other. They have the same shape and the same size. ≅ Now consider figures that have the same shape but not the same size. Such Such figures look 'similar' but in essence are simply proportionate to each other What are similar polygons? Similar polygons are ones that have the same shape, but their sizes may be different. ~ is the symbol that means “similar.” It is a portion of the “is congruent to” sign Two or more polygons are similar only if: The ratios of all pairs of corresponding sides are equal All pairs of corresponding angles are equal Consider the polygons ABCD and LMNO in the figure 1 Figure 1 ≅ Their corresponding angles are equal but their sides are not proportional. Hence they are not similar. Now the sides may be proportional but the angles may not be congruent. For instance we have polygons like PQRS and HIJK (figure 2) Figure 2.Again Again they are not similar. Thus to be similar polygons must satisfy both, the condition of congruent angles and that of proportionate sides. Figure 3 shows some similar polygons. Figure 3 Sizes: Although the size of the two shapes can be different, the sizes of the two shapes must differ by a factor In this case, the factor is x 2 These dodecagons differ by a factor of 6. Enlargements: • When you have a photograph enlarged, you make a similar photograph. Reduction: A photograph can also be shrunk to produce a slide Similar Triangles Two triangles are similar if their corresponding angles are congruent and their corresponding sides are proportional. It is however not essential to prove all 3 angles of one triangle congruent to the other, or for that matter all three proportional to the other sides. Out of these if some particular conditions get satisfied the rest automatically get satisfied. Those particular conditions would be sufficient to ensure similarity. These are the two conditions: 1) In 2 triangles if the corresponding angles are congruent, their corresponding sides are equal. 2) If the sides of 2 triangles are proportional then the corresponding angles are congruent. The following mentioned below, are tests used to show that two triangles are similar. 1)Postulate 1)Postulate 17 (Angle Angle similarity Postulate): Postulate): If two angles of one triangle are congruent congruent with the corresponding two angles of another triangle, the two triangles are similar. similar. HOW? The sum of all three angles of a triangle is 180º. Therefore if two angles are congruent the third is automatically congruent. Therefore the sufficient condition requires only two angles to be congruent. 2)S 2)Side Angle side Test: If two sides of one triangle are proportionate to the two corresponding sides of the second triangle and the angles between the two sides of each triangle are equal the two triangles are similar. HOW? If two sides of a triangle are equal, the angles opposite the sides have equal measures. 3)S 3)Side Side Side Test: If the three sides of one triangle are proportional to the three corresponding corresponding sides of another triangle, triangle, then two triangles are similar. ABC and DEF have a one to one correspondence such that AB =6 DE= 3 AC = 8 DF = 4 BC= 10 EF = 5. Are the two triangles similar ? If so, why ? Solution: AB/DE = AC/DF= BC/EF = 2/1. Yes. The two triangles are similar because their corresponding sides are proportionate. And the measure of their corresponding lenths varies by a factor, 2. Example 1 : Are the triangles shown in the figure similar? Solution: Find the ratios of the corresponding sides. XY 9 3 XZ 15 3 -- = -- = - -- = -- = TS 12 4 SU 20 4 The sides that include angle X and angle S are proportional. Angle X and angle S are congruent (the information is given in the figure). Triangle YXZ and triangle TSU are similar by SideSide-AngleAngle-Side. Side Properties of Similar triangles Perimeters of similar triangles: Theorem 60: Perimeters of similar triangles are in the same ratio as their corresponding sides and this ratio is called the scale factor. If the two triangles have a scale factor of a: a: b, b, then the ratio of their perimeters is a: a: b aswell. Example 1: In figure 3 there are two similar triangles . LMN and PQR. Figure 3 LM/PQ=MN/QR=LN/PR= 1.33 The scale factor is therefore 1.33 1.33 Perimeter of LMN = 8cm + 7cm + 10cm = 25cm Perimeter of PQR = 6cm + 5.25cm + 7.5cm = 18.75cm Perimeter of LMN/ Perimeter of PQR= 25cm/18.75cm = 1.33 Thus, the perimeters of two similar triangles are in the ratio of their scale factor. Areas of similar triangles: The ratio of the areas of two similar triangles is equal to the ratio of the squares of the i.e. the square of the scale factor. corresponding sides, ABC ~ PQR Ratio of corresponding sides (k) = 6/4=3/2 Area of ABC = 1/2ab SinC ½ ×6×3 SinC =Area of ABC/Area of Area of PQR = 1/2pq SinR ½ × 4 ×2 SinQ PQR =18/8 = 9/4= k² If the two figures are similar and the ratio of their corresponding sides is k, then the ratio of their areas will be k² Example 1 Areas of two similar triangles are 144 cm². and 81cm ². If one side of the first triangle is 6 cm then find the corresponding side of the second triangle. Solution : = 4.5 c.m Example 2 The side of an equilateral triangle ABC is 5 cm. Find the length of the side of another equilateral PQR whose area is four times area of ABC. Solution: Since both triangles are equilateral they are similar. Example 3 The corresponding sides of two similar triangles are 4 cm and 6 cm. Find the ratio of the areas of the triangles. Solution: Example 4 ABC ~ PQR such that AB : PQ is 8 : 6. If area of smaller triangle? Solution: ABC is 48cm² what is the area of the 3D Figures For 3D shapes such as Cubes, cuboids, cylinders etc, to be similar, the following conditions apply: 1} The shape must be the same. Eg/ the two shapes must be only cubes or only cylinders, but not a combination of a cube and a cylinder. 2}Their corresponding sides must be in proportion 3}All 3}All pairs pairs of corresponding angles must be equal Surface area: S.A = 2lb + 2lh + 2hb S.A of A = 2 (4cm×3cm) + 2(4cm×2cm) +2(3cm×2cm) = 52cm² S.A of B = 2(12cm×9cm) + 2(12cm×6cm) +2(9cm×6cm) = 468 cm² Ratio of corresponding sides(k) : k = 12cm/4cm = 9cm/3cm = 6cm/2cm = 3 Ratio of S.A’s(x) x = 468 cm²/52cm² = 9 x= k² x= k². Therefore, given two similar 3D shapes, if the ratio of corresponding sides is k, then the ratio of their surface areas is k² Example 1: 1) The ratio of the surface area of two spheres is 3:1. If the surface area of the larger sphere is 540 cm². Find the surface area of the smaller one. Solution: S.A of larger / S.A of smaller Ratio of larger / Ratio of smaller = 540 cm²/ X = 3/1 540/X= 3/1 (cross product multiplication) Therefore X = 180 cm² Example 2: 2 )Water is put into two similar cylinders . The larger one has a surface area of 256cm2 and the ratio of the sides of the two cylinders is 6:3. Find the surface area of the second cylinder. Solution: As we know, the ratio of the surface areas is the square of the ratio of the sides. Ratio of sides : 6/3 Ratio of surface areas : 256 /Y Therefore, {6/3}² = 256/Y 36/9 = 256/Y ( cross product multiplication) Y = 64 cm To prove the theory we can check: (Ratio of sides)² should be equal to Ratio of surface areas. (Ratio of sides)² : Ratio of surface areas = { 6/3}² : 256/64 36/9 :4 4:4. The theory is therefore proved right. Volume: Volume = Base area × height Volume of A = (4cm×3cm)×2cm = 24cm³ Volume of B = (12cm×9cm) ×6cm = 648cm³ Ratio of corresponding sides (k) : k = 12cm/4cm = 9cm/3cm = 6cm/2cm = 3 Ratio of Volume’s(y) y = 648 cm³/24cm³ = 27 y= k². Therefore, given two similar 3D shapes, if the ratio of corresponding sides is k, then the ratio of their volumes is k³ Example 1: Two similar cylinders have a height of 2 cm and 4cm respectively. If the volume of the smaller cylinder is 20 cm², find the volume of the larger cylinder? 1. {4/2}³ = x/20 2. 64/8 = x/20 3. 64 ×20 8 x= 160 cm ³ Example 2: Two similar cones have a surface area in the ratio 4:16. Find the ratio of : a) their lengths As we know, the surface areas of similar 3D figures are proportionate to the square of the ratio of corresponding lengths. Therefore, let the sides be x and y rexpectively. 4/16 = {x/y}² 4/16 =2/4. The lengths of the two similar cones are 2cm and 4cm respectively. b) their volumes As we know, the volumes of similar 3D figures are proportionate to the cube of the ratio of corresponding lengths. Ratio of sides : 2/4 Ratio of volumes : {2/4}³ = 8/64