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Two geometrical objects are called similar if they both have the same shape. More precisely, one is congruent to the result of an enlarging or shrinking of the other. Corresponding sides of similar polygons are in proportion, and corresponding angles of similar polygons have the same measure. One can be obtained from the other by uniformly "stretching" the same amount on all directions, possibly with additional rotation and reflection, i.e., either have the same shape, or one has the same shape as the mirror image of the other. For example, all circles are similar to each other, all squares are similar to each other. SIMILAR POLYGONS Similar Polygons: Two polygons containing vertices that can be paired so that the corresponding angles are congruent and the corresponding sides are in proportion. Example: If you are given two similar polygons ABCD and JKLM, we can write ABCD ~ JKLM which is read as "polygon ABCD is similar to polygon JKLM". The wavy line symbol means 'similar to'. PROPERTIES OF SIMILAR POLYGONS 1. Corresponding angles are the same 2. Corresponding sides are all in the same proportion 3. Corresponding diagonals are in the same proportion. In each polygon the corresponding diagonals are in the same proportion. Their ratio is the same as the ratio of the sides. 4. Area ratio: the ratio of the areas of the two polygons is the square of the ratio of the sides. So if the sides are in ratio 3:1 then the area will be in the ratio 9:1. Although it could be possible that the quadrilaterals are not similar to one another as maybe their corresponding sides are not equal or their corresponding angles. Scale factor: The ratio of the lengths of two corresponding sides of similar polygons Example 1: / A is congruent to / E / B is congruent to / F / C is congruent to / G / D is congruent to / H AB/EF = BC/FG= CD/GH = AD/EH The scale factor of polygon ABCD to polygon EFGH is 10/20 or ½ REFLECTION AND ROTATION Polygons can still be similar even if one of them is rotated, and/or mirror image of the other. In the figure below, all three polygons are similar. Starting with the original polygon on the left, the center polygon is rotated clockwise 90°, the right one is flipped vertically. Mark the angles in the two right polygons that correspond with the angle P on the left. This is illustrated in more depth for triangles in "Similar Triangles", but is true for all similar polygons, not just triangles. TYPE OF QUESTIONS IN WHICH POLYGONS COULD HELP… 1. Problem: Find the value of x, y, and the measure of angle P. Solution: To find the value of x and y, Write proportions involving corresponding sides. Then use cross products to solve. 4/6 = x/9 6x = 36 x=6 4/6 = 7/y 4y = 42 y = 10.5 To find angle P, note that angle P and angle S are corresponding angles. By definition of similar polygons, angle P = angle S = 86o. SIMILAR TRIANGLES The triangle has a couple of special rules dealing with similarity. They are outlined below. 1. Angle-Angle Similarity - If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. Example: Prove triangle ABE is similar to triangle CDE. Solution: Angle A and angle C are congruent (this Information is given in the figure). Angle AEB and angle CED are congruent because vertical angles are congruent. Triangle ABE and triangle CDE are similar by Angle-Angle. 2. Side-Side-Side Similarity - If all pairs of corresponding sides of two triangles are proportional, then the triangles are similar. 3. Side-Angle-Side Similarity - If one angle of a triangle is congruent to one angle of another triangle and the sides that include those angles are proportional, then the two triangles are similar. Example: Are the triangles shown in the figure similar? Solution: Find the ratios of the corresponding sides. UV/KL=9/12 VW/LM=15/20 The sides that include angle V and angle L are proportional. Angle V and angle L are congruent (the information is given in the figure). Triangle UVW and triangle KLM are similar by Side-Angle-Side. PARALLEL LINES AND TRIANGLES You can create similar triangles by drawing a segment parallel to one side of a triangle in the triangle. This is useful when you have to find the value of a triangle's side. If a segment is parallel to one side of a triangle and intersects the other sides in two points, then the triangle formed is similar to the original triangle. Also, when you put a parallel line in a triangle, the sides are divided proportionally. Example: Find PT and PR Solution: 4/7=x/12 because the sides are divided proportionally when you draw a parallel line to another side. 7x = 48 Cross products x = 48/7 PT = 48/7 PR = 12 + 48/7 = 132/7 PERIMETRES AND AREAS OF SIMILAR TRIANGL Perimeters of similar triangles are in the same ratio as their corresponding sides and this ratio is called the scale factor. Example: There are two similar triangles. LMN and PQR. Seg LM/ segPQ = seg MN/seg QR= Seg NL/Seg RP= 1.33 This ratio is called the scale factor. Perimeter of LMN = 8 + 7 + 10 = 25 Perimeter of PQR = 6 + 5.25 + 7.5 = 18.75 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 corresponding sides, i.e. the square of the scale factor. ABC PQR Area of triangle ABC/Area of triangle PQR= line (AB)2/line (PQ)2= line (BC)2/Line (QR)2=Line(CA)2/Line (RP)2. Prove Check; Draw perpendicular from A and P to meet seg.BC and seg.QR at D and S respectively. Since ABC ~ PQR Line AB/ line PQ= line BC/line QR= line CA/Line RP . . . . . . . 1 also B Q In ABD and PQS also B Q and ADB PSQ ABD PQS by Angle to Angle method. Line AB/PQ= line BD/ Line QS= Line DA/line SP………2 = Line BC/ Line QR x Line BC/ line QR = line BC/ Line (QR)2 = Line AB/ Line (PQ)2 = line AC/ Line (PR)2 =Line BC/ Line (PR)2 => (Scale Factor) Thus the areas of two similar triangles are in the same ratio as the square of their scale factors. 2) Areas of two similar triangles are 144 sq.cm. and 81 sq.cm. If one side of the first triangle is 6 cm then find the corresponding side of the second triangle. Solution: = 4.5 c.m. 3) The corresponding sides of two similar triangles are 4 cm and 6 cm. Find the ratio of the areas of the triangles. Solution: 3D shapes Three-dimensional figure is a figure that has depth in addition to width and height. Everyday objects such as a tennis ball, a box, a bicycle, and a redwood tree are all examples of space figures. Some common simple space figures include cubes, spheres, cylinders, prisms, cones, and pyramids. A space figure having all flat faces is called a polyhedron. A cube and a pyramid are both polyhedrons; a sphere, cylinder, and cone are not. VOLUMES Ratio of volumes of similar figures; if the ratio of corresponding sides of figure A and B is a: b, then the ratio of their volume is a3: b3. The volumes of two similar solids are proportional in the ratio k3 when k is the line ratio of the solids. Example: B A 5 12 4 3 Volume of A = 5x3x4 =60 Ratio of volumes: 480/60 :8 = 23 Therefore: Ratio of volume= (ratio of sides)3 5555 10 Volume of B= 12x10x4 =480 4 SURFACE AREAS Ratio of surface areas of similar figures; if the ratio of corresponding sides of figure A and B is a: b, then the ratio of their surface area is a2: b2. Example: B A 4 20 2 6 Scale factor is 1:5 30 10 Surface area of A= 2(lw)+2(lh)+2(wh) = 48+24+16 = 88 Surface area of B=2(lw)+2(lh)+2(wh) = 1200+600+400 =2200 Therefore ratio of surface area = (ratio of sides) 2