Solving Proportions MPM2D: Principles of Mathematics Recap Explain - PDF document

s i m i l a r i t y & p r o p o r t i o n a l i t y s i m i l a r i t y & p r o p o r t i o n a l i t y Solving Proportions MPM2D: Principles of Mathematics Recap Explain why ∆ ABC ∼ ∆ DEF . Properties of Similar Triangles J. Garvin Each side in ∆ ABC is twice as long as that in ∆ DEF . Therefore, ∆ ABC ∼ ∆ DEF by SSS ∼ . J. Garvin — Properties of Similar Triangles Slide 1/15 Slide 2/15 s i m i l a r i t y & p r o p o r t i o n a l i t y s i m i l a r i t y & p r o p o r t i o n a l i t y Solving Proportions Solving Similar Triangles Since the ratios of any two corresponding sides of two similar Example triangles are equal, we can use these ratios to solve for the ∆ ABC ∼ ∆ DEF . Determine | AC | and | EF | . values of unknown sides. Consider the case where ∆ ABC ∼ ∆ DEF . If we know the ratio | AB | | DE | , and we know one of | AC | or | DF | , then we can solve for the other value using the proportion | AB | | DE | = | AC | | DF | . This can be done by inspection, or by using cross-multiplication. Since we know the values of the corresponding sides AB and DE , we can use their ratio to create proportions involving the unknown sides. J. Garvin — Properties of Similar Triangles J. Garvin — Properties of Similar Triangles Slide 3/15 Slide 4/15 s i m i l a r i t y & p r o p o r t i o n a l i t y s i m i l a r i t y & p r o p o r t i o n a l i t y Solving Similar Triangles Solving Similar Triangles Example AC corresponds to DF , whose EF corresponds to BC , whose Determine | AE | . value is known. value is known. | AB | | DE | = | AC | | AB | | DE | = | BC | | EF | | DF | 14 19 14 8 = | AC | 8 = | EF | 12 168 = 8 | AC | 14 | EF | = 152 | AC | = 21 | EF | = 76 7 Since | AE | = | AC | + | CE | , we can find | AC | and add it to | CE | , which is known. J. Garvin — Properties of Similar Triangles J. Garvin — Properties of Similar Triangles Slide 5/15 Slide 6/15

s i m i l a r i t y & p r o p o r t i o n a l i t y s i m i l a r i t y & p r o p o r t i o n a l i t y Solving Similar Triangles Solving Similar Triangles Example AB corresponds with DE , while AC corresponds with CE . Determine | BD | . | DE | = | AC | | AB | | CE | 9 = | AC | 5 13 65 = 9 | AC | | AC | = 65 9 Therefore, | AE | = 13 + 65 9 = 182 m. 9 Since BD is part of a trapezoid rather than a triangle, we cannot use it directly in a proportion. J. Garvin — Properties of Similar Triangles J. Garvin — Properties of Similar Triangles Slide 7/15 Slide 8/15 s i m i l a r i t y & p r o p o r t i o n a l i t y s i m i l a r i t y & p r o p o r t i o n a l i t y Solving Similar Triangles Areas of Similar Triangles Since | BD | = | AD | − | AB | , we can first find | AD | and What happens to the area of a triangle when its dimensions subtract | AB | , which is known. are doubled? | AD | | AB | = | DE | | BC | | AD | = 20 14 12 | AD | = 5 14 3 3 | AD | = 70 | AD | = 70 When the dimensions are doubled, the area is quadrupled. 3 Therefore, | BD | = 70 3 − 14 = 28 3 in. J. Garvin — Properties of Similar Triangles J. Garvin — Properties of Similar Triangles Slide 9/15 Slide 10/15 s i m i l a r i t y & p r o p o r t i o n a l i t y s i m i l a r i t y & p r o p o r t i o n a l i t y Areas of Similar Triangles Areas of Similar Triangles What happens to the area of a triangle when its dimensions In general, a triangle whose dimensions are enlarged by a factor of k will have an area k 2 times greater. are tripled? Areas of Similar Triangles If ∆ ABC ∼ ∆ DEF , and if | AB | = k | DE | , then Area ABC = k 2 · Area DEF . Any ratio of corresponding sides can be used, so choose the one that is easiest to work with. When the dimensions are tripled, the area increases by a factor of nine. J. Garvin — Properties of Similar Triangles J. Garvin — Properties of Similar Triangles Slide 11/15 Slide 12/15

s i m i l a r i t y & p r o p o r t i o n a l i t y s i m i l a r i t y & p r o p o r t i o n a l i t y Areas of Similar Triangles Areas of Similar Triangles Example Determine the scale factor, k , of ∆ DEF . ∆ ABC ∼ ∆ DEF . Determine Area DEF . 12 k = 8 k = 2 3 � 2 � 2 = 4 The area of ∆ DEF will be k 2 = 9 the size of ∆ ABC . 3 Therefore, the area of ∆ DEF is 90 × 4 9 = 40 cm 2 . J. Garvin — Properties of Similar Triangles J. Garvin — Properties of Similar Triangles Slide 13/15 Slide 14/15 s i m i l a r i t y & p r o p o r t i o n a l i t y Questions? J. Garvin — Properties of Similar Triangles Slide 15/15