# Similarity Lesson 8 2 Definition Similar polygons are

• Slides: 11

Similarity Lesson 8. 2

Definition: Similar polygons are polygons in which: 1. The ratios of the measures of corresponding sides are equal. 2. Corresponding angles are congruent.

Similar figures: figures that have the same shape but not necessarily the same size. Dilation: when a figure is enlarged to be similar to another figure. Reduction: when a figure is made smaller it also produces similar figures.

Proving shapes similar: 1. Similar shapes will have the ratio of all corresponding sides equal. 2. Similar shapes will have all pairs of corresponding angles congruent.

Example: ∆ABC ~ ∆DEF D 4 B A 5 8 6 C E 12 10 F Therefore: A corresponds to D, B corresponds to E, and C corresponds to F. 1. The ratios of the measures of all pairs of corresponding sides are equal. = = =

Each pair of corresponding angles are congruent. <B <E <A <D <C <F

∆MCN is a dilation of ∆MED, with an enlargement ratio of 2: 1 for each pair of corresponding sides. Find the lengths of the sides of ∆MCN. C (0, 8) E (0, 4) M (0, 0) D N (3, 0) (6, 0) MC = 8 MN = 6 CN = 10

Given: ABCD ~ EFGH, with measures shown. 1. Find FG, GH, and EH. B 4 C 6 A 7 D FG = 6 F 9 GH = 4. 5 E 3 G H 2. Find the ratio of the perimeter of ABCD to the perimeter of EFGH. EH = 10. 5 PABCD = 20 PEFGH = 30 =2 3

Theorem 61: The ratio of the perimeters of two similar polygons equals the ratio of any pair of corresponding sides.

Given that ∆JHK ~ ∆POM, H = 90, J = 40, m M = x+5, and m O = y, find the values of x and y. First draw and identify corresponding angles. K H M J O <J comp. <K = 50 <K = <M 50 = x + 5 45 = x P <H = <O 90 = 180 = y y

Given ∆BAT ~ ∆DOT OT = 15, BT = 12, TD = 9 Find the value of x(AO). A x AT = BT OT TD O 15 B 12 x + 15 = 12 15 9 T 9 D x=5 Hint: set up and use Means-Extremes Product Theorem.