25 true/false Draw reflections, rotations, dilations Symmetry regarding a regular polygon

Size: px

Start display at page:

Download "25 true/false Draw reflections, rotations, dilations Symmetry regarding a regular polygon"

Nickolas Harper
5 years ago
Views:

1 Geometry Week 27 ch. 12 review 1. hapter 12 Vocabulary: angle of rotation ais of symmetry center of rotation composition dilation enlargement fied point identity transformation image invariance isometry line of reflection line symmetry magnitude of a rotation mapping orientation point symmetry preimage preserved reduction reflection rotation rotational symmetry scale factor similar figures transformation translation ch. 12 review Notes on the test: 25 true/false Draw reflections, rotations, dilations Symmetry regarding a regular polygon 1

2 Section 1.1 Definition: Similar polygons are polygons having corresponding angles that are congruent and corresponding sides that are proportional. If and DEF are similar, the proper notation is ~ DEF. Review Proportions: Remember that a proportion is what we get when we set 2 ratios (fractions) equal to each other. Eample: and are called the etremes 2 and are called the means **The product of the means the product of the etremes ( i.e the cross products are equal ) 1 2 (1)() (2)() 2

3 Important: The means can echange positions with each other or the etremes can echange positions with each other and the proportion remains true. 1 2 or 1 2 or 2 1 Sample Problems: Solve for / onsider: ~ DEF From the notation we know: D E F Ratios of corresponding sides: DE EF DF We say is to DE as is to EF etc.

4 Let s put lengths on the sides and check the ratios. 9 D 2 12 F E DE EF DF or DE EF DF ⅓ ⅓ ⅓ **Unless you are finding the scale factor of a dilation, it does not matter which triangle you start with for your proportion. Sample Problem: If the pairs of figures are similar, find the unknown values

5 (+) y y 10 2 y 0 2/10 y 0/ 1/5 y 15/2 5

6 Sample Problem: Prove that if two triangles are congruent, then they are similar. Given: DEF Prove: ~ DEF E D F Statement 1. DEF 1. Given 2. D, E, F DF, EF, DE Reason 2. Def. of congruent s. DF, EF, DE. Def. of congruent.. Multiplication DF DF DF 1 Property of Equality EF EF EF 1 DE DE DE Transitive DF EF DE. ~ DEF. Def. of similar

7 Section 1.2 Similarity Postulate (1.1): If 2 angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. Remember: We know from an earlier theorem that if 2 angles of a triangle are congruent to 2 angles of another triangle, then the rd pair must also be congruent. SSS Similarity Theorem (1.1): If three sides of one triangle are proportional to the corresponding sides of another triangle, then the triangles are similar. Given: In and XYZ, XY Prove: ~ XYZ YZ XZ E Z X Y D 7

8 E Z X Y D Statement 1. YZ 1. Given 2. Draw segment congruent to by etending XY and call it XD; XD Reason 2. uiliary lines. XD. Def. of congruent seg. XY. Substitution (step XD YZ into 1 5. onstruct DE parallel to YZ 5. uiliary lines. XYZ XDE XZY XED 7. XDE ~ XYZ 7.. orresponding ngle Theorem 8. XY YZ XZ 8. Def. of similar s 9. YZ 9. Transitive (steps DE YZ and XY XD DE XZ XE XZ XZ XE (YZ)() (YZ)(DE) (XZ)() (XZ)(XE) 11. Substitution (steps 8 and 1 into 9 Mult. Prop. of Eq. (cross mult. steps 9 & 10) 8

9 E Z X Y D 12. DE, XE 12. Mult. Prop. of Eq. 1. DE, XE 1. Def. of congruent seg 1. XDE 1. SSS 15. XDE XED 1. XYZ XZY 17. ~ XYZ Def. of congruent s 1. Transitive (see step ) 18. If three sides of one triangle are proportional to the corresponding sides of another triangle, then the triangles are similar. 18. Law of Deduction SS Similarity Theorem (1.2): If two sides of a triangle are proportional to the corresponding two sides of another triangle and the included angles between the sides are congruent, then the triangles are similar. 9

10 Theorem 1.: Similarity of triangles is an equivalence relation. (refleive, symmetric, and transitive) Summary: Ways to Prove Similar Triangles: SSS. SS Note: S and S are not needed because they are covered by. Sample Problem: Given: MN OQ Prove: MNP ~ QOP Statement 1. MN OQ 1. Given 2. PQO PMN POQ PNM. MNP ~ QOP. Reason 2. orresponding ngle Theorem 10

11 Sample Problem: P Given: MN OQ Prove: MNP ~ QOP Q O M N Statement Reason Given

12 Sample Problem: Prove Similarity of triangles is transitive (If ~ LMN and LMN ~ PQR, then ~ PQR.) Given: ~ LMN LMN ~ PQR Prove: ~ PQR Statement Reason Given

13 Solution: Sample Problem: Prove Similarity of triangles is transitive (If ~ LMN and LMN ~ PQR, then ~ PQR.) Given: ~ LMN LMN ~ PQR Prove: ~ PQR Statement 1. ~ LMN LMN ~ PQR 2. L L P M M Q. P Q 1. Given. ~ PQR. Reason 2. Def. of similar s. Transitive property of congruent angles 1

14 Sample Problem: Given: D DE D Prove: ~ ED E D Statement Reason Given

15 Solution: Sample Problem: E Given: D DE D Prove: ~ ED D 1. D DE D Statement 2. DE and are right angles 1. Given Reason 2. Def. of perpendicular. DE. ll right angles are congruent. DE. Vertical ngle Thm. 5. ~ ED 5. 15

16 Section 1. Theorem 1.: n altitude drawn from the right angle to the hypotenuse of a right triangle separates the original triangle into two similar triangles, each of which is similar to the original triangle. D ~ D D ~ D ~ D Proof of the 2 nd case: Given: D is altitude of Prove: D ~ Statement Reason 1. D is altitude of 1. Given 2. D 2. Def. of altitude. D is a right angle. Def. of perpendicular. D. ll rt. angles are Refleive. D ~. 1

17 Look at: When the denominator of one fraction of a proportion is the same as the numerator of the other fraction, that number is called the geometric mean. Eample: Find the geometric mean between and (27) 2 81 ± 81 9 Since we want a number between and 27, we will choose 9 instead of -9. Sample Problem: Find the geometric mean between 12 and ± 15 17

18 Theorem 1.5: In a right triangle, the altitude to the hypotenuse cuts the hypotenuse into two segments. The length of the altitude is the geometric mean between the lengths of the 2 segments of the hypotenuse. D Given: Right D D is an altitude of D Prove: D D Statement 1. Right D D is an altitude of D 1. Given Reason 2. D ~ D 2. The altitude divides a rt. into ~ s.. Def. of similar s D D 18

19 Theorem 1.: In a right triangle, the altitude to the hypotenuse divides the hypotenuse into 2 segments such that the length of a leg is the geometric mean between the hypotenuse and the segment of the hypotenuse adjacent to the leg. Eample: Given, find, y, and z. y 1 z From Thm.1.5 we get: Sometimes it helps to separate the triangles. y From Thm.1. we get: y 20 y y 2 80 y 5 z From Thm.1. we get: z 20 1 z z 2 20 z z y 19

20 Sample Problem: Given right JKL with altitude to the hypotenuse, MK, find KM if LJ 20 and MJ. J M 1 K L MK is the geometric mean of MJ and ML Sample Problem: Given right with altitude to the hypotenuse, D, find if D and. D is the geometric mean of D and 9 20

SHELBY COUNTY SCHOOLS: GEOMETRY PRETEST AUGUST 2015 SEMESTER 1. Created to be taken with the ACT Quality Core Reference Sheet: Geometry.

SHELBY COUNTY SCHOOLS: GEOMETRY PRETEST AUGUST 2015 SEMESTER 1 Created to be taken with the ACT Quality Core Reference Sheet: Geometry. 1. The intersection of plane L and plane M is AB. Point P is in neither