1. 3.3 CPCTC and Circles By: Josie LaCoe and Sarah Parkinson Period 1

2. CPCTC= Corresponding Parts of Congruent Triangles are Congruent

3. Explanation of CPCTC If two triangles are congruent, then all of the corresponding parts of those two triangles are congruent.

4. This means that if COW PIG, then and CO PI. This is also true for all other corresponding parts of the triangles. W O C P I G

5. Circles!!! Point W is the center of this circle. All circles are named by their center, so this circle would be called circle W or O W.. W.

6. A circle is made up of only the outer edge, not the center.. W Circle (rim) Center of circle (Not a part of the circle)

7. Circles!!! Since all of points of a circle are the same distance from the center… THEOREM 19!!!! -All radii of a circle are congruent!

8. Review Formulas AAlthough you probably know these formulas from previous math classes, here’s just a little refresher: AA= r CC=2 r 2 3.141592654

9. Sample Problem #1 Given: OS Prove: IE JO StatementsReasons 1. OS1. Given 2. SJ SO SI SE2.All radii of a O are 3. JSO ISE3. Vertical s are 4. JSO ISE4. SAS (2,2,3) 5. IE JO5. CPCTC. J O S I E.. Solution:

10. Sample Problem #2 Given: OJ Find the perimeter of SJP Solution: SJ and PJ=1/2 PR SJ and PJ=1/2(12) SJ and PJ=6 PS=7 Perimeter of SJP= 6+6+7=19 7 12 S J P R.

11. Practice Problem #1 Given: E A B is the mdpt of AE Prove: C D A B C D E

12. Practice Problem #2 a. Find the coordinates of point S b. Find the circumference of the circle (Round to the nearest tenth). (107, 59). S

13. Practice Problem Solutions 1. StatementsReasons 1. E A1. Given 2. B is mdpt of AE2. Given 3. AB BE3. Mdpt seg into 2 segs 4. ABC EBD4. Vertical s are 5. ABC EBD5. ASA (1, 3,4) 6. C D6. CPCTC

14. Practice Problem Solutions 2a. The coordinates of the center of the circle is (107, 59) (107 being the x-coordinate and 59, the y- coordinate). This means the x- coordinate of S is 107 and since S is on the x-axis, the y-coordinate is 0, making the coordinates of S (107,0)

15. Practice Problem Solutions 2b. C=2 r C=2 59 C 370.7 Because the y-coordinate (the distance from the x-axis to the point) of the center of the circle is 59, this is also the radius of the circle (the distance from the center of a circle to the outside edge). This number is plugged into the equation and the equation is solved for C.

16. Works Cited Rhoad Richard, George Milauskas, Robert Whipple. Geometry for Enjoyment and Challenge. Illinois: McDougal Littell, 1997. Print.