p294 Chapter 4 Review

Math 12AP Circle Geometry Chapter Review

p294 #2, 4, 6, 7a, 9, 12, 15, 17, 22, 25, 27

2. Samantha’s error is that you can’t assume that S is the center of the circle. In fact, it is only the center if the chords are already parallel.

4. The trick is to draw the diameter from diagonal corners of the table base. Once you do, you can use Pythagorean Theorem to find the length is 1.5 m.

6. a) Distance of AB

b) Equation of the Line Joining points A and B

y = mx + b

m =

y = 3x + b. Now plug in the coordinates of either A or B for (x, y)

6 = 3(4) + b

6 = 12 + b

-6 = b

therefore y = 3x -6

c) Midpoint of AB

7a. Show that the diagonals are perpendicular and that the midpoints of the diagonals are in the same spot (this shows that the diagonals bisect each other).

9. The midpoint of AB is . Lets call the midpoint D and the center of the circle C.

The slope of AB is .

The slope of CD is

The slope of AB is the opposite reciprocal of the slope of CD.

12. mÐS = 55°

a) ÐC = 110° because C is the central angle subtended by the same arc as inscribed angle S. Therefore it is double S.

b) ÐCQR = 35° because triangle CQR is isosceles since two of its side are the radius of the circle. Since ÐC = 110°, the other two angles are equal and they all must add up to 180°, they both must be 35°.

c) arc QR = 110° because the are is equal to the measure of the central angle that it subtends.

d) arc RSQ = 250° because the circle must add up to 360°.

15. a) PQ is 20. PQ is a diameter so it is twice the radius CR.

b) ÐQRP = 90° since it is inscribed in a semi-circle. Therefore we can use Pythagorean theorem to solve for QR. QR² + PR² = PQ². QR = 18.33.

c) Area of triangle PQR is (0.5)(18.33)(8) = 73.32 units squared.

17. mÐC = 150°, AD = AB. Arc BAD is 150° since it intercepts central angle ÐBCD. Therefore the rest of the circle is 210°. Since inscribed angle ÐBAD is subtended by this arc, it has an angle of 105°. We can draw line BD to make two isosceles triangles. We know that ÐCDB and ÐCBD are each 15°. We also know that ÐABD and ÐADB are 37.5°. We can add the angles together to find that ÐCDA = ÐCBA = 52.5°. We can check to see that all four angles of this quadrilateral add up to 360°. 150° + 105° + 52.5° + 52.5° = 360°.