Answers to Chapter 2 Review Problems, pp. 105-109

Geometry / Mr. Hansen	Name: _______________________________________
10/3/2007

Warning: For best results, you should peek at these answers only
after attempting the problems on your own, under time pressure.

1.	1.	\| 1. Given
	2.	\| 2. Def.
	3.	\| 3. Def. of rt. angle
	4.	\| 4. Def. comp. (2 angle measures adding to 90)
	(Q.E.D.)

	Note: You may omit step 3 if you wish. In that case, the reason for the final step would be as follows:
		Def. comp. (2 angles adding to a rt. angle)

2.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Add. prop.
	(Q.E.D.)

3.	1.	\| 1. Given
	2.	\| 2. Reflexive prop. (an angle is congruent to itself)
	3.	\| 3. Add. prop.
	(Q.E.D.)

	Note: You may omit step 2 if you wish. Its only purpose is to make the addition of angles more obvious. If you choose to include step 2, it is acceptable to abbreviate the reason as “Refl.”

4.	1.	\| 1. Given
	2. N is the midpt. of	\| 2. Given
	3. O is the midpt. of	\| 3. Given
	4.	\| 4. Div. prop.
	(Q.E.D.)

	Note: It is not correct to say “subtr. prop.” as the reason for step 4. Why not? Because would have to be in evidence, and although true, that statement is not listed in the left column. The easiest way to get into evidence would be to apply the division property, but that would be a waste of time, since you can get the desired conclusion as shown in only 4 steps without placing into evidence at all.

5.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Mult. prop.
	(Q.E.D.)

6.	1.	\| 1. Diagram
	2.	\| 2. Diagram
	3.	\| 3. Given
	4.	\| 4.
	(Q.E.D.)

7.	Let x = one of the imaginary subdivisions
	5x + 2x = 21
	7x = 21
	x = 3
	EF = 2x = 6 cm

8.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Subst. (2 into 1)
	(Q.E.D.)

9.	1.	\| 1. Given
	2.	\| 2. Vert.
	3.	\| 3. Trans. (1, 2)
	(Q.E.D.)

10.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Subtr. prop.
	(Q.E.D.)

11.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3.
	4.	\| 4. Refl.
	5.	\| 5. Subtr. prop.
	(Q.E.D.)

	Note: You may omit step 4 if you wish. Its only purpose is to make the subtraction of angles more obvious.

12.	Let x = meas. of one complementary angle Let 90 – x = meas. of the other x = 6 + 2(90 – x) x = 6 + 180 – 2x 3x = 186 x = 62 Complement has measure 28, but clearly 62 is larger (answer).

13.	Let x = meas. of angle in question Let 180 – x = meas. of angle’s supplement Let 90 – x = meas. of angle’s complement 180 – x = 5(90 – x) 180 – x = 450 – 5x 180 – 450 = –5x – (–x) –270 = –4x x = 67.5 Note: If you stop at x, you would earn the notation RAWQ (right answer, wrong question). The question asks for the measure of the complement, not the angle itself. The correct final answer is 90 – x = 90 – 67.5 = 22.5.

14.	oblique

15.	Since A is mdpt. of segment DE, DE = 24. [We can also see by the number line that DE = 18 – (–6) = 24.] Therefore, DA = AE = 12, forcing the coordinate of A to be –6. Since I and N are trisection pts. of segment DE, we know DI = IN = NE = 8. Thus the coordinate of I must be –10, and the coordinate of N must be –2. The problem asks for AN. Since N has coordinate –2 and A has coordinate –6, the distance between them must be AN = 4.

16.	(a) 97°, 7° (b) 137°44'22", 47°44'22" (c) 83°, undefined (obtuse angle has no complement)

17.	(2, –5) and (6, –11)

18.	(0, 5)

19.	(a) 1 (b) 0.5

20.	(a) B is (0, 8) since y-coordinate must match C. D is (6, –5) since x-coordinate must match C and y-coordinate must match A.
	(b) 5/13

21.	43°43'

22.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Vert.
	4.	\| 4.
	(Q.E.D.)

23.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Def. mdpt.
	4.	\| 4. Trans. (2, 3)
	(Q.E.D.)

24.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Given
	4.	\| 4. Def.
	5.	\| 5. Given
	6.	\| 6. Def.
	7.	\| 7.
	8.	\| 8. Refl.
	9.	\| 9. Subtr. prop. (7 minus 8)
	10.	\| 10. Trans. (1, 2, 9)
	(Q.E.D.)

25.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Given
	4.	\| 4. Div prop.
	5.	\| 5. Vert.
	6.	\| 6. Trans. (4, 5)
	(Q.E.D.)
26.	1.	\| 1. Given
	2.	\| 2. Def. bis.
	3.	\| 3. Diagram
	3.	\| 4. Def. supp. (angles adding to a straight angle)
	4.	\| 5. Def. supp.
	5.	\| 6.
	(Q.E.D.)

	Note: Why would it be a mistake to say “div. prop.” as the reason for step 2? The answer is that the division property requires at least two pairs of angles. Here we have only one angle that is being bisected. Similarly, in step 3 of problem #23, it would be a mistake to say “div. prop.” since there is only one segment being bisected, not two pairs of segments.

27.	1.	\| 1. Given
	2.	\| 2. Given
	3.	\| 3. Given
	4.	\| 4.
	(Q.E.D.)

28.	By vert. angles, 2x = y. Therefore, x = y/2.
	Going across from left to right, 2x + 2y + x = 180.
	By substituting x = y/2, this becomes 2(y/2) + 2y + y/2 = 180.
	By algebra, 3.5y = 180.


29.	RW = WV x + 4 = y + 5 x – 1 = y WV = TV y + 5 = 2x y = 2x – 5 Substituting from earlier equation for y above, we have x – 1 = 2x – 5 –x = –4 x = 4 Therefore, y = x – 1 = 4 – 1 = 3. Perimeter = RW + WV + 2VT + RS = (x + 4) + (y + 5) + 2(2x) + 6y = (4 + 4) + (3 + 5) + 2(2(4)) + 6(3) = 8 + 8 + 16 + 18 = 50.

30.	By vertical angles, x + 10 = y + 3x. Isolate y on one side of the equation to get y = –2x + 10.

31.	(x + 2y) + (x – 2y) = 180 2x = 180 x = 90 By vertical angles, x + 2y = 50 90 + 2y = 50 2y = 50 – 90 2y = –40 y = –20 By how much does x exceed y? The answer would be whatever x – y is. x – y = 90 – (–20) = 110

32.	Let x = meas. of angle in question Let 90 – x = meas. of angle’s complement Let 180 – x = meas. of angle’s supplement 180 – x = 20 + 2(90 – x) 180 – x = 20 + 180 – 2x x = 20 Note: If you stop at x, you would earn the notation RAWQ (right answer, wrong question). The question asks for the measure of half the complement, not the angle itself. Since the complement has measure 70, the measure of half the complement would be 35.

33.	(a) x – 2y = y, which simplifies to x = 3y (b) y = 124 – x (c) x = 3y = 3(124 – x) = 372 – 3x If x = 372 – 3x, then 4x = 372. Therefore, x = 93. =124 – x = 124 – 93 = 31

34.	10 – 3x = x² – 6x 10 = x² – 3x x² – 3x – 10 = 0 (x + 2)(x – 5) = 0 x + 2 = 0 or x – 5 = 0 x = –2 or x = 5 We must reject x = 5, since that would make angle MOP’s measure equal to 10 – 3x = 10 – 15 = –5, which is impossible. Therefore, x = –2, from which angles MOP and POE each have measure 16. Therefore, = 32.

35.	(a) A = bh/2, a formula you should have learned in previous classes. (Here, b = base, h = height of triangle.) A = BD · DE/2 = 7 · 8/2 = 28 square units
	(b) Just as the area of is 28 square units, the area of is 28 square units since the same base and height are involved as before. By subtracting the area of from each, we conclude that the areas of and are equal. (We subtracted the same value from 28 in each case.)

36.	(a) Net rotation was 180°, meaning that B is at (–1, –2). (b) (1, 2) is starting point, in Quadrant I A 100° clockwise rotation is more than a right angle, hence placing us in Quadrant IV. Then, an 80° counterclockwise rotation places us back in Quadrant I. The third rotation moves us to the opposite side altogether. The fourth rotation leaves us a net of 180° shifted from our original starting point in Quadrant I.

37.	There are 24 different times that could be displayed when the cuckoo sounds. Of these, only 1:00, 2:00, 3:30, 4:30, 5:30, 6:30, 7:30, 8:30, 10:00, and 11:00 produce acute angles. The probability is therefore 10/24, making the odds 10:14 in favor or 14:10 against. Bippy is betting $10 on the acute angle, which is less likely to occur than an obtuse or right angle. Therefore, Tippy must bet more than $10 to make this a fair bet. Because the odds ratio is 14:10, the correct amount for Tippy to put up is $14.

38.	By the first given statement, we know Divide this equation by 2 to get However, we know an angle whose measure is half that of angle ABD, namely angle CBD (by def. of bisection). Similarly, we know an angle whose measure is half that of angle EDB, namely angle BDC (again, by def. of bisection). Therefore, by substitution, This last statement is true iff angles CBD and BDC are complementary. (Q.E.D.)