Download Fall Semester Exam Review Problems Worked Out

1st Semester Review Problems
1. Triangles I and IV are clearly isosceles because only 2 sides are congruent on each. Triangle II is also
technically isosceles because it has AT LEAST 2 congruent sides (even though it is ALSO equilateral). So,
the best answer is C.
2. By definition of a parallelogram, BC is parallel to AD, which means angle B and angle A are SAME SIDE
INTERIOR angles. Same side interior angles are always SUPPLEMENTARY.
So, (3x+7) + (2x+18) = 180
5x + 25 = 180
5x = 155
X = 31, which means angle A = 2x + 18 = 2(31) + 18 = 80 degrees.
3. All 3 interior angles of any triangle always add up to 180 degrees. This is easy!
4. Angle R + supplement = 180
6z+50 + supplement = 180
Supplement = 180 – 50 – 6z
Supplement = 130 – 6z or switching the order gives: -6z + 130
5. The two smaller segments add up to equal the larger segment.
So, (4x+8) + 27 = 6x
4x + 35 = 6x
35 = 2x
17.5 = x
To find CE, plug “x” back in. 6x = 6(17.5) = 105.
6. Use Pythagorean theorem: 72 + 𝑥 2 = 122
49 + 𝑥 2 = 144
𝑥 2 = 95
√𝑥 2 = √95
𝑥 = √95 𝑜𝑟 9.7
7. Use Pythagorean theorem: leg2 + leg2 = hypotenuse2
122 + 162 = 𝑐 2
400 = 𝑐 2
20 = c
8. Use the 45-45-90 sample triangle on the E.O.C. formula sheet and set up a proportion:
𝑥
√2
=
15
1
Cross multiply and get 𝑥 = 15√2
9. Use the trigonometric functions (see E.O.C. formula sheet) and choose which one to use to solve for
the angle. In this case, you could use any of them. REMEMBER: SOH CAH TOA and put your calculator
in “degree” mode.
30
sin 𝑥 = 60 or cos 𝑥 =
30√3
60
30
30
or tan 𝑥 = 30
30√3
)
60
Hit sin−1 (60) or Hit cos −1 (
√3
30
or Hit tan−1 (30 3)
√
Any of these gives: x= 30⁰
10. Choose cos(x) since you have adjacent and hypotenuse on this triangle.
cos 𝑥 =
6
14
6
14
then hit cos −1 ( ) so x = 64.62⁰
11. Use distance formula (on E.O.C. formula sheet): 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
𝑑 = √(0 − 4)2 + (6 − 3)2
𝑑 = √16 + 9
𝑑 = √25
12. Use midpoint formula (on E.O.C. formula sheet): midpoint =
=
8+(−6)
2
,
8+2
2
=
2
2
,
10
2
d=5
𝑥1 +𝑥2 𝑦1 +𝑦2
, 2
2
= (1,5)
𝑦 −𝑦
5−0
5
13. Slope is rise over run: use slope formula to calculate: 𝑚 = 𝑥2 −𝑥1 = 0−(−3) = 3
2
1
14. Use slope formula to find slope of line (see #13).
m= – 1 then plug one of the points into the slope-intercept form
y = mx+b
becomes 3 = 0(x) + b
solving for b leaves us with b=3
So, the equation is y= -x+3
15. Perpendicular lines have slopes that are negative reciprocals of each other, so the answer is (c)
because 1/3 and -3/1 are negative reciprocal slopes.
16. Same-side interior angles are supplementary, so angles 2 and 5, 4 and 8 are possibilities.
17. The two angles are corresponding angles, and corresponding angles are always congruent. So,
3x = 90 and x = 30
18. Angles 3 and 6 are alternate interior angles, and alternate interior angles are always congruent. So
find “x” first.
x + 25 = 4x – 8
33 = 3x
11 = x, then plug “x” back into angle 6:
4x – 8
4(11) – 8 = 44 – 8 = 36 means angle 6 = 36⁰
BUT, angle 5 is supplementary to angle 6, so angle 5 is 180 – 36 = 144⁰
19. Corresponding angles
20. Maria’s angle is a same-side interior with the 70⁰ angle, so those two angles are supplementary.
X + 70 = 180 means Maria’s angle is 110⁰
21. Converse means switch the first and last parts of the sentence.
Inverse means put “not” in front of both halves of the sentence.
Contrapositive means do both of the above.
22. Choice (d)
23. The terms go up by 3, so the next two terms are: 13, 16
24. Add one more diameter through the center of the circle, gives us answer (d).
25. (see # 21)
26. Right triangles and regular triangles are both specific kinds of triangles so these both need to be
inside the category of “triangles”. Isosceles triangle are NOT even mentioned in this problem, so don’t
use them.
27. The only possible answer is (b) because there is no information given about prices of desserts or
about how healthy they are.
28. Choice (c) is the only one that is certain all the time. If it is only going to rain when Maria washes
her dog, then if it never rained, she never washed her dog!
29. The only information explicitly given is that AB divides angle A into half, so it must be the angle
bisector. AB does not end in a right angle so it is NOT perpendicular.
30. Label the diagram, and you’ll see that XY + YZ = XZ, so 2x + (5x-9) = (33 + x)
Solve for x: 7x – 9 = 33 + x
6x = 42
x = 7. Plug “x” back into each expression.
XY = 2x = 2(7) = 14
YZ = 5x-9 = 5(7)-9 = 35-9 = 26
XZ = 33 + x = 33 + 7 = 40
31. Two of the angles are vertical angles, and vertical angles are always congruent. So, (122 – x) = x
122 – x = x
122 = 2x
61 = x
32. Remember, the two interior angles on the far side of a triangle add up to equal the exterior angle
on the other side of the triangle. So, x + y = 100. But you also need to know that this triangle is
isosceles. So two sides are congruent and two angles are congruent. So, the third angle in the triangle is
also “y”. All three angles in a triangle add up to 180⁰. So x + y + y = 180 which simplifies to x + 2y = 180.
You have two equations and two variables.
x + y = 100
x + 2y = 180
Solve using substitution. Solve for “y” in the top equation and substitute it into the bottom one.
y = 100 – x substituted into the bottom equation is: x + 2(100-x) = 180.
Solve for x: x + 200 – 2x = 180
-x = -20
x = 20 and plug back into the first equation to solve for “y”
(20) + y = 100 means y = 80
33. Supplementary angles add up to 180.
34. All 3 angles add up to 180, so:
4x + 5x + 24 + 60 – x = 180
8x + 84 = 180
8x = 96
x = 12 , so substitute back into each angle expression
4x = 4(12) = 48
60 – x = 60 – 12 = 48
5x + 24 = 5(12) +24 = 84
35. As “b” increases, the tray gets taller so the angles “a” and “c” are getting smaller.
36. Y-intercept is 3 and slope is rise over run, which is 1 over -3 so slope = -1/3
Y = -1/3x + 3
37. (d)
38. Two congruent angles is ISOSCELES and all of the angles are LESS THAN 90⁰ so this is also ACUTE.
39. This moves 6 units left and nothing in the y-direction. So, choice (c).
40. Draw a diagram if it helps to see this. Since H is the midpoint, that means the two halves are equal.
So, GH = HI
7x = 4x + 21
3x = 21
x = 7, plug into GH gives us 7x = 7(7) = 49.
41. This is an equilateral triangle since all 3 sides are congruent. Also, all angles are congruent. All 3
angles add up to 180⁰ so we can use this equation below:
(16x + 5) + (16x + 5) + (16x + 5) = 180
48x + 15 = 180
48x = 165
x = 3.44
42. Use midpoint formula just like in # 12.
5+4 −9+3
,
2
2
43.
𝟗
𝟐
= ( , -3)
Use trigonometry (look at E.O.C. formula sheet if you need to). The height of the tree is the
“opposite” side of angle 29⁰. Label that side “x”. Now use the sin function.
𝑥
sin(29) = 67
𝑥
.48 = 67
Put your calculator into “degree” mode and hit sin(29)
Next, cross multiply
(.48)(67) = x
32 = x
So, x = 32 ft.
44. Two pairs of parallel sides looks like
>
↑
↑
So, it’s a rectangle.
>
45. Even if all the students who have an “A” are boys, there are still only 11 boys in the class (25-14). So
if there are 12 A’s, then at least one of those A’s must be a girl.
11 boys with an “A” + 1 girl with an “A” = 12 A’s total