Download Solutions_for_Semester_study_guide

1.
2.
3.
4.
5.
6.
7.
A triangle with three
acute angles and no
congruent sides is acute
scalene triangle.
A triangle with side
lengths of 5 cm, 3 cm, and
5 cm is isosceles triangle.
The triangle shows 3
different tick marks which
means 3 different
measures and it shows a
90° angle so it is D-right
scalene.
Two 35° angles are shown
so the third angle has to
be 110°. Therefore the
answer is C-obtuse
isosceles.
Because the directions say
the triangle is isosceles,
the two legs are
congruent. Therefore
3x+2=5x-14. So solving x =
8. Plugging in 8 for x the
legs measure 26 each. So
to find the base
26 + 26 + x = 82 and the
third side (base) is 30.
<BCD is exterior so it
equals the sum of the two
remote interior angles. So
3x+10 =40+2x+20. X
calculates to be 50.
Plugging in m<BCD is 160,
choice D.e
Looking at the triangle I
notice it is isosceles. So
base angles are equal. So
40 + x + x = 180. X equals
B-70°.
8. This is a bowtie so the
vertical angles are
congruent. In the triangle
to the left that has 90° and
60° labeled, the third
angle has to be 30°. That
means that we have 30°
plus 65° given in the
triangle on the right.
When I subtract 95 from
180 the measure of <1 is
E-85°.
9. If ABC congruent to XYZ,
order matters on stating
congruence. So the
answer is C – CBA cannot
be congruent to ZXY.
10. Since EFG congruent to
HIJ, then m<E=m<H,
m<F=m<I so m<F=65 and
now we can find m<E =80.
And we can use
substitution to find m<H is
D-80°.
11. Since EFG congruent to
HIJ, then EG = HJ.
12. Since X congruent to N,
m<N is 20°. So
20+75+5x-10=180. Solving
we find x=19.
13. Since m<Z = m<0, m<Z is
5x – 10. Substituting 10
for x, m<Z is 85°.
14. Since K congruent to C,
M<K is 3x – 5. So
3x-5+40+2x = 180. X
calculates to be 29°.
15. Using Choice B we could
get SSS, Using choice C
we could get SAS. So
answer is E- B or D.
16. Using Choice C we would
have SAS.
17. For the triangles to be
congruent we could use
choice A or Choice C. That
means our answer is
Choice D – A and C.
18. Knowing that side AB is
congruent by reflexive
property, we would have
C- ASA.
19. Knowing that the vertical
angles are congruent, we
would have D-AAS.
20. We have one marked pair
of congruent angles and
we can mark the
congruent side BD
because of reflexive.
Therefore we need
another angle to use AAS.
So answer is B.
21. We have congruent
vertical angles and one
pair of congruent sides.
So we need another angle
to use ASA.
22. Since M is midpoint of KQ,
then we know KM and MQ
are congruent. Adding the
congruent vertical angles,
we would have SSA which
is not valid. Looking closer
I see parallel lines KL and
QP. That means that <K
and <Q are alternate
interior angles as well as
<P and <L. This gives me
enough information to use
ASA, Choice C.
23. Looking at the two
congruent angles lets me
know that their opposite
sides are congruent. So
3x+1 = x+11 and x = 5,
Choice B.
24. This is a 45-45-90 triangle,
making it isosceles. So the
legs are congruent!
Therefore 3x+4=7x-8.
X = 3. Plugging in 3 for x
means my legs are each
13, choice D.
25. The markings on this
triangle indicate it is
isosceles, so the base
angles are congruent.
That means y = 35. Then
35 + 35 + x = 180. X = 110.
Choice E.
26. When I look at the triangle
I see it is equilateral which
automatically means it is
equiangular. That means
x + x + x = 180. X = 60,
Choice B.
27. This problem takes a few
steps to complete. First of
all look at the reflexive
side that is shared. It is
congruent to itself. Then
the unmarked leg on the
lower triangle is the same
as the left side of the
triangle because the base
angles are marked as
congruent. That gives us
Choice A-SSS.
28. The triangle on the left
shows all three sides
congruent. That means
each angle has a measure
of 60°. X is the angle that
is supplementary to one of
those 60° angles so x has
to be 120. Looking at the
right side angle, it is
isosceles. That means
since one base angle is y,
so is the other. So
120+y+y=180. Y = 30.
Choice D
29. Looking at the lower
triangle, I see it is
isosceles. So both base
angles are 35°.
35 + 35 + y = 180. This
means y = 110. X is
supplementary to 110 so
x = 70. Choice C.
30. Looking at the entire big
triangle (ignore internal
segment) you wee that
z+35 + 90 = 180. Z = 55.
And w + 35 = 180. This
means w = 145. Choice D.
31. Since we are told that YW
is perpendicular bisector
then we know that YB has
same measure as YA. That
means 7x-4=5x+6. X-5.
Choice E.
32. Only III has enough
information to say that AP
and PB have the same
measure. So choice C.
33. PQ is marked congruent to
PO so PQ is 10, choice B.
34. Because M is on the
perpendicular bisector the
measure of LM is the
same as the measure of
MN which is 8.
35. From choices shown, the
only one that is correct is
Choice B.
36. This distance from P to B
is the same as the
distance from P to C so
15 = 3x+6. That means x is
3, choice B.
37. This problem uses
Pythagorean theorem and
answer is 3 Choice C.
38. With QK = 20, we can use
Pythagorean theorem to
find NQ = 15. That is the
same for NP. Choice E.
39. Remember that I told you
with centroids the
distance from the median
to the centroid is half the
distance from the centroid
to the vertex. This means
that since CW is 4, then CX
is 8, choice D.
40. VZ = 4 + 8 so Choice E –
12.
41. UY goes from vertex to
opposite side at a right
angle, making it an
altitude, a perpendicular
bisector, a median and an
angle bisector. So Choice
E.
42. UV is midsegment so XY is
10. W is midpoint of XY so
WX is half of 10 or 5,
choice A.
43. WU is half of UZ so it is
half of 8 – which is 4.
44. KL is twice BP so it is 50.
KM is 30. Using
Pythagoren Theorem LM
is 40. So 50 + 30 + 40 =
120, choice B.
45. FG is half of KM. so (5x+7)
+5x+7) = 12x + 8 and x
comes out to be 3.
Plugging in 5(3)+7 gives us
22.
46. Since H is smallest angle, I
is middle sized angle, and J
is largest angle, then the
opposite sides are
proportionally sized.
So JI < HJ<HI, choice C.
47. Since the third angles
measures 70, the smallest
angle is B and the shortest
side is AC, choice A.
48. Since <G is 80, the largest
side down to smallest are
EF, GF, and EF, choice D.
49. The shortest
measurement would be
14-8 or 6, while the largest
side could be 8+14 which
is 22. So all sides possible
are between 6 and 22.
That omits the side that is
exactly 22, Choice D.
50. 3x+2x+2 = 42. X = 8 Then
plugging in 8 for x, LM is
3(8) or 24.
51. DE is half of AC so
2(y+6) = 3y+4 and y =8.
Sub in 4 for y and AC is
3(8) + 4 or 28.
52. Since the triangle is
isosceles, the base angles
are congruent. So
x + x + 50 = 180, or x = 65.
Choice J.
53. Each time we divide by 2
so next number is 4.
54. H, Median
55. Choice G – because MP is
shown as a segment that
connects midpoints of two
sides of the triangle it is a
midsegment.
56. Triangle is isosceles,so
base angles congruent. So
x+x+110 = 180. X = 35
Choice D.
57. Using congruent vertical
angles we have SAS,
choice G.
58. Choice D – point B.
59. Because the base angles
are congruent, ABD is
isosceles triangle – Choice
H.
60. Lines are parallel so we
have SSI interior angles
that are supplementary.
So 3x – 45 + 90 = 180. X =
45, Choice A.
61. If <2 congruent to <6 the
lines would be parallel
because <2 and <6 are
corresponding. Choice J
62. X + 42 + 36 = 180. So x =
102. And y + 36 + 42 =
180. So y =102 also!
63. OMIT THIS PROBLEM.
64. <A, <C, <B, Choice B
65. Inverse negates the
hypothesis and the
conclusion so : If the sky
is NOT blue, then it IS
raining. Choice A.
66. The two angles together
form a right angle so they
add up to 90. So
16y+2+13y+1=90 and y = 3
Choice H.
67. 22 = ½ (4x) or 22=2x, = 11,
Choice G.
68. Using the congruency
statement TV is congruent
to LP, choice G.
69. This is an isosceles
triangle so base angles
are congruent. Therefore
x + x + 48 = 180 and x =
66, choice A.
70. All right triangles have two
acute angles – Choice G.
71. 70 + x = 115 so x = 45,
choice C.