1. No category

This worksheet is from Study Island. All answers, as well as explained solutions, are
in the last few pages. Enjoy!
Generation Date: 03/31/2014
Generated By: Paige Cardaci
1. In the figure below, AB
DE, BC
EF, and CD
FA.
Note: picture not drawn to scale
If m AGF = m BGC = m DGC = m EGF = 69°, what is m AGB?
A. 35°
B. 111°
C. 69°
D. 42°
2. In the figure below, trapezoid WXYZ is isosceles, with WZ
Note: picture not drawn to scale
If m XWZ = 124° and m WXZ = 28°, what is m XZY?
XY.
A. 35°
B. 20°
C. 28°
D. 56°
3. In the figure below, triangle JPK is an equilateral triangle, and quadrilateral KNML is
a parallelogram.
Note: picture not drawn to scale
If m MLK = 134°, what is m PKN?
A. 88°
B. 60°
C. 44°
D. 74°
4. In the figure below, trapezoid JKLM is an isosceles trapezoid, with JM
Note: picture not drawn to scale
If m KNL = 70° and m JMN = 15°, what is m HJM?
A. 90°
B. 85°
C. 70°
KL.
D. 95°
5. In the figure below, quadrilateral PQRV and quadrilateral RSTU are divided into two
congruent halves by line PT, and PQ RQ and RS TS.
Note: picture not drawn to scale
If m PQR = 38° and m RST = 68°, what is m QRS?
A. 74°
B. 106°
C. 53°
D. 127°
6. In the figure below, each side of hexagon MNPQRS is 2 cm, and each angle measures
120°.
Note: picture not drawn to scale
If triangle NQS is an equilateral triangle, with each side 3.5 cm long, what is m MNS?
A. 30°
B. 45°
C. 60°
D. 25°
7. In the figure below, m LJK = 39° and m
KLJ = 70°.
Note: picture not drawn to scale
What is m JKH?
A. 19°
B. 39°
C. 9°
D. 90°
8. In the figure below, m RSP = 60°, m SPR = 45°, and m QRS = 111°.
Note: picture not drawn to scale
What is m QRP?
A. 36°
B. 46°
C. 90°
D. 136°
9. In the figure below, quadrilateral JKLM is a trapezoid that is not isosceles.
Note: picture not drawn to scale
If m KJM = 130°, m JML = 50°, and m KLN = 152°, what is m LKJ?
A. 152°
B. 50°
C. 162°
D. 130°
10. In the figure below, DG is parallel to EF and
FDG
DFE.
Note: picture not drawn to scale
If m FDG = 67° and m EDG = 101°, what is m FED?
A. 90°
B. 101°
C. 74°
D. 79°
11. In the figure below, quadrilateral EFHJ is a parallelogram, and triangle FGH is a
scalene right triangle.
Note: picture not drawn to scale
If m JEF = 140°, what is m HFG?
A. 30°
B. 60°
C. 40°
D. 50°
12. In the figure below, MN
MQ, RN
RQ, NO
QP, and m SON = m SPQ.
Note: picture not drawn to scale
If m RMN = 34° and m MNO = 128°, what is m OPQ?
A. 124°
B. 144°
C. 108°
D. 128°
13. In the figure below, quadrilateral GHKL is a rectangle, and triangle HJK is isosceles,
with HJ KJ.
Note: picture not drawn to scale
If m HJK = 36°, what is m JKL?
A. 172°
B. 72°
C. 162°
D. 144°
14. In the figure below, triangle JKL is isosceles, with JK
equilateral.
LK, and triangle LMN is
Note: picture not drawn to scale
If m JKL = 28°, what is m KLM?
A. 54°
B. 44°
C. 39°
D. 49°
15. In the figure below, triangle ABC is a right triangle, and quadrilateral CDEF is a
square.
Note: picture not drawn to scale
If m CAB = 57°, what is m ACF?
A. 157°
B. 147°
C. 140°
D. 150°
16. In the figure below, m PSQ = 69° and m SRQ = 27°.
Note: picture not drawn to scale
What is m PST?
A. 132°
B. 48°
C. 42°
D. 96°
17. In the figure below, triangle PQR is an isosceles triangle, PR
37°.
QR, and m PQR =
Note: picture not drawn to scale
What is m QRS?
A. 106°
B. 74°
C. 143°
D. 53°
18. In the figure below, AB is parallel to DC, and BC is parallel to ED.
Note: picture not drawn to scale
If m DCB = 90°, m DCE = 34° and m BAC = 47°, what is m ECA?
A. 81°
B. 9°
C. 103°
D. 13°
19. In the figure below, m PDE = 40°, m APB = 67°, and m AED = 106°.
Note: picture not drawn to scale
What is m AEB?
A. 33°
B. 123°
C. 213°
D. 73°
20. In the figure below, quadrilateral VWXY is a rectangle and has congruent diagonals
that bisect each other.
Note: picture not drawn to scale
If m VWY = 28°, what is m VZY?
A. 46°
B. 56°
C. 124°
D. 62°
Answers
1. D
2. C
3. D
4. B
5. C
6. A
7. A
8. A
9. A
10. D
11. D
12. C
13. C
14. B
15. B
16. B
17. B
18. D
19. A
20. B
Explanations
1. It is given that m AGF = m BGC = m DGC = m EGF = 69°. Since
DGE are vertical angles, m AGB = m DGE.
AGB and
The sum of the angle measures surrounding a single point always equals 360°.
m AGF + m AGB + m BGC + m DGC + m DGE + m EGF = 360°
(m AGF + m BGC + m DGC + m EGF) + (m AGB + m DGE) = 360°
4(m AGF) + 2(m AGB) = 360°
4(69°) + 2(m AGB) = 360°
276° + 2(m AGB) = 360°
2(m AGB) = 84°
m AGB = 42°
Therefore, m AGB = 42°.
2. Since trapezoid WXYZ is isosceles and WZ XY, both sets of base angles are
congruent. Therefore, m XWZ = m WXY and m WZY = m XYZ.
The sum of the angle measures in a quadrilateral always equals 360°.
m XWZ + m WXY + m WZY + m XYZ = 360°
2(m XWZ) + 2(m WZY) = 360°
2(124°) + 2(m WZY) = 360°
248° + 2(m WZY) = 360°
2(m WZY) = 112°
m WZY = 56°
The sum of the angle measures in a triangle always equals 180°.
m XWZ + m WXZ + m XZW = 180°
124° + 28° + m XZW = 180°
m
XZW = 28°
Since XZW and XZY are adjacent angles and form WZY, the sum of m XZW
and m XZY is equal to m WZY.
m XZW + m XZY = m WZY
28° + m XZY = 56°
m
XZY = 28°
Therefore, m XZY = 28°.
3. Since triangle JPK is equilateral, each angle is equal to 60°; therefore, m JKP = 60°.
Since quadrilateral KNML is a parallelogram, opposite angles are congruent. Therefore,
m MLK = m KNM and m LKN = m NML.
The sum of the angle measures in a quadrilateral always equals 360°.
m MLK + m KNM + m LKN + m NML = 360°
2(m MLK) + 2(m LKN) = 360°
2(134°) + 2(m LKN) = 360°
268° + 2(m LKN) = 360°
2(m LKN) = 92°
m
Since
180°.
JKP,
PKN, and
LKN = 46°
LKN form a straight angle, the sum of their measures is
m JKP + m PKN + m LKN = 180°
60° + m PKN + 46° = 180°
106° + m PKN = 180°
m
PKN = 74°
Therefore, m PKN = 74°.
4. Since KNL and JNM are vertical angles, m KNL = m JNM = 70°.
The sum of the angle measures in a triangle always equals 180°.
m JNM + m JMN + m MJN = 180°
70° + 15° + m MJN = 180°
85° + m MJN = 180°
m MJN = 95°
Since MJN and HJM form a linear pair, they are supplementary angles and the sum
of their measures is 180°.
m MJN + m HJM = 180°
95° + m HJM = 180°
m
HJM = 85°
Therefore, m HJM = 85°.
5. Since PQ RQ, triangle PQR is isosceles, which means the base angles are congruent.
So, m QRP = m QPR.
The sum of the angle measures in a triangle always equals 180°.
m PQR + m QRP + m QPR = 180°
m PQR + 2(m QRP) = 180°
38° + 2(m QRP) = 180°
2(m QRP) = 142°
m QRP = 71°
Since RS TS, triangle RST is isosceles, which means the base angles are congruent.
So, m SRT = m STR.
The sum of the angle measures in a triangle always equals 180°.
m RST + m SRT + m STR = 180°
m RST + 2(m SRT) = 180°
68° + 2(m SRT) = 180°
2(m SRT) = 112°
m SRT = 56°
Since
180°.
QRP,
QRS, and
SRT form a straight angle, the sum of their measures is
m QRP + m QRS + m SRT = 180°
71° + m QRS + 56° = 180°
127° + m QRS = 180°
m
QRS = 53°
Therefore, m QRS = 53°.
6. Since triangle NQS is equilateral, each angle is equal to 60°; therefore, m SNQ =
60°. Since each angle of hexagon MNPQRS measures 120°, m MNP = 120°.
Since their side lengths are the same, triangle SMN triangle NPQ. These triangles are
also isosceles, which means their base angles have the same measure. So, m MSN =
m MNS = m PNQ = m PQN.
Since MNS, SNQ, and PNQ are adjacent angles and form MNP, the sum of
m MNS, m SNQ, and m PNQ is equal to m MNP.
m MNS + m SNQ + m PNQ = m MNP
(m MNS + m PNQ) + m SNQ = m MNP
2(m MNS) + 60° = 120°
2(m MNS) = 60°
m MNS = 30°
Therefore, m MNS = 30°.
7. To determine m JKH, first calculate m JKL.
The sum of the angle measures in a triangle always equals 180°.
m LJK + m KLJ + m JKL = 180°
39° + 70° + m JKL = 180°
m JKL = 71°
Since LKH forms a linear pair with a right angle, it is also a right angle. Also, since
JKL and JKH are adjacent and form LKH, the sum of their measures is 90°.
m JKL + m JKH = 90°
71° + m JKH = 90°
m
JKH = 19°
Therefore, m JKH = 19°.
8. To determine m QRP, first calculate m PRS.
The sum of the angle measures in a triangle always equals 180°.
m RSP + m SPR + m PRS = 180°
60° + 45° + m PRS = 180°
m
PRS = 75°
Since PRS and QRP are adjacent angles and form QRS, the sum of m PRS and
m QRP is equal to m QRS.
m PRS + m QRP = m QRS
75° + m QRP = 111°
m QRP = 36°
Therefore, m QRP = 36°.
9. Since KLN and MLK form a linear pair, they are supplementary angles and the
sum of their measures is 180°.
m KLN + m MLK = 180°
152° + m MLK = 180°
m MLK = 28°
The sum of the angle measures in a quadrilateral always equals 360°.
m KJM + m JML + m MLK + m LKJ = 360°
130° + 50° + 28° + m LKJ = 360°
208° + m LKJ = 360°
m LKJ = 152°
Therefore, m LKJ = 152°.
10. Since FDG and EDF are adjacent angles and form EDG, the sum of m FDG
and m EDF is equal to m EDG.
m FDG + m EDF = m EDG
67° + m EDF = 101°
m EDF = 34°
Since
FDG
DFE, m FDG = m DFE = 67°.
The sum of the angle measures in a triangle always equals 180°.
m EDF + m DFE + m FED = 180°
34° + 67° + m FED = 180°
101° + m FED = 180°
m FED = 79°
Therefore, m FED = 79°.
11. Since quadrilateral EFHJ is a parallelogram, opposite angles are congruent.
Therefore, m JEF = m FHJ = 140° and m EFH = m HJE.
The sum of the angle measures in a quadrilateral always equals 360°.
m JEF + m FHJ + m EFH + m HJE = 360°
2(m JEF) + 2(m EFH) = 360°
2(140°) + 2(m EFH) = 360°
280° + 2(m EFH) = 360°
2(m EFH) = 80°
m EFH = 40°
Since EF is parallel to JG and angle FGH is a right angle, angle EFG is also a right angle.
Therefore, m EFH + m HFG = 90°.
m EFH + m HFG = 90°
40° + m HFG = 90°
m
Therefore, m HFG = 50°.
12. Since QRM is a right angle,
90°.
HFG = 50°
NRM is also a right angle; therefore, m
NRM =
The sum of the angle measures in a triangle always equals 180°.
m NRM + m RMN + m MNR = 180°
90° + 34° + m MNR = 180°
124° + m MNR = 180°
m MNR = 56°
Since MNR and RNO are adjacent angles and form MNO, the sum of m
and m RNO is equal to m MNO.
m MNR + m RNO = m MNO
MNR
56° + m RNO = 128°
m RNO = 72°
Since m NRM = 90°, m NRS = 90°. Also, since
90°.
RSO is a right angle, m RSO =
The sum of the angle measures in a quadrilateral always equals 360°.
m RNO + m NRS + m RSO + m SON = 360°
72° + 90° + 90° + m SON = 360°
252° + m SON = 360°
m SON = 108°
Therefore, since m SON = m SPQ, and SPQ is the same angle as OPQ, m OPQ
= 108°.
13. Since triangle HJK is isosceles, with HJ KJ, then m JKH = m JHK.
The sum of the angle measures in a triangle always equals 180°.
m HJK + m JKH + m JHK = 180°
m HJK + 2(m JKH) = 180°
36° + 2(m JKH) = 180°
2(m JKH) = 144°
m JKH = 72°
Since quadrilateral GHKL is a rectangle, each of its angles is a right angle; therefore,
m HKL = 90°.
Since JKH and HKL are adjacent angles and form JKL, the sum of m
m HKL is equal to m JKL.
m JKL = m JKH + m HKL
JKH and
= 72° + 90°
= 162°
Therefore, m JKL = 162°.
14. Since triangle JKL is isosceles and JK
LK, m KLJ is equal to m KJL.
The sum of the angle measures in a triangle always equals 180°.
m JKL + m KJL + m KLJ = 180°
28° + 2(m KLJ) = 180°
2(m KLJ) = 152°
m KLJ = 76°
Since triangle LMN is equilateral, each angle is equal to 60°; therefore, m
MLN = 60°.
In the figure, KLJ, KLM, and MLN form a straight angle; therefore, the sum of
their measures is 180°.
m KLJ + m KLM + m MLN = 180°
76° + m KLM + 60° = 180°
136° + m KLM = 180°
m KLM = 44°
Therefore, m KLM = 44°.
15. Since triangle ABC is a right triangle, with angle ABC as a right angle, m ABC =
90°.
The sum of the angle measures in a triangle always equals 180°.
m CAB + m ABC + m BCA = 180°
57° + 90° + m BCA = 180°
m BCA = 33°
Since BCA and ACF form a linear pair, they are supplementary angles and the sum
of their measures is 180°.
m BCA + m ACF = 180°
33° + m ACF = 180°
m ACF = 147°
Therefore, m ACF = 147°.
16. Start with triangle PSQ. Since
SQP is denoted as a right angle, its measure is 90°.
The sum of the angle measures in a triangle always equals 180°.
m SQP + m PSQ + m QPS = 180°
90° + 69° + m QPS = 180°
m QPS = 21°
Next, determine m PSR.
m QPS + m SRQ + m PSR = 180°
21° + 27° + m PSR = 180°
m PSR = 132°
Since PSR and
measures is 180°.
PST form a linear pair, they are supplementary and the sum of their
m PSR + m PST = 180°
132° + m PST = 180°
m PST = 48°
Therefore, m PST = 48°.
17. Since triangle PQR is an isosceles triangle and PR
= 37°.
QR, then m PQR = m RPQ
The sum of the angle measures in a triangle always equals 180°.
m RPQ + m PQR + m QRP = 180°
37° + 37° + m QRP = 180°
m
QRP = 106°
Since QRP and QRS form a linear pair, they are supplementary angles and the sum
of their measures is 180°.
m QRP + m QRS = 180°
106° + m QRS = 180°
m QRS = 74°
Therefore, m QRS = 74°.
18. The sum of the angle measures in a triangle always equals 180°.
m CBA + m BAC + m ACB = 180°
90° + 47° + m ACB = 180°
137° + m ACB = 180°
m ACB = 43°
Since DCE, ECA, and ACB are adjacent angles and form DCB, the sum of
m DCE, m ECA, and m ACB is equal to m DCB.
m DCE + m ECA + m ACB = m DCB
34° + m ECA + 43° = 90°
77° + m ECA = 90°
m
ECA = 13°
Therefore, m ECA = 13°.
19. Since m APB = 67°, then m DPE = 67° because they are vertical angles.
The sum of the angles in a triangle always equals 180°.
m PDE + m DPE + m PED = 180°
40° + 67° + m PED = 180°
m PED = 73°
Since PED and AEB are adjacent angles and form AED, the sum of m
m AEB is equal to m AED.
m PED + m AEB = m AED
PED and
73° + m AEB = 106°
m AEB = 33°
Therefore, m AEB = 33°.
20. Since the diagonals are congruent and bisect each other, triangle VWZ is isosceles,
with m WVZ = m VWZ. Notice that VWY is the same angle as VWZ.
The sum of the angle measures in a triangle always equals 180°.
m WVZ + m VWZ + m WZV = 180°
2(m VWZ) + m WZV = 180°
2(28°) + m WZV = 180°
56° + m WZV = 180°
m
WZV = 124°
Since WZV and VZY form a linear pair, they are supplementary angles and the sum
of their measures is 180°.
m WZV + m VZY = 180°
124° + m VZY = 180°
m VZY = 56°
Therefore, m VZY = 56°.