Centre for Education in Mathematics and Computing
Faculty of Mathematics
University of Waterloo
Waterloo, ON Canada N2L 3G1
GEOMETRY
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Education in Mathematics and Computing, University of Waterloo.
Answers:
1.
a. opposite: a and c, b and d, e and g, f and h
supplementary: a and b, b and c, c and d, d and a, e and f, f and g, g and h, h and e
b. opposite: i and k, j and l, m and o, n and p, q and s, r and t
supplementary: i and j, j and k, k and l, l and i, m and n, n and o, o and p, p and m,
q and r, r and s, s and t, t and q
2.
a. a = 130º,
b. a = 59º,
b = 50º,
b = 75º,
c = 50º
c = 59º,
3.
a. 76º
b. 33º
c. 9º
4.
a. 46º
b. 93º
c. 78º
5.
a. 75º, 105º
b. 30º, 60º
6.
a. corresponding
d. alternate
b. alternate
e. corresponding
7.
a. a = 122º, b = 58º, c = 58º,
b. a = 84º, b = 84º, c = 96º
c. a = 55º, b = 125º, c = 55º,
8.
Both angles are 45º.
9.
a. 56º
10.
a. a = 59º,
b. a = 62º,
c. a = 46º,
c. corresponding
f. corresponding
d = 122º, e = 58º
d = 55º,
b. Both angles are 34º.
b = 70º,
b = 74º
b = 46º,
d = 46º
e = 125º, f = 55º,
g = 125º
c. 40º
c = 37º
c = 32º,
d = 116º
X
11.
40°
c = 60º, d = 47º
a
b
c = 53º
c = 106º, d = 49º
c
d
c = 70º, d = 110º, e = 40º
W
1
∆WXY is an isosceles, so b = c = (180 ! 40) = 70º. Then note
e
2
a + 40º = b = 70º, because ∆XYZ is an isosceles. So we find a = 30º.
Z
Then e = 180o – 2(70o )= 40o . Also, c and d are supplementary, so d = 110º.
1
For more activities and resources from the University of Waterloo’s Faculty of Mathematics, please visit www.cemc.uwaterloo.ca.
a.
b.
c.
d.
a = 73º,
a = 101º,
a = 106º,
a = 30º,
b = 47º,
b = 79º,
b = 25º,
b = 70º,
Y
Centre for Education in Mathematics and Computing
Faculty of Mathematics
University of Waterloo
Waterloo, ON Canada N2L 3G1
GEOMETRY
This resource may be copied in its entirety, but is not to be used for commercial purposes without permission from the Centre for
Education in Mathematics and Computing, University of Waterloo.
V
e. a = 42º, b = 53º, c = 85º, d = 53º, e = 42º, f = 138º
a
b is the supplementary angle of !VWX, so b = 180º – 127º = 53º.
d and b are alternate angles and VW and YZ are parallel, so d = b = 53º.
c and !VXW are opposite angles, so c = 85º.
a + b + c = 180º, so a = 42º.
Note that a and e are alternate, so e =a = 42º.
d
Y
f and e are supplementary, so f = 138º.
W
b 127°
c
X
85°
M
f. a = 25º, b = 70º, c = 85º, d = 85º, e = 29º, f = 66º, g = 49º
∆MRQ is a right triangle, and a = 90º – 65º = 25º.
Since b = 180º – 110º = 60º, c = 180º – (25º + 70º) = 85º
c and d are opposite, so d = 85º. Then e = 180º – (66º + 85º) = 29º.
Also, we see f = 66º. (opposite angles)
Then g = 180º – (65º + 66º) = 49º.
a
O
N
e
b c d
S 110° 66°
P
f
g
R
12.
Since the pentagon may be divided into 3 non-overlapping triangles,
the interior angle sum is 3 ! 180 = 540 o .
13.
Since !QOT is a straight angle, !POQ + !POU + !UOT = 180º.
!POU = !ROS (opposite angles)
Therefore !POQ + !ROS + !UOT = 180º.
The sum of the angles in a triangle is 180º. Then we have
!P + !Q + !POQ + !U + !T + !UOT + !R + !S + !ROS
= 3 ! 180° = 540o.
Subtracting (1) from (2) gives
!P + !Q + !U + !T + !R + !S
= 540 o ! 180 o = 360 o.
14.
Let !BAC = t, !BCA = w, !ABE = u, as shown.
Then t = 180 ! w ! 2u
= 180 ! (180 ! 2x) ! 2u
= 2x ! 2u = 2(x ! u)
Hence t = 2y.
The answer is c.
B
t
A
65° Q
H
O
E
U
S
…….(1)
…….(2)
C
x x
u
u
f
Z
e
D
y
E
For more activities and resources from the University of Waterloo’s Faculty of Mathematics, please visit www.cemc.uwaterloo.ca.
2