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CONCEPT
5 ★3
polygons |
ANGLES
M ain T heory & C oncept
Tip from Top
➊ Sum of interior angles
➊ Sum of the interior angles
n sides  n +1 sides
➧ Number of triangles in the polygon s 180°
Hexagon
Triangle
Quadrilateral
Pentagon
+1 side
Increased by 180°
······
1 triangle
180°
2 triangles
3 triangles
4 triangles
2 s 180° = 360° 3 s 180° = 540° 4 s 180° = 720°
b
α
c
β
γ
α+β+γ
= 360°
2
a
4
= 360°
1
≪Each interior/ exterior angle≫
a
θ6
θ4
5
······
θ5
1 + 2 + 3 + 4 θ1 +θ2 +θ3 +θ4
+ 5 = 360°
+θ5 +θ6 = 360°
Sum of the exterior angles in n-sided polygon
➌ Regular
θ1
θ3
3
d
a+b+c+d
θ2
360°
n-sided polygon
Side
3
Name
Triangle
Interior
Exterior
60°
120°
4
Quadrilateral
90°
90°
5
Pentagon
108°
72°
6
Hexagon
7
Heptagon
60°
51 3 °
7
45°
8
Octagon
120°
128 4 °
7
135°
9
Nonagon
140°
40°
10
Decagon
11
Undecagon
12
Dodecagon
144°
147 3 °
11
150°
36°
22 8 °
11
30°
360°
n
Exterior angle in
regular n-sided
polygon
One(each) interior angle ➧ θ= [(n 2) s 180°] ÷ n
= 180° 360°
n
b
At a angle of a polygon
Interior angle + Exterior angle = 180°
➌ Regular n-sided polygons
Sum of the interior angles in n-sided polygon (n 2) s 180°
➋ Sum of exterior angles
➋ Sum of exterior angles
360°
O
n
180° 360°
n
One(each) exterior angle ➧ α = 360° ÷ n
Simple
Sample
Essence
1. If the sign is a regular octagon as shown,
1. Since the sum of eight interior
then what is each angle of the sign?
(A) 150°
(B) 144°
(D) 120°
(E) 108°
(C) 135°
2. In rhombus ABCD, A = (10x + 6)° and
B = (5x 6)°. What is the measure of C?
(A) 96°
(B) 108°
(D) 126°
(E) 132°
(C) 112°
(D) Twenty
(E) Twenty four
2. Since AD || BC, A + B = 180°
< (10x + 6)° + (5x 6)° = 180°
D
B
C
162°, then how many sides does the polygon have?
(B) Sixteen
each angle of the regular
octagon is 1080° = 135°
8
Ans (C)
A
3. If the measure of each interior angle of a regular polygon is
(A) Twelve
angles is (8 2) s 180° = 1080°,
STOP
(C) Eighteen
< 15x° = 180°
12
< x = 12°
< C = A = 10 s 12° + 6°= 126°
Ans (D)
3. Since each interior angle of
the regular polygon is 162°, each
exterior angle is 180° 162° = 18°
<The number of sides of the
polygon is 360° = 20 Ans
(D)
18°
5. ANGLES
209
pattern guide | Ace of Base
➊ Sum of interior angles
Sum of the interior angles
in n-sided polygon
(n 2) s180°
In hexagon ABCDEF, what is the average
(arithmetic mean) of a, b, c, d, e, and f ? B
(A) 60
(B) 90
(C) 120
(D) 150
C
c°
b°
f°
a°
d°
e°
F
A
D
E
(Sol) Since the polygon ABCDEF is a hexagon, the sum
of the six interior angles is a° + b° + c° + d ° + e° + f °
= (6 2) s 180° = 4 s 180° = 720°
< The average (arithmetic mean) of a, b, c, d, e,
➋ Special quadrilateral
Supplementary angle
Sum of co-interior angles = 180°
a°
b°
a°
b°
a°
b°
b°
a°
Parallelogram
Rhombus
a°
c°
d°
b°
Trapezoid
a° + b° = c° + d° = 180°
➌ Interior/Exterior Angles
Interior angle + Exterior angle
at a vertex point = 180°
β
α
α+β= 180°
Sum of the exterior angles of
any-sided polygon = 360°
➍ Regular polygons
Each interior angle
in n - sided regular polygon
θ = (n 2) s 180° = 180° 360°
n
n
and f is 720 ÷ 6 = 120
Ans (C)
In parallelogram ABCD, C = 120° and BDC = 24°.
B
A
What is the measure of DAE?
(A) 24°
(B) 36°
(C) 48°
(D) 54°
D
E
C
(Sol)
In parallelogram ABCD, AD || BC. < C + D = 180°
< D = 180° C = 180° 120° = 60° (kCo-interior angle)
< ADE = D BDC = 60° 24° = 36°
Ans (D)
In right triangle ADE, DAE = 90° 36° = 54°
C
100°
In pentagon ABCDE, what is the measure
of the exterior angle at E ?
(A) 60°
(C) 120°
(B) 80°
(D) 240°
D
B 110°
60°
E
A
(Sol) If the exterior angle at E is x°, the sum of C 80°
100°
the exterior angles is 60° + (180 110)°
70°
+ (180 100)° + (180 90)° + x°
= 60° + 70° + 80° + 90° + x° = 360°
B 110°
60°
< 300° + x° = 360° < x° = 360° 60° = 300°
A
D
90°
x° E
Ans (A)
What is the measure each exterior angle of a regular
polygon with 15 sides?
(A) 144°
(C) 36°
(B) 156°
(D) 24°
(Sol) Since the sum of exterior angles of any-sided polygon
is 360°, each exterior angle of the polygon is
360° ÷ 15 = 24°
210
5. ANGLES
Ans (D)
BASE
pattern drill | Ace of Base
answers ... p. 103
(1) What is the sum of the interior angles of a trapezoid?
Sum of interior angles
(A) 180°
Special quadrilateral
3
In / Exterior angles
Regular polygon
4
(C) 360°
(D) 480°
(2) In the pentagon ABCDE, B = 108°, C = 100°, D = 112° and
(A) 90
(B) 100
(C) 110
E
A
D
E = 120°. What is the degree measure of A?
(D) 120
C
B
(3) What is the average(arithmetic mean) of the interior angles of a nonagon?
(A) 114°
(B) 135°
(C) 140°
(D) 150°
(4) If the sum of interior angles of a polygon is 2,700°, what is the number of sides
of the polygon?
(A) 15
2
(B) 240°
(B) 17
(C) 19
(D) 21
A
(1) In isosceles trapezoid ABCD, AB || CD and AD = BC.
What is the sum of measures of A and C?
(A) 90°
(B) 100°
(C) 180°
D
(D) 200°
(2) In rhombus ABCD, B = 130°. What is the value of x ?
(A) 30
(B) 50
(C) 60
(B) 25
130°
(D) 70
D
x°
What is the value of y ?
3x°
(C) 24
(D) 15
C
B
x°
A
(3) In parallelogram ABCD, A = x°, B = 5y° and D = 3x°. A
(A) 27
B
C
B
5y°
C
D
A 60°
(1) In quadrilateral ABCD, A = 60°, B = 100°, D = 50°.
100°
B
what is the measure of exterior angle of C ?
(A) 30°
(B) 80°
(C) 260°
C
(D) 330°
50°
(2) What is the average(arithmetic mean) of the exterior
D
angles of an octagon?
(A) 45°
(B) 40°
(C) 22.5°
100°
(3) In hexagon ABCDEF, what is the value of x ?
(A) 50
(B) 60
(C) 70
B 60°
C
140°
40°
(D) 16°
(D) 80
A
D
x°
F
E
70°
(1) Which of the following is the measure of an interior angle in a regular decagon
(10-sided polygon)?
(A) 36°
(B) 48°
A
(C) 135°
(D) 144°
(2) In regular pentagon ABCDE, what is the measure of AFC?
(A) 108°
(B) 114°
➊ (1) C (2) B (3) C (4) B ➋ (1) C (2) B (3) A ➌ (1) A (2) A (3) C ➍ (1) D (2) A (3) 18
1
(C) 120°
(D) 136°
E
B
F
D
C
(3) If each exterior angle of a regular polygon is 20°, what is the number of sides
of the polygon?
5. ANGLES
211
Polygons
n - sided polygon ➧
a
Sum of the interior angles = (n 2) s180 °
b
Sum of the exterior angles = 360 °
Special regular polygons
spark !
A
① Regular hexagon
➧ 6 equilateral triangles
F
a
B
60° 60°
60° a
60°
60° 60°
a
A a B√2
② Regular octagon
H
C
a
E
a
D
F
In the figure, if two parallel lines l and k
pass two vertices of a regular pentagon ABCDE,
(D) 22
E
x°
l
B
92°
D
(C) 20
135° D
a 45°
135°
45°
E
A
F
then what is the value of x ?
(B) 18
(E) 28
a
G
CAP
(A) 12
135° 45°
a 45°
135° C
➧ Square 4 isosceles right triangles
k
C
One exterior angle of regular n-sided polygon = 360°
Accent
n
In the figure, since an exterior angle of a regular pentagon is 360° = 72°,
5
A
CDG = 72°
Simple
Solution
E
Since two lines l and k are parallel,
F
D
B
92°
72° x
°
G 92°
H
C
l CGH = BFD = 92° (k Corresponding angles)
In triangle CDG, DCG + CDG = CGH
k <x° + 72° = 92°
<x° = 92° 72° = 20°
H
(1) In pentagon HOUSE, H, U and S are right angles. If O and E
are x° respectively, then what is the value of x?
(B) 115
(D) 135
(E) 145
x° O
E x°
(C) 125
U
S
E
F
(2) In the figure, what is the sum of the measures, in degrees,
of the six angles A, B, C, D, E and F?
212
(A) 270
(B) 360
(D) 540
(E) 630
5. ANGLES
(C) 450
(C)
answers ... p. 104
T R A I N I N G
(A) 95
Ans
A
D
B
C
➊
Sum of interior angles
In the figure,
a°
what is the value of
b°
a + b + c + d + e + f + g + h + i?
(A) 720
(B) 810
(C) 900
(D) 1080
➧ (n 2) s180°
c° d °
b°
c° d ° In the figure,
x° e°
a°
e° a°+ b°= x°+ y°
<a°+ b°+ c°+ d °+ e°+ f °+ g°+ h°+ i °
y°
= x°+ y°+ c°+ d °+ e°+ f °+ g°+ h°+ i °
f°
i
°
i°
f°
g
=
(
x
°+
c
°+
d
°+
e
°)
+
(
y
°+
f
°+
g
°+
h
°+
i
°)
°
h °
h° g°
quadrilateral
pentagon
= (4 2) s180° + (5 2) s180° = 2·180° + 3·180°
= 360° + 540° = 900°
Ans (C)
➋
Sum of interior angles
➧ (n 2) s180°
The interior angles of a polygon, measured
If the polygon has n sides, the sum of interior
in degrees, form an arithmetic progression.
angles of the polygon is
If the smallest angle is 126° and the greatest
n (126° + 174°) = (n 2) s 180°
2
5
3
< n s300° = (n 2) s180°
2
< n s5 = (n 2) s3 < 5n = 6(n 2)
2
< 5n = 6n 12 < n = 12
Ans
angle is 174°, then what is the number of
sides of the polygon?
(A) 10
(B) 12
(C) 14
(D) 16
(B)
➌
Sum of an interior angle and its exterior angle
➧ Interior angle + Exterior angle = 180°
If the measures of the angles of a pentagon are If the five interior angles are 2k °, 3k °, 4k °, 5k °
and 6k °,
in the ratio of 2 : 3 : 4 : 5 : 6, then what is
2k ° + 3k ° + 4k ° + 5k ° + 6k ° = (5 2) s180°
the smallest exterior angle of the pentagon?
27
(A) 18°
(B) 20°
(C) 24°
(D) 27°
<20k ° = 3 s180° = 540° <k ° = 27°
<Since the largest interior angle is 6k°
= 6 s27° = 162°, the smallest exterior angle
Ans
is 180° 162° = 18°
(A)
➍
Special quadrilateral
➧ Rhombus and Parallelogram Sum of two consecutive angles = 180°
A
Since the quadrilateral MATS M 63°
63°
=
is an isosceles trapezoid,
SMA = 63°, what is the measure of HMS?
T
H
M
A <A = AMS = 63°
(A) 42°
(B) 54°
63°
S
Since MA || ST, A + T = 180°
(D) 76°
(C) 63°
<T = 180° 63° = 117° <M = T = 117°
H
T
S
Ans (B)
<HMS = 117° 63° = 54°
In parallelogram MATH, MS = AT and
=
5. ANGLES
213
THEMELTINGZONE
1
answers ... p. 105
SHOW CASE
In the figure, what is the average (arithmetic mean) of the measures A
of acute angles of A, B, C, D and E?
(A) 24°
(D) 40°
(B) 30°
(E) 54°
B
E
(C) 36°
C
D
A
I sea!
J v
v
u
I u
z
E
D
2
3
4
5
6
7
214
H
From the five exterior triangles of pentagon FGHIJ,
A + B + C + D + E + 2(x + y + z + u + v) = 5 s 180°
x F
x
y
yG
z
B
Since the sum of the exterior angles of a pentagon is 360°,
x + y + z + u + v = 360°
< A + B + C + D + E + 2 s 360° = 5 s 180°
C
< A + B + C + D + E = 900° 720° = 180°
< The average of five angles A, B, C, D and E is 180° ÷ 5 = 36°
Ans (C)
What is the number of sides of a polygon for which the sum of the measures of its
interior angles is 2160°?
(A) 12
(B) 14
(D) 18
(E) 20
(C) 16
In parallelogram ABCD, A = 3x° 20°, C = 7x° 100°
E
A
B
and DE bisects ADC. What is the measure of DEB?
(A) 110°
(B) 120°
(D) 140°
(E) 150°
(C) 130°
C
D
A
a°
In the figure, if two rays AF and DE are parallel, then what is the
B
average (arithmetic mean) of a, b, c and d ?
b°
C c°
(C) 135
(A) 120
(B) 125
(D) 150
(E) 165
D
In the figure, what is the sum of the measure of ten angles
C
d°
of A + B + C + D + E + F + G + H + I + J ?
(A) 720°
(B) 900°
(D) 1,260°
(E) 1,440°
B
(C) 1,080°
F
E
(D) 36°
(E) 40°
(D) 36°
(E) 48°
5. ANGLES
(C) 30°
A
B
E
(C) 32°
what is the measure of BED?
(B) 24°
F
A
A'
A
In the regular decagon ABCDEFGHIJ (10-sided polygon),
(A) 20°
E
G
measure of Aa BC?
(B) 28°
I
J
If the regular pentagon ABCDE is folded as shown, then what is the
(A) 24°
D
H
D
B
C
J
I
D
H
E
G
F
C
JUMP
answers ... p. 105
1
5
In the figure, what is the measure of the
greatest angle in quadrilateral QUAD?
U
(A) 100°
(B) 110°
(C) 120°
(D) 130°
and E = 80°. What is the value of x?
A
(x 20)°
Q (x 40)°
(E) 140°
In the figure, A = 110°, B = 50°, D = 40°
50°
110°
C x°
40°
E 80°
(x +30)°
D
A
6
B
D
A
An equilateral triangle and
a square have a common side
as shown. What is the measure
2
E
B
D
C
of DAC?
In the figure, what is the average(arithmetic
mean) of a, b, c, d, e and f ?
(A) 105
(B) 84
a°
c°
b°
(C) 75
(D) 70
(E) 55
7
d°
e°
120°
f°
In the figure, a regular pentagon ABCDE and
an isosceles triangle CFD have a
A
common side CD as shown.
F
B
If FC = FD and CFD = 48°,
E
48°
what is the measure of ADF?
D
C
3
In regular hexagon ABCDEF as shown, what
is the degree measure of FBE?
A
(A) 15
Math Mania!
B
(B) 17.5
(C) 20
F
C
(D) 22.5
(E) 30
E
8
C
B
150°
In the figure, what is the
F
A x°
value of x?
80°
130° D
G
D
E
H
40°
30°
I
J
70°
4
Which type of polygon has an interior angle
that is always four times the measure of its
exterior angle?
9
K
In regular octagon ABCDEFGH, what is the
sum of a + b + c + d + e + f ?
D
(A) A regular pentagon
C
(C) A regular octagon
(D) A regular decagon
(E) A regular dodecagon
c°
B
E
d°
(B) A regular hexagon
e°
b°
A
F
f°
G
a°
H
5. ANGLES
215
All-Round Checks
EXAMINATION
1
In the polygon ABCD as shown, what is
5
the average(arithmetic mean) of a, b
and c ?
B
(A) 6
b°
(B) 14
(A) 50°
(B) 70°
(C) 110°
A
(E) 42
angles of a convex polygon is 2450°, what
polygon?
a° 84° c°
(D) 28
If the sum of all but one of the interior
is the value of the remaining angle in the
D
(C) 21
answers ... p. 106
(D) 130°
C
(E) 150°
2
In a quadrilateral ABCD, A = 3x° + 10°,
6
B = x° 15°, C = 4x° 45° and
D = 2x° + 50°. What is the measure of
In the figure, A = 58°, B = 84°, C = 64°
and D = 78° and E : F = 3 : 1. What is
the measure of E?
the greatest angle in quadrilateral ABCD?
(A) 48°
(A) 118°
(B) 51°
(B) 122°
(C) 54°
(C) 128°
(D) 57°
(D) 135°
(E) 60°
B 84°
E
In an isosceles trapezoid ABCD with
7
AD = BC, A = (4x + 6)°, C = (2x 18)°
In five sided-polygon ABCDE, A = 4x°,
B = 7x°, C = 6x°, D = 3x° and
and BE bisects B. What is the measure
DEA = 5x°. What is the value of x?
of BED?
(A) 12
A
B
(4x +6)°
(C) 111°
(C) 15
D
E
(D) 110°
(2x 18)°
(D) 16
C
(E) 18
(E) 109°
If the rectangular sheet is folded as
shown, then what is the value of x ?
(A) 140
(B) 130
(E) 100
5. ANGLES
B
5x° E
3x°
6x°
C
measure of an exterior angle of a regular
polygon?
(C) 32°
40°
D
7x°
Which of the following cannot be the
(B) 24°
(C) 120
(D) 110
8
(A) 18°
x°
A 4x°
(B) 12.5
(B) 112°
216
D
A
(A) 113°
4
78°
58°
(E) 140°
3
C
64°
F
(D) 45°
(E) 72°
TEST
9
In the figure, O is the center of a circle
and a quadrilateral OABC is inscribed in
13
In the figure, if F = 40°, what is the
average (arithmetic mean) of degree
the sector. If AOC = 62°, then what is
measures of A, B, C, D and E?
the measure of ABC?
(A) 60
A
(A) 128°
I
40°
(D) 75
O 62°
(E) 162°
(E) 80
C
H
G
(C) 72
(C) 148°
(D) 149°
A
(B) 64
B
(B) 132°
E
D
B
F
C
10
In regular pentagon ABCDE, triangle FCD
is an equilateral triangle. What is the
14
(arithmetic mean) of a, b, c, d and e?
(A) 27
measure of AFB ?
A
(A) 60°
B b°
(B) 36
E
(C) 78°
B
F
D
C
15
D
F
(E) 60
(E) 86°
c°
d°
a°
(D) 54
(D) 84°
E
A regular pentagon ABCDI and a regular
hexagon IDEFGH have a common side ID
B
(A) 18°
measures of the interior angles of a
(B) 20°
n - sided polygon is x°, what is the sum of
the interior angles of a n + 3 sided
C
A
(C) 24°
I
(D) 32°
polygon?
(E) 36°
(A) 3x ° + 270°
D
E
H
(B) 3x ° + 360°
G
(C) nx ° + 270°
16
(E) nx ° + 1,080°
e°
as shown. What is the measure of DCE?
If the average(arithmetic mean) of the
(D) nx ° + 540°
C
A
(C) 42
(B) 74°
11
In the figure, what is the average
F
In the figure, A, B, C, D and E are vertices
of a regular polygon. Two sides AB and
DE are extended to meet at P as shown.
12
In regular pentagon ABCDE, AOD = 30°.
If BPC = 140°, how many sides does the
What is the value of x ?
regular polygon have?
B
(A) 84
(B) 72
(C) 68
(D) 66
(E) 64
(A) 12
A
O
30°
E
(B) 16
C
x°
D
A
B
P
C
D
E
(C) 20
F
(D) 24
(E) 27
5. ANGLES
217
TEST
17
In parallelogram ABCD, A = 11x°,
B = 4x°, C = 6z° and D = 3y°. What is
21
the value of y + z?
A
D
that have integer degrees as their interior
angles?
B
4x°
11x°
3y°
How many regular polygons are there
22
6z° C
(1) In the figure, what is the value of x?
100°
20°
18
30°
In the figure, what is the value of x + y?
B
110°
x°
(2) In the figure, what is the value of
a + b + c + d + e + f?
96°
124°
A
19
x°
a°
y°
c°
b°
C
d°
g°
(1) An equilateral triangle ABC is
f°
inscribed in regular pentagon ADEFG.
e°
If BC || EF, then what is the value of x?
A
D
x°
23
G
E
and DG are parallel and FAE = 3x° and
CDG = x°. What is the value of x?
C
B
In regular pentagon ABCDE, two lines FA
F
F
3x°
A
B
(2) Point P is inside regular octagon
E
ABCDEFGH so that triangle ABP is
equilateral. How many degrees are
D
in angle BCP?
20
A parallelogram sheet ABCD is folded
as shown. If A = 110° and GFE = 25°,
what is the measure of C′HG?
A
110°
x°
G
C′
H
24
C
G
In the figure, what is the sum of the
measures of seven angles A, B, C, D,
E, F, and G?
B
E
A
B
G
C
F
25°
D
F
C
E
D
M T C
218
5. ANGLES
18. In triangle EBH, BHC = 26° +A36° = 62°
In triangle HGC,
x°
HGF = 34° + 62° = 96°
J
B
F
34°+62°
x° + y° + AGD = 180° E 36°
G
26°+36°=62°
I
< x° + y° + 96° = 180°
34°
y° H
C
< x° + y° = 180° 96° = 84°
D
Since x : y = 4 : 3,
12
if x = 4k° and y = 3k°, x° + y° = 7k° = 84°
<k° = 12° < x° = 4k° = 4 s 12° = 48°
19.
(kTwo straight angles)
2a° + 2b° = 360° (x° + y°)
And BEC = ABE = 4x°
ADB = A = 62°.
<3x° + AEB + 4x° = 180°
<AEB + 7x° = 180°
= 180° 124° = 56°.
Since AB d DC,
3x°
D
4x°
E
= 180° 126° = 54°
Ans
C
④ H
= 180° (a° + b° + d ° + f °) + (c° + e°)
= 180° 250° + 100° = 30°
21.
In the figure, DAB = ABC
=
and DAB = CAB
56°
56°
x° 48°
in triangle ADB, x° + y° + 134 = 180° A
x° x°
In triangle ABC,
2x° + 2y° + C = 180°
134° D
<2(x° + y°) + C = 180°
<C = 180° 2(x° + y°)
B
= 180° 92° = 88°
P.211
5★3
Ans
pattern drill
➊ (1) Since a trapezoid is a quadrilateral, the sum
of the interior angles is (4 2) s180°
angles of the pentagon is
= (5 2) s180°
< Triangle CAB is an isosceles triangle.
Ans
(C)
E
120°
D 112°
C
A
100° 108° B
<A + 108° + 100° + 112° + 120° = 3 s 180°
< x° = 180° 2 s48° = 180° 96° = 84°
Ans
88
polygons
A + B + C + D + E
< CAB = CBA
C
y°
y°
= 2 s180° = 360°
B
C
<x° + y° = 180° 134° = 46°
(2) Since the sum of five interior
=
C
30°
D
A
(kAlternate angles)
(kSymmetric angles)
Ans
D
=
In the figure, if BAD = x ° and ABD = y°,
= 180° 2 s 46°
f° F
x°
In triangle FGH,
G
(a° + b° c° + d ° e°) + f ° + x° = 180°
< G = x° = 180° (a° + b° c° + d ° e°) f °
62°
DBC = 180° 2 s 56° = 180° 112° = 68°
Ans 68°
24.
a° b° B
① a°+ b°
An exterior angle is equal
c
°
a°+ b° c °
C
to the sum of the other two
② d°
D
interior angles
③
E e° a°+ b° c °+ d °
a°+ b° c °+ d °e°
B
Since triangle BCD is isosceles
54°
A
62°
63°
BDC = ABD = 56° (kAlternate angles)
<AEB = 180° 7x° = 180° 7 s 18°
20.
A
<ABD = 180° 2 s 62°
B
4x°
2x°
(kAlternate angles)
Ans
23. Since triangle ABD is isosceles,
A
F
D
=
<x° = 18°
E
= 180° 117° = 63°
In right triangle AED, 2x° + 3x° = 90°
<5x° = 90°
54°
y° C
b°b°
x°
B a°
a°
= 360° 126° = 234° < a° + b° = 234° = 117°
2
In triangle BCD, BDC = 180° (a° + b°)
48
In rectangle ABCD, D = 90°
18
A
x° + y° = 180° 54° = 126°
If CBD = a° and BCD = b°,
(2a° + x°) + (2b° + y°) = 360°
26°
Ans
In the figure, if ABC = x° and ACB = y°,
in triangle ABC,
=
In triangle ADG,
22.
84°
< A + 440 = 540 < A = 540° 440° = 100°
Ans (B)
SOLUTIONS· EXPLANATIONS· ANSWERS
103
(3) Since a nonagon is a 9-sided polygon,
(3) Since the sum of the all
the average (arithmetic mean) of the
exterior angles of the
interior angles is [(9 2) s180°] ÷ 9
hexagon is 360°,
20
= 7 s180° = 7 s20° = 140°
9
Ans
B 60°
40°
A 140° C40°
100°
80°
D
(180° 100°) + (180° 140°)
(C)
x°
+ 60° + 40°+ 70° + x° = 360°
F
E
<80° + 40° + 60° + 40°+ 70° + x° = 360°
(4) If the polygon has n sides, the sum of
interior angles is (n 2) s180° = 2,700°,
30
n 2 = 2,700° = 30 = 15
180° 2
2
Ans (B)
<n = 15 + 2 = 17
<290° + x° = 360° < x° = 360° 290° = 70°
Ans (C)
➍ (1) An interior angle in a regular decagon is
180° (360° ÷ 10) = 180° 36° = 144°
Ans
=
is isosceles trapezoid,
A = B and C = D
A
B
D
D = 180° (360° ÷ 5)
C
Since AB || DC, A + D = 180°
(C)
B
(2) In rhombus ABCD, BC || AD.
x°
130°
=
(3) Since the sum of all exterior angles of
D
<x° + 130° = 180°
Ans
< x° = 180° 130° = 50°
AB || DC <A + D = 180°
x°
n-sided polygon is 360°, each exterior angle
(B)
of the polygon is 360° ÷ n = 20°
18
< n = 360° = 18
20°
B
5y°
3x°
(kSum of co-interior angles) D
C
<x° + 3x° = 180° < 4x° = 180° 45 < x° = 45°
Since D = B, 3x° = 3 s 45° = 135° = 5y°
27
< y° = 135° = 27°
Ans (A)
5
P.212
1.
➌ (1) Since the sum of interior angles of
quadrilateral ABCD is 360°, A 60° 100°
60° + 100° + C + 50° = 360°
210° + C = 360°
< The exterior angle of C is
Since the sum of the interior
(5 2) s180° = 540°,
Ans
C
(A)
H
2.
sided polygon is 360°, the average
Ans
F
104
Ans
SOLUTIONS· EXPLANATIONS· ANSWERS
(D)
E
In triangle AEH,
A + E = AHC.
In triangle FDG,
F + D = DGB
A
G
B
D
H
C
= (A + E) + B + C + (D + F)
= AHC + B + C + DGB (kSum of
(arithmetic mean) of exterior angles of an
octagon is 360° ÷ 8 = 45°
U
S
<A + B + C + D + E + F
(2) Since the sum of the exterior angles of any
x° O
E x°
< 2x° = 540° 270° = 270°
135
<x = 270 = 135
2
D
18
angles in a pentagon HOUSE is
B
50°
<C = 360° 210° = 150°
Ans
T R A I N I N G
2 sx° + 3 s90° = 540°
180° 150° = 30°
B
In triangle CDF, AFC = 36° + 72° = 108°
Ans (A)
C
(kSum of co-interior angles)
(3) In parallelogram ABCD, A
36° F 36°+72°
=
Ans
<A + C = 180°
A
36°
36° 72°
36°
In isosceles triangle CDE,
C
D
DCE = 1 (180° 108°) = 1 s72° = 36°
2
2
FDC = 108° 36° = 72°
(kSum of co-interior angles)
<A + B = 180°
E
= 180° 72° = 108°
(D)
A
(2) In regular pentagon ABCDE,
=
➋ (1) Since quadrilateral ABCD
70°
(A)
interior angles of quadrilateral GHCB)
= (4 2) s180° = 360°
Ans
(B)
THEMELTINGZONE
12
(n 2) s180° = 2160° <n 2 = 216
18
<n 2 = 12 <n = 12 + 2 = 14
Ans
A
3x°20°
E
D
E
s180° 36
(B)
=
Ans
= 108° 72° = 36°
7x°100°
7.
C
20
< A = 3x° 20° = 3 s20° 20° = 40°
Since AB || DC, A + D = 180°
(kSum of co-interior angles)
C
(D)
B
A
C
J
I
D
H
E
G
F
since quadrilateral BCDE is an isosceles
<D = 180° 40° = 140°
trapezoid, 2x° + 2 s144 = 360°
1
s140° = 70°
2
Since DE bisects ADC, EDC =
Since EDC + DEB = 180°
<2x° = 360° 288° = 72°
<BED = x° = 72° = 36°
2
(kSum of co-interior angles),
DEB = 180° EDC = 180° 70° = 110°
Ans
Since AF || DE, F + E = 180°
B
1.
F
f°
a°
e°
D
JUMP
U
Q + U + A + D = 360°
+ 90° + (x + 30)° = 360°
d°
(x 20)°
Q (x 40)°
(x +30)°
D
<3x° + 60° = 360°
E
(D)
In quadrilateral QUAD,
<(x 40)° + (x 20)°
b°
C c°
Ans
P.215
(A)
A
(kSum of co-interior angles)
a° + b° + c° + d ° + (e° + f °)
= a° + b° + c° + d ° + 180°
each interior angle is
[(10 2) s180°] ÷ 10
= 8 s180° = 8 s18° = 144°
10
If BED = x°,
<3x° 20° = 7x° 100° < 4x° = 80° <x° = 20°
In hexagon ABCDEF,
In regular decagon ABCDEFGHIJ,
36° B
36°
=
108°
= 3 s36° = 108° and
5
A'
▲AaBE is an isosceles triangle,
D
Aa BE = ABE = 1 s(180° 108°)
2
1
=
s72° = 36°
2
< Aa BG = ABC ABAa= 108° 2 s 36°
B
In parallelogram ABCD, A = C
A
< 3x° = 360° 30° = 300° < x° = 100°
= (6 2) s 180°
Since (x 40)° < (x 20)° < (x + 30)°,
< a + b + c + d = 4 s 180° 180°
the greatest angle in quadrilateral QUAD is
= 720° 180° = 540°
D = x° + 30° = 100° + 30° = 130°
< The average(arithmetic mean) of a, b, c and
d is 540 ÷ 4 = 135
5.
=
= 3
n-sided polygon is (n 2) s180°,
4.
108°
Aa = A = [(5 2) s180°] ÷ 5
2. Since the sum of the interior angles in
3.
A
6. In regular pentagon ABCDE,
=
P.214
Ans
(C)
2.
Ans
(D)
Since the sum of interior angles of a pentagon
is 180° s(5 2) = 180° s3 = 540°,
In the figure, JGK + GJK = KAF + KFA.
a° + b° + c° + d° + e° + f ° = 540° 120° = 420°
< The average (arithmetic mean) of a, b, c, d,
Ans (D)
e and f is 420 ÷ 6 = 70.
The sum of 10 angles of A + B + C + D +
E + F + G + H + I + J
is equal to the sum of interior
and quadrilateral GHIJ.
B
< 180° s(6 2) + 180° s(4 2)
H
G
A
I
K
E
J
F
= 180° s4 + 180° s2 = 720° + 360° = 1080°
Ans (C)
3.
Since an interior angle of the
regular hexagon is
30°
[(6 2) s180°] ÷ 6 = 4 s180°
6
= 4 s30° = 120°, A = 120°
A
120°
=
angles of hexagon ABCDEF
D
C
=
B
C
F
SOLUTIONS· EXPLANATIONS· ANSWERS
D
E
105
Since ▲ CFD is an isosceles triangle,
CDF = 1 (180° 48°) = 1 s132° = 66°
2
2
Since D = CDF + ADF + EDA = 108°,
Since triangle ABF is isosceles,
ABF = 1 s(180° 120°)
2
1
=
s60° = 30° and ABE = 1 s120° = 60°
2
2
< FBE = ABE ABF = 60° 30° = 30°
Ans (E)
4. In the figure, a : b = 4 : 1
If a = 4k°, b = k° <4k° + k° = 180°
a°
Since the sum of exterior angles of
< n = 10
H
< The polygon is a regular decagon.
Ans
5.
A
Since the figure ABCDE is
40°
E 80°
110° + 50° + (360° x°) + 40° + 80°
B
< HGI = 100° 40° = 60°
< AGF = HGI = 60° (kVertical angles)
D
In pentagon CDEFL,
= 540° (kSum of interior angles in a pentagon)
Ans
< CLF = 540° 390° = 150° (kStraight angle)
100
< BLF = 180° 150° = 30°
=
=
=
=
Ans
C
30°
A
F
=
the regular pentagon ABCDE is
B
60°
=
7. In the figure, an interior angle of
< x° + 240° = 360°
=
E
<BAC = BCA = EAD = EDA D
= 1 (180° 150°) = 1 s30° = 15°
2
2
< DAC = A 2BAC = 60° 2 s15°
= 60° 30° = 30°
(kSum of interior angles in a quadrilateral)
A
< ABC = 60° + 90° = 150°
B
[(5 2) s180°] ÷ 5
108° E
36°
= 48° = =
3
s180°
=
= 3 s36° = 108°
36°
5
66°
D
▲ AED is an isosceles triangle. C
1
1
(180° 108°) =
s72° = 36°
<EDA =
2
2
106
SOLUTIONS· EXPLANATIONS· ANSWERS
120
x° + 150° + 30° + 60° = 360°
triangle and quadrilateral BCDE is a square,
ABE = 60° and EBC = 90°.
Ans
In quadrilateral ABLG,
In the figure, since ▲ ABE is an equilateral
triangles.
J
CLF + 80° + 130° + 90° + 90° = CLF + 390°
<x° = 640° 540° = 100°
▲ABC and ▲AED are isosceles
30°
70°
30° + 70° = HGI + 40°. < 100° = HGI + 40°
< 640° x° = 180° s(5 2) = 3 s180° = 540°
Since AB = BC and AE = ED,
E
60°
40°
I
130° D
K
= 280° + (360° x°) = 640° x°
6.
60°
In the figure, since J + K = HGI + H,
C x°
a pentagon, the sum of the
interior angles is
(D)
50°
110°
L C
F 90°
A x°
=
G
(k10-sided polygon)
6°
80°
30°150°
150°
B
any sided polygon is 360°,
360° ÷ n = 36°
Ans
<ADF = 108° 102° = 6°
8.
=b°
<5k° = 180° 36° < k° = 36° < b° = 36°
66° + ADF + 36° = 108° <ADF + 102° = 108°
9.
< x° = 360° 240° =120°
Since the regular octagon ABCDEFGH is
inscribed in a circle, a°, b°, c°, d°, e°, and f ° are
the same. (kInscribed angles)
G = [(8 2) s180°] ÷ 8
C
D
E
d°
3
45°
e°
c°
= 6 s180°
= 3 s45° = 135° B
F
4
8
b°
In isosceles triangle HFG,
f°
1
A
G
45°
a°
f°=
s(180° 135°) =
2
2
H
< a° = b° = c° = d° = e° = f ° = 45°
2
3
45°
< a° + b° + c° + d° + e° + f ° = 6 s
2
= 3 s45° = 135°
Ans 135
P.216
E· X· A· M· ·I N· A· T· ·I O· N
B
1. In quadrilateral ABCD,
a° + b° + c° + d ° = 360° ······①
{ 84° + d ° = 360° ······②
5.
b°
d°
① ②; a° + b° + c° 84° = 0
<a° + b° + c° = 84°
a° 84°
A
< (n 2) s 180 > 2450
< n 2 > 2450 = 13.6······
180
< n > 2 + 13.6······ < n > 15.6······
c°
< The average (arithmetic mean) of
a, b and c is (a + b + c) ÷ 3 = 84 ÷ 3 = 28
2.
(16 2) s 180° 2450° = 14 s 180° 2450°
(D)
= 2520° 2450° = 70°
Ans
Since the sum of interior angles in a
quadrilateral ABCD is 360°,
(3x° + 10°) + (x° 15°) + (4x° 45°) + (2x° + 50°)
= 360°
36
6.
<x° = 36°
<10x° = 360°
64°
F
B 84°
< A = 3 s36° + 10° = 72° + 10° = 82°,
E
B = 36° 15° = 21°,
C = 4 s36° 45° = 144° 45° = 99°,
D
E
< (4x + 6)° + (2x 18)° = 180°
< In quadrilateral ABCD,
(x° + 58°) + 84° + 64° + (78° + y°) = 360°
(k Sum of interior angles in quadrilateral
ABCD)
< x° + y° + 284° = 360°
(2x 18)°
C
< x° + y° = 360° 284° = 76°
32°
67° + BED = 180° < BED = 180° 67° = 113°
Ans (A)
E
A
In the figure,
ABC = EBC = x° D
(k Symmetric angles)
B
x°
<k° = 19°
a° + b° = 3k° + k° = 4k° = 76°
< x° = 32°
Since BE bisects B, ABE = EBC = 1 B
2
= 1 A = 1 (4 s 32 + 6)° = 1 s 134° = 67°
2
2
2
Since ABE + BED = 180°(kCo-interior angles),
< a° + b° = 76°
If a° = 3k° and b° = k°,
< 6x° 12° = 180° < 6x° = 180° + 12° = 192°
4.
78°
G y° D
and GDA = y°, a° + b° = x° + y°.
(B)
=
=
(k Co-interior angles)
< A + C = 180°
(4x +6)°
b°
a°
In the figure, if E = a°, F = b° GAD = x°
= D and
3. In isosceles trapezoid ABCD, C
B
A
A + D = 180°
C
58°x°
and D = 2 s36° + 50° = 72° + 50° = 122°
Ans
< a° = 3k° = 3 s19° = 57°
7.
< E = 57°
Ans
In pentagon ABCDE, the sum
of the interior angles is
A
7x° B
4x° + 7x° + 6x° + 3x° + (360° 5x°) 5x°
E
D
3x°
= (5 2) s180°
y
y
C
12
< 15x° = 540° 360° = 180°
F
40°
G
If FCB = DCB = y°(k Symmetric angles),
y° + y° + 40° = 180°(k Straight angle)
< 2y° = 180° 40° = 140° 70° < y° = 70°
6x°
Ans
< x = 12°
Ans
C
(A)
8. Since the sum of the exterior angles of any
sided regular polygon is 360°, an exterior
angle of a regular polygon is a factor of 360°
In trapezoid ABCD,
x° + y° = 180° (k Co-interior angles)
<x° + 70° = 180° < x = 180° 70° =110°
(D)
4x°
< 15x° + 360° = 3 s180° = 540°
x°
(B)
A
<122° (= D) is the greatest angle in the
quadrilateral ABCD.
< n = 16
< The remaining angle of the polygon is
C
Ans
If the convex polygon has n sides, the sum of
all interior angles is (n 2) s 180°.
(A) 18° = 360° ÷ 20
(B) 24° = 360° ÷ 15
(C) 32° ≠ 360° ÷ n
(D) 45° = 360° ÷ 8
(E) 72° = 360° ÷ 5
Ans
(D)
SOLUTIONS· EXPLANATIONS· ANSWERS
107
(C)
9.
13.
A
a°
In the figure, two triangles OAB
and OBC are isoscelces triangles.
E
A
I
=
if OAB = OBA = a°
In triangle AHI, HIG = A + (40° + E)
a° B
b°
=
In quadrilateral OABC,
In triangle EFH, IHA = 40° + E°
O
2(a° + b°) + 62° = 360°
A+40°+E
G 40°
b° C
=
and OBC = OCB = b°,
62°
D
(k Sum of interior angles in a quadrilateral)
40°+E
H
B
F
< 2(a° + b°) = 360° 62° = 298°
C
< a° + b° = 298° ÷ 2 = 149°
Ans
< ABC = a° + b° = 149°
In quadrilateral BCDI,
(D)
B + C + D + (A + 40° + E) = 360°
< B + C + D + A + E = 360° 40°
= 320°
In the figure, ▲FCD is an equilateral triangle.
< The average (arithmetic mean) of
<FCD = 60° and DC = FC = BC.
A, B, C, D and E is
A
54°
< ▲ BCF is an isosceles triangle.
Since C = [(5 2) s180°] ÷ 5
36
E
= 3 s180° = 3 s36° = 108°,
5
FCB = 108° 60° = 48°
42° B
66°
F
=
(A + B + C + D + E) ÷ 5 = 320° ÷ 5
= 64°
=
=
14.
Since the sum of the interior angles of a
= 360° 90° = 270°
E
Ans
A
I
= 4 s180° = 4 s30° = 120° H
6
<CDE = 360° (108° + 120°)
= 360° 228° = 132° (kPerigon) G
F
pentagon ABCDE is [(5 2) s180°] ÷ 5
36°
= 3 s180° = 3 s36° = 108°
5
In triangle OBF, BFO = 180° (30° + 108°)
< x ° = 108° 42° = 66°
P
16.
B
A
Ans
SOLUTIONS· EXPLANATIONS· ANSWERS
(D)
=
108°
120°
C
D
=
E
F
Since CD = DE, ▲DEC is an isosceles triangle.
DCE = 1 s(180° 132°) = 1 s48° = 24°
2
2
Ans (C)
= 180° 138° = 42°
In triangle CDF, x ° + 42° = 108°
(D)
=
D
x° 42°
e°
36°
IDC = [(5 2) s180°] ÷ 5 = 3 s180°
5
B
= 3 s36° = 108°
30°
Since B is an interior angle of a regular
108
D
<IDE = [(6 2) s180°] ÷ 6
108°
C
108°
E
a°
In regular pentagon ABCDI,
In regular hexagon IDEFGH,
B
30°
c°
d°
The average(arithmetic mean) F
of a, b, c, d and e is 270 ÷ 5 = 54
15.
C
=
n-sided polygon is n sx°, the sum of the
interior angles of a (n + 3)-sided polygon is
n sx° + 3 s180° = nx° + 540°
Ans (D)
O
B b°
a° + b° + c° + d ° + e° + 90° = 360°
A
<a° + b° + c° + d ° + e°
AEB = 180° ( 1 s108° + 42°)
2
= 180° (54° + 42°) = 180° 96° = 84°
Ans (D)
A
Since the sum of all exterior angles
of the polygon is 360°,
In triangle ABF,
12.
(B)
=
48°
60°
C
D
In isosceles triangle CBF,
1
1
CBF =
(180° 48°) =
s132° = 66°
2
2
< ABF = 108° 66°= 42°
11.
Ans
=
10.
140°
C
x°
x° D
180°x°
180°x°
E
In the figure, if PDC = PBC = x°,
CDE = 180° x° (k exterior angle)
Ans
H
B
60° 60°
= 75°
60°
=
=
= 27
A
(2)
=
In quadrilateral BPDC, x° + 140° + x°
= 180° x° < 3x° = 40° < x° = 40°
3
Since an exterior angle of the regular
polygon is 40° , the number of sides of the
3
polygon is 360° ÷ 40° = 360° s 3 = 9 s 3
3
40°
C
P
D
G
E
F
(E)
In the figure, since triangle ABP is
equilateral, ABP = 60°
A
17.
D 3y°
Since AD
4x°
11x°
In regular octagon ABCDEFGH,
3
45
B = [(8 2) s180°] ÷ 8 = 6 s 180°
84
= 3 s 45° = 135°
B
6z° C
< PBC = B ABP = 135° 60° = 75°
|| BC, A + B = 180°
(kSum of co-interior angles)
Since BP = BC, ▲ BCP is an isosceles
triangle, BCP = 1 (180° 75°)
2
1
=
s 105° = 52.5°
2
Ans (1) 48 (2) 52.5°
12
<11x° + 4x° = 180° < 15x° = 180° <x° = 12°
Since B = D, 3y° = 4x° = 4 s12° = 48°
< y° = 48° = 16°
3
Since A = C, 6z° = 11x° = 11 s12° = 132°
< z° = 132° = 22° < y° + z° = 16° + 22° = 38°
6
Ans 38
18. In quadrilateral DEFG,
E
if DEF = a° and EFG = b°,
a°
a° + b° + 124° + 96° = 360°
< a° + b° + 220° = 360°
B
x°
110°
F
G
25°
25°
C′
H
70°B
45° E
45°
110° C
In the figure,
124° b°
G
A
A
D
D 96°
(kSum of the interior
angles in quadrilateral)
20.
EFC = GFE = 25° (kSymmetric angles)
y°
Since C = A = 110°, in triangle FEC,
C
F
FEC = 180° (25° + 110°) = 45°
< a° + b° = 360° 220° = 140° ······ ①
Since C′EF = CEF = 45°(kSymmetric angles),
Since a° + x° = 180 and b° + y° = 180°
HEB = 180° 2 s 45° = 180° 90° = 90°
(kStraight angles).
(kStraight angle)
< a° + b° + x° + y° = 360° ······ ②
Since AD || BC, B = 180° 110° = 70°
① m ②; 140° + x° + y° = 360°
< x° + y° = 360° 140°= 220°
Ans
(kCo-interior angles)
220
In triangle HBE, BHE = 180° (90° + 70°)
= 180° 160° = 20°
< C′HG = BHE = 20° (kVertical angles)
Ans 20°
19. (1) In regular pentagon AGFED,
3 s 180°36
F = [(5 2) s180°] ÷ 5 =
5
= 3 s 36 = 108°
A
Since BC || EF,
GCB = F = 108°
D
(kCorresponding angles) B
x° = 108° 60° = 48°
E
60°
108°
H
x°
C
F
G
21.
If an interior angle in a polygon has an
integer degree, its exterior angle has an
integer degree too. Since the sum of the
exterior angles in any sided polygon is 360°
and the positive factors of 360 are 1, 2, 3, 4,
······, 180 and 360, each exterior angle of a
SOLUTIONS· EXPLANATIONS· ANSWERS
109
regular polygon could be 1°, 2°, 3°, 4°, ······,
120° (kExcept 180° and 360°).
A
24.
a°
B
H °
b
b°+d°
I a°+b°+d°
G
Since the number of positive factors of 360
is (3 + 1)(2 + 1)(1 + 1) = 4 s 3 s 2 = 24
3
2
[k360 = 2 ·3 ·5], the number of possible
c° C
exterior angles of regular polygons is 24 2
F
= 22.
< 22 regular polygons have integer degrees
as their interior angles.
22.
Ans
22
(1) In triangle ABC, B = 180° (20° + 30°)
B
= 180° 50° = 130°
D
20° 100° 130°x° E
30°
130° + x° + 110° + 90 + 100°
F
C
110°
= (5 2) s 180°
H
G
In pentagon BEGHD, A
< 430° + x° = 3 s 180° = 540
A
GBA = c° + g°
In pentagon ABDEF,
G
c°+g° B
a° b°
J
g°
H
a° + [(c° + g°) + b°] + d ° + e° + f ° F
c° C
I
d° D
f°
e° E
is the sum of all interior angles.
< a° + b° + c° + d ° + e° + f ° + g° = (5 2) s 180°
= 3 s 180° = 540°
Ans
F
23.
D
In the figure, the sum of interior angles of
two triangles ADF and BEG is 2 s180° = 360°
In triangle BDI, BIA = b° + d °
In triangle IAH, IHC = a° + (b° + d °)
In triangle HCE, (a° + b° + d °) + e° + c ° = 180°
< A + B + C + D + E + F + G is the
sum of interior angles of the three triangles
ADF, BEG and HCE = 3 s180° = 540°
540°
6 ★ 1 perimeter of polygons
P.223
pattern drill
➊ (1) The perimeter of pentagon ABCDE is
AB + BC + CD + DE + EA
Ans
= 3 + 4 + 5 + 2 + 4 = 18
(C)
(2) In the figure, CD + EF = AB and BC + DE = AF
3
B
AB + BC + CD + DE + EF + AF
D
= AB + (BC + DE) + (CD + EF) + AF 4
B
E 108°
108°
D
< The perimeter of the polygon is A
3x°
H
(1) 110 (2) 540
A
72°
3x°
d°
Ans
< x° = 540° 430° = 110°
(2) In triangle CGB,
E
e°
= AB + AF + AB + AF = 2(AB + AF)
x°
= 2(3 + 4) = 2 s7 = 14
C
G
Since FA || DG, AHD = FAH = 3x°
F
E
Ans
(B)
(3) Since the sum of lengths of 9 sides is 54,
(k Alternate angles)
the average(arithmetic mean) of the lengths
In regular pentagon ABCDE, AED = EDC
36°
= [(5 2) s180°] ÷ 5= 3 s180° = 3 s36°
5
= 108°
of 9 sides is 54 ÷ 9 = 6
<HED = 180° 108° = 72°
In triangle EDH, EDG = 3x° + 72°
< 108° + x° = 3x° + 72°
110
< x° = 18°
SOLUTIONS· EXPLANATIONS· ANSWERS
(B)
➋ (1) Since the 3 sides of an equivalent triangle
are the same, each side is
2 1 feet ÷ 3 = 5 feet ÷ 3 = 5 feet
2
6
2
2
Ans (A)
= 5 s12 inches = 10 inches
6
(k Straight angles)
< 2x° = 108° 72° = 36°
Ans
Ans
18°
C