1 problem 1 problem 3 problem 2 problem 4 problem 5 problem 8problem 7 problem 6 standard 13...
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1
PROBLEM 1PROBLEM 3
PROBLEM 2PROBLEM 4
PROBLEM 5
PROBLEM 8PROBLEM 7PROBLEM 6
STANDARD 13
SUPPLEMENT AND COMPLEMENT: NUMERIC
PROBLEM 10PROBLEM 9
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SUPPLEMENT: GENERAL
COMPLEMENT: GENERAL
SUPPLEMENT AND COMPLEMENT: GENERAL
WORD PROBLEMS:
SUPPLEMENTARY AND COMPLEMENTARY ANGLES
END SHOW
2
STANDARD 7:
Students prove and use theorems involving the properties of parallel lines cut by a transversal, the properties of quadrilaterals, and properties of circles.
ESTÁNDAR 7:
Los estudiantes prueban y usan teoremas involucrando las propiedades de líneas paralelas cortadas por una transversal, las propiedades de cuadriláteros, y las propiedades de círculos.
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3
STANDARD 13
Find the measure of the COMPLEMENT and SUPPLEMENT of the angle below:
90° – 65° = 25°65°
115°
Supplement:
180° – 65° = 115°
Calculating the complement:
Calculating the supplement:
25°
Complement
65°
65°
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4
Find the measure of the COMPLEMENT and SUPPLEMENT of the angle below:
90° – 55° = 35°55°
125°
Supplement:
180° – 55° = 125°
Calculating the complement:
Calculating the supplement:
STANDARD 13
35°
Complement
55°
55°
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5
X°
Find the measure of the SUPPLEMENT of the angle below:
180° – X°
(180 – X)°
Supplement:
STANDARD 13
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6
Find the measure of the COMPLEMENT of the angle below:
90 – X°
Complement
(90-X)°
X°
STANDARD 13
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7
Find COMPLEMENT AND SUPPLEMENT of any given angle:
90° – X°
(180 – X)°
Supplement:
180° – X°
Calculating the complement:
Calculating the supplement:
(90 – X)°
Complement
X°
X°
Let’s call our angle X:
X°
STANDARD 13
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8
X is the angle
180 – X is the supplement
180 – X = X – 20
The supplement of an angle is twenty less than the angle, find this angle and the supplement.
+ X + X
180 = 2X – 20 +20 +20
200 = 2X2 2
X = 100 Then the supplement:
180 – X = 180 – 100
= 80
The angle is 100° and the supplement is 80°.
X180 –X
STANDARD 13
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9
X is the angle
90 – X is the complement
90 – X = 2X
The complement of an angle is two times the angle, find this angle and the complement.
+ X + X
3 3
X = 30 Then the complement:
90 – X = 90 – 30
= 60
The angle is 30° and the complement is 60°.
90 = 3X
X
90 - X
STANDARD 13
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10
X is the angle
90 – X is the complement
90 – X = X + 10
The complement of an angle is 10 more than the angle, find this angle and the complement.
+X +X
90 = 2X + 10 -10 -10
80 = 2X2 2
X =40 Then the complement:
90 – X = 90 – 40
= 50
The angle is 40° and the complement is 50°.
X
90 - X
STANDARD 13
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11
X is the angle
180 – X is the supplement
180 – X =
The supplement of an angle is a third of the angle, find this angle and the supplement.
540 = 4X4 4X = 135
Then the supplement:
180 – X = 180 – 135
= 45
The angle is 135° and the supplement is 45°.
X180 –X
X13
180 – X = X13
33
540 – 3X = X
+3X +3X
STANDARD 13
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12
X is the angle
180 – X is the supplement
180 – X =
The supplement of an angle is a fourth of the angle, find this angle and the supplement.
720 = 5X5 5X = 144
Then the supplement:
180 – X = 180 – 144
= 36
The angle is 144° and the supplement is 36°.
X180 –X
X14
180 – X = X14
44
720 – 4X = X
+4X +4X
STANDARD 13
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13
X is the angle
90 – X is the complement
90 – X = 3X – 10
The complement of an angle is 10 less than three times the angle, find this angle and the complement.
+X +X
90 = 4X – 10 +10 +10
100 = 4X4 4
X = 25 Then the complement:
90 – X = 90 – 25
= 65
The angle is 25° and the complement is 65°.
X
90 - X
STANDARD 13
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14
X is the angle
180 – X is the supplement
180 – X = 3X + 40
The supplement of an angle is forty more than the triple of the angle, find this angle and the supplement.
+ X + X
180 = 4X + 40 -40 -40
140 = 4X4 4
X = 35 Then the supplement:
180 – X = 180 – 35
= 145
The angle is 35° and the supplement is 145°.
X180 –X
STANDARD 13
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15
X is the angle
90 – X is the complement
90 – X = 2X – 30
The complement of an angle is thirty less than twice the angle, find this angle and the complement.
+X +X
90 = 3X – 30 +30 +30
120 = 3X3 3
X = 40 Then the complement:
90 – X = 90 – 40
= 50
The angle is 40° and the complement is 50°.
X
90 - X
STANDARD 13
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16
X is the angle
180 – X is the supplement
5(180 – X) = (90 – X) + 630
Five times the supplement of an angle is 630 more than its complement. Find the angle, the supplement and the complement.
+5X +5X
900 = 4X + 720 -720 -720
180 = 4X4 4
X = 45
Then the supplement:
180 – X = 180 – 45
= 135
The angle is 45°, the supplement is 135° and the complement is 45°.
900 – 5X = 720 – X
Then the complement:
90 – X = 90 – 45
= 45
STANDARD 13
90 – X is the complementX180 –X
X
90 - X
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17
(180 – X) + 20 = 3(90 – X)
Twenty more than the supplement of an angle is three times its complement. Find the angle, the supplement and the complement.
+X +X
200 = -2X + 270 -270 -270
-70 = -2X-2 -2
X = 35
Then the supplement:
180 – X = 180 – 35
= 145
The angle is 35°, the supplement is 145° and the complement is 55°.
200 –X = 270 – 3X
Then the complement:
90 – X = 90 – 35
= 55
STANDARD 13
X is the angle
180 – X is the supplement
90 – X is the complementX180 –X
X
90 - X
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