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Sagot :
x + y = 13
y = 13 - x
Representation:
x = total number of sides of first polygon
13 - x = total number of sides of second polygon.
Number of diagonals in a polygon:
[tex] \frac{n(n-3)}{2} [/tex]
where n = number of sides of a polygon
Number of diagonals in first polygon with x sides.
Substitute x for n:
= x (x-3)
2
= x² - 3x
2
Number of diagonals in second polygon with 13-x sides.
Substitute 13 - x for n:
= 13 - x (13 - x - 3)
2
= 13 - x (10 - x)
2
= 130 -13x - 10x + x²
2
= x² - 23x + 130
2
The sum of diagonals of the two polygons is 25.
x² - 3x + x² - 23x + 130 = 25
2 2
x² + x² -3x - 23x + 130 = 25
2
2 (2x² - 26x + 130 = 25) 2
2
2x² - 26x + 130 = 50
2x² - 26x + 130 - 50 = 0
2x² - 26x + 80 = 0
Factor out 2 (GCF)
2 (x² - 13x + 40) = 0
Factor the quadratic equation, then solve for the roots (x):
x² - 13x + 40 = 0
(x - 8) (x - 5) = 0
x - 8 = 0 x - 5 = 0
x = 8 x = 5
The number of sides of each polygon:
First polygon:x ⇒ x = 8 sides (octagon)
Second polygon: 13 - x ⇒ 13 - 8 = 5 sides (pentagon)
To check if 8 and 5 sides are correct:
The sum of the sides of the two polygons is 13
8 + 5 = 13
13 = 13
The sum of the diagonals of the two polygons is 25:
Diagonals of first polygon with 8 sides, where n = sides:
n (n-3) = 8 (8-3) = 8(5) = 40 = 20 diagonals
2 2 2 2
Diagonals of the second polygon with 5 sides, where n = sides:
n(n-3) = 5 (5-3) = 5(2) = 10 = 5 diagonals
2 2 2 2
Add the diagonals:
20 + 5 = 25
25 = 25
FINAL ANSWER: The number of sides of the two polygons are 8 and 5 sides.
y = 13 - x
Representation:
x = total number of sides of first polygon
13 - x = total number of sides of second polygon.
Number of diagonals in a polygon:
[tex] \frac{n(n-3)}{2} [/tex]
where n = number of sides of a polygon
Number of diagonals in first polygon with x sides.
Substitute x for n:
= x (x-3)
2
= x² - 3x
2
Number of diagonals in second polygon with 13-x sides.
Substitute 13 - x for n:
= 13 - x (13 - x - 3)
2
= 13 - x (10 - x)
2
= 130 -13x - 10x + x²
2
= x² - 23x + 130
2
The sum of diagonals of the two polygons is 25.
x² - 3x + x² - 23x + 130 = 25
2 2
x² + x² -3x - 23x + 130 = 25
2
2 (2x² - 26x + 130 = 25) 2
2
2x² - 26x + 130 = 50
2x² - 26x + 130 - 50 = 0
2x² - 26x + 80 = 0
Factor out 2 (GCF)
2 (x² - 13x + 40) = 0
Factor the quadratic equation, then solve for the roots (x):
x² - 13x + 40 = 0
(x - 8) (x - 5) = 0
x - 8 = 0 x - 5 = 0
x = 8 x = 5
The number of sides of each polygon:
First polygon:x ⇒ x = 8 sides (octagon)
Second polygon: 13 - x ⇒ 13 - 8 = 5 sides (pentagon)
To check if 8 and 5 sides are correct:
The sum of the sides of the two polygons is 13
8 + 5 = 13
13 = 13
The sum of the diagonals of the two polygons is 25:
Diagonals of first polygon with 8 sides, where n = sides:
n (n-3) = 8 (8-3) = 8(5) = 40 = 20 diagonals
2 2 2 2
Diagonals of the second polygon with 5 sides, where n = sides:
n(n-3) = 5 (5-3) = 5(2) = 10 = 5 diagonals
2 2 2 2
Add the diagonals:
20 + 5 = 25
25 = 25
FINAL ANSWER: The number of sides of the two polygons are 8 and 5 sides.
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