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Find the number of sides of each of the two polygons if the total number of sides of the polygons is 13, and the sum of the number of diagonals of the polygon is 25.

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.

x
² - 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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