SOLUTION: The perimeter of a quadrilateral is 160, the sides are (4m^2+1), (3m-4), (2m+6), (4m^2+2m+1)
Please help, thanks in advance!
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Question 789771: The perimeter of a quadrilateral is 160, the sides are (4m^2+1), (3m-4), (2m+6), (4m^2+2m+1)
Please help, thanks in advance!
Found 2 solutions by mananth, stanbon:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
(4m^2+1), (3m-4), (2m+6), (4m^2+2m+1)
(4m^2+1)+(3m-4)+ (2m+6)+ (4m^2+2m+1)=160
add up like terms
4m^2+4m^2+3m+2m+2m+1-4+6+1=160
8m^2+7m+4=160
8m^2+7m-156=0
a= 8 , b= 7 , c= -156
b^2-4ac= 49 + 4992
b^2-4ac= 5041
x1=( -7 + 71 )/ 16
x1= 4.00
x2=( -7 -71 ) / 16
x2= -4.88
the sides cannot be negative
so x=4
Plug 4 in all the sides to get the lengths
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The perimeter of a quadrilateral is 160, the sides are
(4m^2+1), (3m-4), (2m+6), (4m^2+2m+1)
-----
Add those to get 8m^2+ 7m + 4 = 160
-----
8m^2 +7m - 156 = 0
m = 4
--------------
4m^2+1 = 65
3m-4 = 8
2m+6 = 14
4m^2+2m+1 = 73
==================
Cheers,
Stan H.
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