How do I solve this?

Answers

Answer 1

Answer: area=LW
perimiter=2(L+W)

aera=36
P=25

36=LW
25=2(L+W)

25=2(L+W)
divide both sides by 2
12.5=L+W
minus W
12.5-W=L

sub for L
36=W(12.5-W)
36=12.5W-W^2
minus (12.5W-W^2) both sides
0=W^2-12.5W+36
use quadratic formula

if you have
ax^2+bx+c=0
x= $\frac{-b+/- \sqrt{b^(2)-4ac} }{2a}$

a=1
b=-12.5
c=36

W= $\frac{-(-12.5)+/- \sqrt{(-12.5)^(2)-4(1)(36)} }{2(1)}$
W= $(12.5+/- √(156.25-144) )/(2)$
W= $(12.5+/- √(12.25) )/(2)$
aprox
W=8 or 4.5

sub
12.5-W=L

12.5-8=L=4.5
12.5-4.5=L=8
either way

the dimentions are 4.5cm by 8 cm

Answer 2

Answer: $\sf\nP=2L+2W=25 \sf\nA=LW=36\n\nFind\ the\ value\ of\ one\ of\ the\ variables\ in\ terms\ of\ the\ other. \sf\n36=LW\nW= (36)/(L)\n\nSubstitute.\nP=25\n=2L+2W\n=2L+2( (36)/(L) )\n=2L+ (72)/(L) \n\nMake\ them\ have\ a\ common\ denominator.\n2L+(72)/(L)\n= (2L)/(1) +(72)/(L)\n=(2L^2)/(L)+ (72)/(L) \n= (2L^2+72)/(L) \n\nMultiply\ L\ on\ the\ other\ side.\n25L=2L^2+72\n0=2L^2-25L+72\n\nUse\ the\ quadratic\ formula.\n L=\frac{-b+/- \sqrt{b^(2)-4ac} }{2a}\nax^2+bx+c$
$a=2\nb=-25\nc=72\n\n L=\frac{-(-25) +/- \sqrt{(-25)^(2)-4(2)(72)} }{2(2)} \n =(25+/- √(625-576) )/(4) \n =(25+/- √(49) )/(4)\n =(25(+/-)7)/(4) \n\n\sf\ We\ now\ have\ two\ options.\n(1) L=(25+7)/(4)= (32)/(4) =8\n(2)L=(25-7)/(4)= (18)/(4) =4.5\n\n\sf\ Either\ W\ is\ 8\ and\ L\ is\ 4.5\ or\ W\ is\ 4.5\ and\ L\ is\ 8.\ It\ doesn't\ matter.\n\n{\boxed{The\ dimensions\ are\ 8\ cm\ by\ 4.5\ cm.}$

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