# SOLUTION: The bottom on Johns rectangular box is 3 inches longer than it is wide. The diagonal is 15 inches. What is the area of the bottom of the box? Thanks

Algebra ->  Systems-of-equations -> SOLUTION: The bottom on Johns rectangular box is 3 inches longer than it is wide. The diagonal is 15 inches. What is the area of the bottom of the box? Thanks      Log On

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 Question 65616: The bottom on Johns rectangular box is 3 inches longer than it is wide. The diagonal is 15 inches. What is the area of the bottom of the box? ThanksAnswer by ptaylor(2058)   (Show Source): You can put this solution on YOUR website!Letx=width Then x+3=length Using the Pythagorean theorm, we have: x^2+(x+3)^2=15^2 simplifying we get x^2+x^2+6x+9=225 2x^2+6x+9-225=0 2x^2+6x-216=0 using the quadratic formula we get: x=(-b+or-sqrt(b^2-4ac))/2a x=(-6+or-sqrt(36+1728))/4 x=(-6+or-sqrt(1764))/4 x=(-6+or-42)/4 x=(-6+42)/4 = 36/4=9 x=(-6-42)/4=-48/4=-12 we'll discount the negative value for x Therefore the width is 9 inches and the length is 12 inches AREA OF BOTTOM =(l)(w)=9x12=108 sq inches Ck 9^2+12^2=15^2 81+144=225 225=225 also x=9 x+3=12 Hope this helps----ptaylor