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

Click here to see ALL problems on Systems-of-equations

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?

Thanks
Answer by ptaylor(2198) (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