anonymous
  • anonymous
the length of a rectangle box is one inch greater than its width. the height is three times the length. the diagonal box exceeds the height by one inch. Find the volume of the box.
Mathematics
chestercat
  • chestercat
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Mertsj
  • Mertsj
|dw:1328539986535:dw|
Mertsj
  • Mertsj
\[AB=\sqrt{w^2+(w+1)^2}=\sqrt{2w^2+2w+1)}\]
Mertsj
  • Mertsj
\[AC=\sqrt{2w^2+2w+1+9(w+1)^2}=\sqrt{11w^2+20w+10}\]

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Mertsj
  • Mertsj
Now AC is one more than the height which is 3w+3 so AC = 3w+4
Mertsj
  • Mertsj
\[\sqrt{11w^2+20w+10}=3w+4\]
Mertsj
  • Mertsj
Let's square both sides.
anonymous
  • anonymous
now im really confused lol
Mertsj
  • Mertsj
What about?
anonymous
  • anonymous
uhhh everything lol
anonymous
  • anonymous
like idk what the main answer is lol
Mertsj
  • Mertsj
That might be because we haven't found the answer yet. We have to find w based on the information they gave us. Then we will be able to find the dimensions of the box and then the volume.
Mertsj
  • Mertsj
As i said, let's square both sides of the equation.
Mertsj
  • Mertsj
\[11w^2+20w+10=9w^2+24w+16\]
anonymous
  • anonymous
is 16 the square root ?
Mertsj
  • Mertsj
\[2w^2-4w-6=0\]
Mertsj
  • Mertsj
Divide by 2
Mertsj
  • Mertsj
\[w^2-2w-3=0\]
Mertsj
  • Mertsj
Factor
Mertsj
  • Mertsj
\[(w-3)(w+1)=0\]
Mertsj
  • Mertsj
\[w=3 or w=-1\]
Mertsj
  • Mertsj
Discard -1 since the width cannot be negative. So w = 3. That means : |dw:1328541450706:dw|
Mertsj
  • Mertsj
so the box is 3 by 4 by 12 and the volume is 3(4)(12) or 144 cubic inches

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