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A rectangular piece of metal is 10 in longer than it is wide. Squares with sides 2 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 918 incubed3​, what were the original dimensions of the piece of​ metal?

Answer :

Kberriosreyes

Answer:

The dimensions of the piece of metal can be represented by x and x+10.

 

Now, 2 in squares are being cut out of each corner.

 

So the new dimensions are x-4 (2in from each side) and x+10-4 or x+6.

 

When you fold it up, the height becomes 2 and the base has dimensions x-4 and x+6. Now plug this into the volume formula.

 

V=l*w*h

 

1302 = (x+6)(x-4)(2)

651   = x2+2x-24

0      = x2 + 2x-675

0      = (x+27)(x-25)

x=-27 reject since lengths cannot be negative or x=25.

 

So your original dimension for the piece of metal are 25 by 35.  

 

Hope this helped.

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