Al den is 12 feet longer than it is wide. If the den's area is 405 square feet, what are the dimensions of the room? (_,_)?

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Let, width of the room be x feet

Then its length = (x+12) feet

According to the given problem:

den's area = 405 square feet

`rArr x*(x+12)=405`

`rArrx^2+12x=405`

`rArrx^2+12x-405=0`

`rArrx=(-12+-sqrt(12^2-4*1*-405))/2`

`rArrx=(-12+-sqrt(144+1620))/2`

`=(-12+-sqrt(1764))/2`

`=(-12+-42)/2` =30/2=15 or -54/2=-27

Since dimensions cannot be negative, we get x=15

width=15 feet and length=(15+12) or 27 feet.

**Therefore, the dimensions of the room are width=15 feet, length=27 feet**

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