SOLUTION: I have a large rectangle with a smaller rectangle inside. The smaller rectangle is exactly in the center of the larger 1. The smaller rectangle's width is 20 and length is 40. The

Question 41209: I have a large rectangle with a smaller rectangle inside. The smaller rectangle is exactly in the center of the larger 1. The smaller rectangle's width is 20 and length is 40. The space between the Big and Small rectangle is the deck. The smaller rectangle is the pool.
What is the area of the deck?
What is the combined area of the deck and the pool?
---i tried solving it like this---
The length of the large rectangle is 20+2d
The width of the large rectangle is 40+2d
A=(20+2d)(40+2d)
800+40d+80d+4d^2
800+120d+4d^2
4d^2+120d+800 <------is this the right answer?
Thanks
Answer by psbhowmick(878) (Show Source): You can put this solution on YOUR website!
What you have found out is the combined area of the deck and the pool.
The area of only deck = (combined area of the deck and the pool) - (area of pool)
=
=

Note this is true only when the length of the two rectangles are parallel.
This means that the deck is of constant width 'd'.
Else, if the length (longer side) of the outer rectangle is not parallel to the length (longer side) of the inner rectangle for then you cannot write
"The length of the large rectangle is 20+2d
The width of the large rectangle is 40+2d"

Understand?
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