The width of a room is $4$ feet shorter than its length, and its height is $3$ feet less than its length. The area of $4$ walls is larger than the sum of the areas of the floor and ceiling by $134$ square feet. What is the length of the room?
This is what i got from the question
I don't get about the area thing? how would I solve this problem?
On
The first sentence of the problem translates mathematically to \begin{align*} w &= l-4 & h&=l-3\tag{1} \end{align*} The second sentence translates to $$ 2\,lh+2\,wh=134+2\,lw\tag{2} $$ Dividing equation $(2)$ through by $2$ gives $$ \frac{2\,lh+2\,wh}{2}=\frac{134+2\,lw}{2} \implies lh+wh=67+lw\tag{3} $$ Substituting the relations in $(1)$ into equation $(3)$ gives $$ l(l-3)+(l-4)(l-3)=67+l(l-4)\tag{4} $$ Simplifying $(4)$ gives $$ l^2-6\,l-55=0\tag{5} $$ Can you solve equation $(5)$ for $l$?
So you know that the Area of the Ceiling and Floor together is $$A_{f,c} = 2L(L-4)$$ you also know that the sum of the other four walls is $$A_w = 2L(L-3) + 2(L-4)(L-3)$$ and by the constraint you laid out we have that $$A_w-A_{f,c}=134$$ so then we have to solve the quadratic $$2L(L-3)+2(L-4)(L-3)-2L(L-4)-134=0$$ which can be simplified down to $$2L^2-6L +2(L^2-7L+12)-2L^2+8L-134=2L^2-12L-110=0$$ so you just need to solve $$L^2-6L-55=0$$ and that factors to $$(L-11)(L+5)=0$$ and you get that $$L=11$$ because length can't be negative.