Question 250148
<font face="Garamond" size="+2">


I presume by the phrase "has given" you mean "is given."  I also have to presume that the room is in the shape of a right rectangular prism. (otherwise the problem cannot be solved with the information given).


Let *[tex \Large x] represent the length of the floor.  Perforce *[tex \Large x] is also the horizontal dimension of one pair of walls. Let *[tex \Large y] represent the width of the floor, and then must also represent the horizontal dimension of the other set of walls.  Let *[tex \Large z] represent the height of the room which is the vertical dimension of all four walls.


Let *[tex \Large A_1] represent the area of the first pair of walls.  Let *[tex \Large A_2] represent the area of the other pair of walls.  And let *[tex \Large A_3] represent the area of the floor.  Since each of these surfaces is a rectangle, we know the area is calculated by multiplying the length times the width, so:


1.*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_1\ =\ xz]


2.*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_2\ =\ yz]


3.*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_3\ =\ xy]


Solve 1 for *[tex \Large x] and solve 2 for *[tex \Large z]


4.*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{A_1}{z}]


5.*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \frac{A_2}{y}]


Substitute the RHS expression from 5 into 4:


6.*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{A_1}{\frac{A_2}{y}}]


Substitute the RHS expression from 6 into 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_3\ =\ \left(\frac{A_1}{\frac{A_2}{y}}\right)y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_3\ =\ \left(\frac{A_1y}{A_2}\right)y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_3\ =\ \left(\frac{A_1y^2}{A_2}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ =\ \frac{A_3A_2}{A_1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sqrt{\frac{A_3A_2}{A_1}}]


From here using the value of *[tex \Large y] you just determined, you can calculate *[tex \Large z] using 5, and then using the value of *[tex \Large z], you can calculate *[tex \Large x] using 4.


And please take a class in English before you post again.  Your question was nearly undecipherable gibberish.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>