Question

A rectangular room is 3 meters longer than it is wide, and its perimeter is 30 meters. Find the dimension ofthe room.The length is :meters and the width ismeters.

Answer 1

We have a rectangular room with the following dimensions:

`\begin{gathered} \text{Length = L} \\ \text{Side = w} \end{gathered}`

The dimensions of the rectangular room is related to each other by the following statement:

" 3 meters longer than it is wide "

We will go ahead an decrypt the above statement and express it into a mathematical form in terms of the dimensions ( defined above ) of the rectangular room:

`L\text{ = w + 3 }\ldots\text{ }\text{\textcolor{#FF7968}{Eq 1}}`

The next statement relates the perimeter ( P ) of the rectangular room in terms of its previously defined dimensions.

" perimeter is 30 meters "

We will go ahead an decrypt the above statement and express it into a mathematical form in terms of the dimensions ( defined above ) of the rectangular room:

`P\text{ = 30 }`

The perimeter of a rectangle ( P ) is defined as the sum of all the boundary sides which can be expressed in terms of its length ( L ) and width ( w ) as follows:

`\begin{gathered} P\text{= 2}\cdot(L+w)\text{ } \\ P=\text{ 2}\cdot(L\text{ + w ) = 30} \\ L\text{ + w = }(30)/(2) \\ \\ L\text{ + w = 15 }\ldots\text{\textcolor{#FF7968}{ Eq 2}} \end{gathered}`

Now we have two equations ( Eq1 and Eq2 ) which are just mathematical decryption of the statements provided in the question and these equations are expressed in terms of two variables ( L and w ).

We have two equations and two unknowns we can solve these equations simultaneously.

Step 1: Substitute Eq1 into Eq2.

`\begin{gathered} L\text{ = w + 3 }\ldots\text{\textcolor{#FF7968}{ Eq1}} \\ L\text{ + w = 15 }\ldots\text{ }\text{\textcolor{#FF7968}{Eq2}} \\ =========== \\ (\text{ w + 3 ) + w = 15 }\ldots\textcolor{#FF7968}{Eq3} \end{gathered}`

Step 2: Solve the Eq3 for the variable ( w ).

`\begin{gathered} 2\cdot w\text{ = 15 - 3} \\ w\text{ = }(12)/(2) \\ \textcolor{#FF7968}{w}\text{\textcolor{#FF7968}{ = 6 meters}} \end{gathered}`

Step 3: Back-substitution of the result of one variable ( w ) into Eq1.

`\begin{gathered} L\text{ = w + 3 , w = 6 } \\ L\text{ = 6 + 3} \\ \textcolor{#FF7968}{L}\text{\textcolor{#FF7968}{ = 9 meters}} \end{gathered}`

The solution to the two equations are given as:

`\begin{gathered} \textcolor{#FF7968}{L}\text{\textcolor{#FF7968}{ = 9 meters}} \\ \textcolor{#FF7968}{w}\text{\textcolor{#FF7968}{ = 6 meters}} \end{gathered}`

Answer:

Length ( L ) = 9 meters , Width ( w ) = 6 meters