In this article, we explore how to calculate the dimensions of a room using algebra and quadratic equations. Specifically, we will solve a problem where the area of a room is given, and one dimension is related to the other through a simple linear equation. We will walk you through the process step by step, using real examples to illustrate the application of these mathematical tools.
Example 1: Finding Room Dimensions with an Area and Length-Width Relationship
Let's start with a specific problem: If the area of a room is 24 square meters and the length is 2 meters more than the width, what are the dimensions of the room?
Solution:
Step 1: Define the variables.
Let ( w ) represent the width of the room in meters. Since the length is 2 meters more than the width, the length ( l ) is ( w 2 ) meters. The area ( A ) of the room is given as 24 square meters.Step 2: Use the area formula.
The area of the room can be expressed as:
$$ A l times w $$Substituting the given values and expressions:
$$ 24 (w 2) times w $$Step 3: Simplify and solve the quadratic equation.
Expanding the equation:
$$ 24 w^2 2w $$Rearrange to standard quadratic form:
$$ w^2 2w - 24 0 $$Step 4: Solve the quadratic equation using the quadratic formula:
$$ w dfrac{-b pm sqrt{b^2 - 4ac}}{2a} $$Here, ( a 1 ), ( b 2 ), and ( c -24 ).
$$ w dfrac{-2 pm sqrt{2^2 - 4(1)(-24)}}{2(1)} $$ $$ w dfrac{-2 pm sqrt{4 96}}{2} $$ $$ w dfrac{-2 pm sqrt{100}}{2} $$ $$ w dfrac{-2 pm 10}{2} $$Calculating the two possible solutions:
$$ w dfrac{8}{2} 4 $$ $$ w dfrac{-12}{2} -6 $$Since width cannot be negative, ( w 4 ) meters.
Substitute ( w 4 ) back into the expression for length:
$$ l w 2 4 2 6 $$Therefore, the length of the room is 6 meters, and the width is 4 meters.
Example 2: Another Room Area and Dimensions Problem
In this second example, the area of the room is given as 168 square feet, and the relationship between the length and width is also given.
Solution:
Step 1: Define the variables.
Let ( x ) represent the width of the room in feet. Since the length is twice the width, the length is ( 2x ) feet. The area ( A ) of the room is given as 168 square feet.Step 2: Use the area formula.
The area of the room can be expressed as:
$$ A x times 2x 168 $$Step 3: Simplify and solve the quadratic equation.
$$ 2x^2 168 $$ $$ x^2 2x - 168 0 $$Step 4: Solve the quadratic equation using the quadratic formula:
$$ x dfrac{-b pm sqrt{b^2 - 4ac}}{2a} $$Here, ( a 1 ), ( b 2 ), and ( c -168 ).
$$ x dfrac{-2 pm sqrt{2^2 - 4(1)(-168)}}{2(1)} $$ $$ x dfrac{-2 pm sqrt{4 672}}{2} $$ $$ x dfrac{-2 pm sqrt{676}}{2} $$ $$ x dfrac{-2 pm 26}{2} $$Calculating the two possible solutions:
$$ x dfrac{24}{2} 12 $$ $$ x dfrac{-28}{2} -14 $$Since width cannot be negative, ( x 12 ) feet.
Substitute ( x 12 ) back into the expression for length:
$$ l 2x 2 times 12 24 $$Therefore, the dimensions of the room are 24 feet long by 12 feet wide.
Conclusion
We explored two examples of solving room dimensions using algebra and quadratic equations. By setting up and solving quadratic equations, we were able to find the exact dimensions of the rooms. The algebraic methods are not only useful for practical applications in construction and planning but also reinforce our understanding of algebra and quadratic equations.