Sure! Let's solve the problem step by step.
1. Understand the Problem:
- The problem gives us the perimeter of a rectangular bedroom: 60 meters.
- It also provides the total area of the four walls: 144 square meters.
- We are tasked to find the height of the room.
2. Determine the Width and Length of the Room:
- The given perimeter [tex]\( P \)[/tex] is 60 meters.
- For a rectangle, the perimeter [tex]\( P \)[/tex] is given by:
[tex]\[
P = 2 \times (\text{length} + \text{width})
\][/tex]
From the value provided:
[tex]\[
60 = 2 \times (\text{length} + \text{width})
\][/tex]
[tex]\[
\text{length} + \text{width} = 30 \quad \text{meters}
\][/tex]
3. Use the Relationship between Length and Width:
- Suppose the length is [tex]\( L \)[/tex] and the width is [tex]\( W \)[/tex]. Therefore:
[tex]\[
L + W = 30
\][/tex]
- Since the perimeter is evenly distributed, the width at the ceiling and floor can be written as half the total perimeter width:
[tex]\[
W = \frac{30}{2} = 15 \quad \text{meters}
\][/tex]
4. Calculate the Height Using the Area of the Walls:
- The total area of the four walls is given as 144 square meters.
- The area of the four walls can be thought of as twice the sum of the heights times the length and width (since there are two sets of opposite walls):
[tex]\[
2 \times (\text{height} \times L + \text{height} \times W)
\][/tex]
- We know that:
[tex]\[
\text{Area of four walls} = 144 \quad \text{square meters}
\][/tex]
Substituting the area formula in:
[tex]\[
2 \times (\text{height} \times L + \text{height} \times W) = 144
\][/tex]
- We already know [tex]\( W \)[/tex], and notice that [tex]\( W \)[/tex] and [tex]\( L \)[/tex] are equal in this problem:
[tex]\[
2 \times (\text{height} \times W + \text{height} \times W) = 144
\][/tex]
This simplifies to:
[tex]\[
4 \times (\text{height} \times W) = 144
\][/tex]
Solving for height:
[tex]\[
\text{height} \times W = 36 \quad \text{(because \( 144 / 4 = 36 \shortrightarrow 36 / 15 = 2.4 can be wrong need to calculate properly)}
\][/tex]
[tex]\[
\text{height} = \frac{36}{15} = 2.4 \quad \text{meters}
\][/tex]
Therefore, the height of the room is [tex]\( 2.4 \)[/tex] meters.
