To solve the problem, we need to find the dimensions of the house given that it is twice as long as it is wide and has an area of $ 200 \, \mathrm{m}^2 $.

Let's denote:
- the width of the house as $ w $
- the length of the house as $ l $

According to the problem, we have the following relationships:
1. The length is twice the width: 
   \[
   l = 2w
   \]
2. The area of the house is given by the formula:
   \[
   \text{Area} = l \times w
   \]
   Substituting the area value:
   \[
   l \times w = 200
   \]

Now, we can substitute the expression for $ l $ from the first equation into the area equation:
\[
(2w) \times w = 200
\]

This simplifies to:
\[
2w^2 = 200
\]

Next, we can solve for $ w $:
\[
w^2 = \frac{200}{2} = 100
\]
\[
w = \sqrt{100} = 10 \, \mathrm{m}
\]

Now that we have the width, we can find the length using the first equation:
\[
l = 2w = 2 \times 10 = 20 \, \mathrm{m}
\]

Thus, the dimensions of the house are:
- Width: $ 10 \, \mathrm{m} $
- Length: $ 20 \, \mathrm{m} $

In conclusion, the measurements of the house are $ 10 \, \mathrm{m} $ wide and $ 20 \, \mathrm{m} $ long.

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Question

$A house is twice as long as it is wide. It has an area of $ 200 \mathrm{~m}^{2} $. Find the measurement of the house.$

A house is twice as long as it is wide. It has an area of $ 200 \mathrm{~m}^{2} $. Find the measurement of the house.

Ask by Conner Cook.

Feb 17,2025 14:15

UpStudy AI Solution

Tutor-Verified Answer

Answer

The house is 10 meters wide and 20 meters long.

Solution

To solve the problem, we need to find the dimensions of the house given that it is twice as long as it is wide and has an area of $ 200 \, \mathrm{m}^2 $. Let's denote: - the width of the house as $ w $ - the length of the house as $ l $ According to the problem, we have the following relationships: 1. The length is twice the width: \[ l = 2w \] 2. The area of the house is given by the formula: \[ \text{Area} = l \times w \] Substituting the area value: \[ l \times w = 200 \] Now, we can substitute the expression for $ l $ from the first equation into the area equation: \[ (2w) \times w = 200 \] This simplifies to: \[ 2w^2 = 200 \] Next, we can solve for $ w $: \[ w^2 = \frac{200}{2} = 100 \] \[ w = \sqrt{100} = 10 \, \mathrm{m} \] Now that we have the width, we can find the length using the first equation: \[ l = 2w = 2 \times 10 = 20 \, \mathrm{m} \] Thus, the dimensions of the house are: - Width: $ 10 \, \mathrm{m} $ - Length: $ 20 \, \mathrm{m} $ In conclusion, the measurements of the house are $ 10 \, \mathrm{m} $ wide and $ 20 \, \mathrm{m} $ long.

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Simplify this solution

Mind Expander

To find the measurements of the house, we start by letting the width be $ w $ meters. Since the house is twice as long as it is wide, the length can be expressed as $ 2w $ meters. The area of the house can be calculated using the formula for the area of a rectangle, which is: \[ \text{Area} = \text{Length} \times \text{Width} = (2w) \times (w) \] Setting this equal to the given area: \[ 2w^2 = 200 \] Dividing both sides by 2 gives: \[ w^2 = 100 \] Taking the square root of both sides, we find: \[ w = 10 \text{ m} \] Thus, the length, being twice the width, will be: \[ \text{Length} = 2w = 20 \text{ m} \] So, the dimensions of the house are 20 meters in length and 10 meters in width. Now the house is not only spacious but also has a delightful balance between its measurements, making it ideal for cozy family gatherings or creative home projects!