The perimeter of a square is 40 m. A rectangle, whose perimeter is 9 m, has an area equal to the area of ​​a square. The length of the rectangle will be:

The problem provides the following information:

1. The perimeter of the square is 40 meters.
2. The perimeter of the rectangle is 9 meters, and its area is equal to the area of the square.

### Step 1: Calculate the side length of the square

The formula for the perimeter of a square is:

\[
\text{Perimeter of square} = 4 \times \text{side length of square}
\]

Given that the perimeter is 40 meters:

\[
4s = 40 \implies s = \frac{40}{4} = 10 \text{ meters}
\]

So, the side length of the square is \( s = 10 \) meters.

### Step 2: Calculate the area of the square

The area \( A \) of the square is:

\[
A = s^2 = 10^2 = 100 \, \text{square meters}
\]

### Step 3: Use the area of the rectangle

Since the area of the rectangle is equal to the area of the square, the area of the rectangle is also 100 square meters.

Let the length of the rectangle be \( l \) and the width be \( w \). The area of the rectangle is:

\[
A = l \times w = 100 \, \text{square meters}
\]

### Step 4: Use the perimeter of the rectangle

The formula for the perimeter of a rectangle is:

\[
\text{Perimeter of rectangle} = 2(l + w)
\]

Given that the perimeter is 9 meters:

\[
2(l + w) = 9 \implies l + w = \frac{9}{2} = 4.5
\]

### Step 5: Solve the system of equations

Now we have two equations:

1. \( l \times w = 100 \)
2. \( l + w = 4.5 \)

We can solve these equations using substitution. From the second equation, solve for \( w \):

\[
w = 4.5 - l
\]

Substitute this into the first equation:

\[
l \times (4.5 - l) = 100
\]

Expanding this:

\[
4.5l - l^2 = 100
\]

Rearranging it into a quadratic equation:

\[
l^2 - 4.5l + 100 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
l = \frac{-(-4.5) \pm \sqrt{(-4.5)^2 - 4(1)(100)}}{2(1)}
\]
\[
l = \frac{4.5 \pm \sqrt{20.25 - 400}}{2}
\]
\[
l = \frac{4.5 \pm \sqrt{-379.75}}{2}
\]

Since the discriminant is negative, there is no real solution for this case. This suggests there might be an issue with the given perimeter and area values, as no real length satisfies the problem under these constraints.

Would you like to revisit the problem or explore a different aspect of it? 

Here are 5 related questions:

1. What is the relationship between area and perimeter for a rectangle?
2. How do you interpret negative discriminants in a quadratic equation?
3. Could the dimensions of the rectangle change if we allow different types of units?
4. What are the different methods to solve quadratic equations?
5. How do you apply algebraic techniques to geometry problems?

**Tip:** When solving geometry problems involving shapes, always ensure the given constraints (like area and perimeter) are consistent to avoid contradictions.

Solve the problem of finding the length of a rectangle with a given perimeter and an area equal to that of a square. Learn to apply geometric formulas and quadratic equations, suitable for students in Grades 8-10.

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