Rectangular Field Dimensions Solving Area Perimeter Problems

Jul 26, 2025 by ADMIN 61 views

Unveiling the Dimensions of a Rectangular Field: A Mathematical Journey

Hey there, math enthusiasts! Today, we're diving into a classic problem involving the area and perimeter of a rectangular field. We'll use our algebraic superpowers to unravel the mystery of its dimensions. So, buckle up, and let's get started!

Problem Statement: Cracking the Code of the Rectangle

We're given a rectangular field with an area of 150 square meters and a perimeter of 50 meters. Our mission, should we choose to accept it, is twofold:

a) First, we need to determine the length and breadth of this field by wielding the mighty quadratic equation. b) Second, we'll explore how many equal parts we should subtract from both the length and breadth to shrink the field's area.

Part A: Decoding the Dimensions with Quadratic Equations

1. Setting the Stage: Variables and Equations

Let's start by assigning variables to our unknowns. Let 'l' represent the length of the rectangular field and 'b' represent its breadth. Now, we can translate the given information into mathematical equations.

Area: We know that the area of a rectangle is given by the product of its length and breadth. So, we have the equation:
```
l * b = 150  (Equation 1)
```
Perimeter: The perimeter of a rectangle is the total length of its boundary, which is twice the sum of its length and breadth. This gives us the equation:
```
2 * (l + b) = 50
```
Simplifying this, we get:
```
l + b = 25   (Equation 2)
```

2. The Quadratic Equation Enters the Scene

Now, we have two equations with two unknowns. Our goal is to transform these equations into a single quadratic equation. To do this, we can use a clever trick. Let's solve Equation 2 for one variable, say 'b':

b = 25 - l

Now, we can substitute this expression for 'b' into Equation 1:

l * (25 - l) = 150

Expanding and rearranging this equation, we get a quadratic equation:

l^2 - 25l + 150 = 0

3. Solving the Quadratic Equation: Unveiling the Roots

We've arrived at a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = -25, and c = 150. To solve this, we can use the quadratic formula:

l = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values, we get:

l = (25 ± √((-25)^2 - 4 * 1 * 150)) / (2 * 1)

l = (25 ± √(625 - 600)) / 2

l = (25 ± √25) / 2

l = (25 ± 5) / 2

This gives us two possible solutions for 'l':

l1 = (25 + 5) / 2 = 15
l2 = (25 - 5) / 2 = 10

4. Finding the Breadth: Completing the Picture

Now that we have two possible values for the length, we can use Equation 2 (l + b = 25) to find the corresponding breadths:

If l = 15, then b = 25 - 15 = 10
If l = 10, then b = 25 - 10 = 15

So, we have two possible sets of dimensions: length = 15 meters and breadth = 10 meters, or length = 10 meters and breadth = 15 meters. Since length is typically considered the longer side, we can conclude that the length of the field is 15 meters and the breadth is 10 meters.

Part B: Shrinking the Field: A Subtraction Puzzle

1. Setting the Stage: Subtracting Equal Parts

Now, let's tackle the second part of the problem. We want to subtract an equal amount, let's call it 'x', from both the length and breadth of the field. This will change the dimensions of the rectangle, and we're interested in how this affects the area.

The new length will be (15 - x) meters, and the new breadth will be (10 - x) meters.

2. The New Area: An Expression Emerges

The new area of the rectangle will be the product of the new length and breadth:

New Area = (15 - x) * (10 - x)

Expanding this expression, we get:

New Area = 150 - 25x + x^2

3. The Question of Subtraction: A Deeper Dive

The problem asks how many equal parts should be subtracted to shrink the field's area, but it doesn't specify the desired new area. This is where we need to make an assumption or be given more information to solve for 'x'.

Scenario 1: If we want to reduce the area by a specific amount:

Let's say we want to reduce the area by 50 square meters. This means the new area should be 150 - 50 = 100 square meters. We can set up the equation:

150 - 25x + x^2 = 100

Rearranging this, we get another quadratic equation:

x^2 - 25x + 50 = 0

We can solve this quadratic equation using the quadratic formula again to find the value(s) of 'x'.

Scenario 2: If we want to reduce the area to a specific fraction of the original area:

Let's say we want to reduce the area to half of the original area. This means the new area should be 150 / 2 = 75 square meters. We can set up the equation:

150 - 25x + x^2 = 75

Rearranging this, we get another quadratic equation:

x^2 - 25x + 75 = 0

Again, we can solve this quadratic equation to find the value(s) of 'x'.

In Conclusion:

To determine the exact number of equal parts to subtract, we need a target area or a specific reduction amount. Once we have that, we can set up a quadratic equation and solve for 'x'.

Wrapping Up: The Beauty of Mathematical Problem-Solving

Wow, guys, we've really put our math skills to the test today! We successfully navigated the world of rectangles, areas, perimeters, and quadratic equations. We learned how to translate word problems into mathematical equations, solve those equations, and interpret the results. Remember, the key to solving math problems is to break them down into smaller, manageable steps. Keep practicing, keep exploring, and most importantly, keep having fun with math!