Math Problem Statement

I

જેની પરિમિતિ 80 મી અને ક્ષેત્રફળ 400 મીટર હોય તેવી લંબચોરસ બનાવી શકાય છે? જો તે શક્ય હોય, તો તેની લંબાઈ અને પહોળાઈ શોધો.

Solution

The question in the image is in Gujarati and translates to:

Question: If the perimeter of a rectangle is 80 meters and the area is 400 square meters, can such a rectangle be constructed? If possible, find its length and breadth.

Let's solve this step by step:

Given:

  • Perimeter P=80P = 80 meters
  • Area A=400A = 400 square meters

Formulae:

  • The perimeter of a rectangle:
    P=2(l+b)P = 2(l + b) where ll is the length and bb is the breadth.

  • The area of a rectangle:
    A=l×bA = l \times b

Step 1: Solve for the perimeter

From the perimeter equation P=2(l+b)P = 2(l + b), we have: 80=2(l+b)80 = 2(l + b) l+b=40(Equation 1)l + b = 40 \quad \text{(Equation 1)}

Step 2: Solve for the area

The area equation gives us: l×b=400(Equation 2)l \times b = 400 \quad \text{(Equation 2)}

Step 3: Solve the system of equations

From Equation 1, we can express ll as: l=40bl = 40 - b Substitute this into Equation 2: (40b)×b=400(40 - b) \times b = 400 40bb2=40040b - b^2 = 400 Rearranging this equation: b240b+400=0b^2 - 40b + 400 = 0

This is a quadratic equation, which can be solved using the quadratic formula: b=(40)±(40)24(1)(400)2(1)b = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(1)(400)}}{2(1)} b=40±160016002b = \frac{40 \pm \sqrt{1600 - 1600}}{2} b=40±02b = \frac{40 \pm \sqrt{0}}{2} b=402=20b = \frac{40}{2} = 20

Step 4: Find the length

Using b=20b = 20 in Equation 1: l+20=40l + 20 = 40 l=4020=20l = 40 - 20 = 20

Step 5: Conclusion

Both the length and breadth of the rectangle are 20 meters, meaning the rectangle is a square.

Thus, the rectangle can be constructed, and its length and breadth are both 20 meters.

Would you like more details or further clarifications?


Here are five related questions:

  1. What is the difference between the perimeter and area of a rectangle?
  2. Can a rectangle always be a square if its length equals its breadth?
  3. How do you solve quadratic equations in geometry problems?
  4. What are other ways to find the dimensions of a rectangle given the perimeter and area?
  5. How do the dimensions of a rectangle change when the perimeter is constant but the area changes?

Tip: Always check for consistency between the perimeter and area equations when solving geometry problems involving rectangles.

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Math Problem Analysis

Mathematical Concepts

Geometry
Algebra
Quadratic Equations

Formulas

Perimeter of rectangle: P = 2(l + b)
Area of rectangle: A = l × b
Quadratic equation: ax^2 + bx + c = 0

Theorems

Quadratic Formula Theorem

Suitable Grade Level

Grades 8-10