Math Problem Statement

The perimeter of the irregular figure is _____.

The area of the irregular figure is _____.

I know this because .

The perimeter of the irregular figure is _____. The area of the irregular figure is _____. I know this because .

Solution

Let's solve for the perimeter and area of the given figure based on its dimensions:

The figure:

  • The shape consists of a rectangle in the center with two semicircles on either end.
  • The length of the rectangle is 100m100 \, \text{m}, and the width is 60m60 \, \text{m}.
  • The diameter of each semicircle is equal to the width of the rectangle, 60m60 \, \text{m}.

1. Perimeter of the irregular figure:

  • The perimeter of the figure includes:

    1. The two semicircles (together, they form a full circle of diameter 60m60 \, \text{m}).
    2. The two lengths of the rectangle (100m+100m100 \, \text{m} + 100 \, \text{m}).
  • The circumference of a full circle with diameter dd is: Circumference=πd\text{Circumference} = \pi d Substituting d=60md = 60 \, \text{m}: Circumference of the full circle=π×60=60πm.\text{Circumference of the full circle} = \pi \times 60 = 60\pi \, \text{m}.

  • Adding the two lengths of the rectangle (100m100 \, \text{m} each): Perimeter=100+100+60π.\text{Perimeter} = 100 + 100 + 60\pi. Approximating π3.14\pi \approx 3.14: Perimeter200+60(3.14)=200+188.4=388.4m.\text{Perimeter} \approx 200 + 60(3.14) = 200 + 188.4 = 388.4 \, \text{m}.

Final Perimeter: 388.4m\boxed{388.4 \, \text{m}}


2. Area of the irregular figure:

  • The area of the figure includes:

    1. The area of the rectangle.
    2. The area of the two semicircles (which together form a full circle).
  • Area of the rectangle: Area=Length×Width=100×60=6000m2.\text{Area} = \text{Length} \times \text{Width} = 100 \times 60 = 6000 \, \text{m}^2.

  • Area of the full circle: The area of a circle with diameter dd (or radius r=d/2r = d/2) is: Area=πr2.\text{Area} = \pi r^2. Substituting r=60/2=30mr = 60/2 = 30 \, \text{m}: Area=π×(30)2=π×900=900πm2.\text{Area} = \pi \times (30)^2 = \pi \times 900 = 900\pi \, \text{m}^2. Approximating π3.14\pi \approx 3.14: Area of the circle900×3.14=2826m2.\text{Area of the circle} \approx 900 \times 3.14 = 2826 \, \text{m}^2.

  • Total Area: Total Area=Area of rectangle+Area of circle=6000+2826=8826m2.\text{Total Area} = \text{Area of rectangle} + \text{Area of circle} = 6000 + 2826 = 8826 \, \text{m}^2.

Final Area: 8826m2\boxed{8826 \, \text{m}^2}


Conclusion:

  • The perimeter is approximately 388.4m388.4 \, \text{m}.
  • The area is approximately 8826m28826 \, \text{m}^2.

Let me know if you'd like further clarification or if you'd like me to break it down further!


5 Related Questions:

  1. How does the formula for the circumference of a circle change if only a portion of the circle is used (e.g., semicircles)?
  2. What is the formula to calculate the area of any irregular composite shape?
  3. Why do we use π3.14\pi \approx 3.14 and how does using more accurate approximations of π\pi affect calculations?
  4. Can the area of this shape be verified using integration methods for curves?
  5. How do we handle perimeter calculations for figures involving arcs of a circle?

One Tip:

Always break composite shapes into smaller basic shapes (rectangles, circles, triangles, etc.) when solving for perimeter or area—it makes calculations easier and more systematic!

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Math Problem Analysis

Mathematical Concepts

Geometry
Perimeter
Area
Composite Figures

Formulas

Circumference of a circle: C = πd
Area of a circle: A = πr^2
Area of a rectangle: A = length × width

Theorems

Properties of Circles
Properties of Rectangles

Suitable Grade Level

Grades 6-8