Math Problem Statement

Find the perimeter of the given shapes (E, O, I) and the area of the given shapes (E, S) based on the side lengths and dimensions provided in the image.

Solution

Let's break down and solve the problems shown in the image. Here's the process step-by-step:


Part 1: Find the perimeter

The perimeter of a shape is the sum of all its side lengths.

  1. Figure E (Triangle): Side lengths are 3x+83x + 8, 4x14x - 1, and 5x+25x + 2.
    Perimeter=(3x+8)+(4x1)+(5x+2)\text{Perimeter} = (3x + 8) + (4x - 1) + (5x + 2)
    Simplify: Perimeter=3x+4x+5x+81+2=12x+9\text{Perimeter} = 3x + 4x + 5x + 8 - 1 + 2 = 12x + 9

  2. Figure O (Triangle): Side lengths are 9x109x - 10, 2x22x^2, and x2+4x^2 + 4.
    Perimeter=(9x10)+(2x2)+(x2+4)\text{Perimeter} = (9x - 10) + (2x^2) + (x^2 + 4)
    Simplify: Perimeter=2x2+x2+9x10+4=3x2+9x6\text{Perimeter} = 2x^2 + x^2 + 9x - 10 + 4 = 3x^2 + 9x - 6

  3. Figure I (Rectangle): Side lengths are 2x32x - 3 and 4x74x - 7.
    The perimeter of a rectangle is: Perimeter=2×((2x3)+(4x7))\text{Perimeter} = 2 \times \left((2x - 3) + (4x - 7)\right)
    Simplify: Perimeter=2×(2x+4x37)=2×(6x10)=12x20\text{Perimeter} = 2 \times (2x + 4x - 3 - 7) = 2 \times (6x - 10) = 12x - 20


Part 3: Find the area

  1. Figure E (Rectangle): Dimensions are 3x3x and 8x58x - 5.
    The area of a rectangle is: Area=length×width\text{Area} = \text{length} \times \text{width} Substitute: Area=3x×(8x5)=24x215x\text{Area} = 3x \times (8x - 5) = 24x^2 - 15x

  2. Figure S (Rectangle): Dimensions are 2x22x^2 and x2+7x+4x^2 + 7x + 4.
    The area is: Area=length×width\text{Area} = \text{length} \times \text{width}

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomials
Geometry

Formulas

Perimeter of a triangle = sum of all side lengths
Perimeter of a rectangle = 2 × (length + width)
Area of a rectangle = length × width

Theorems

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Suitable Grade Level

Grades 7-10