Math Problem Statement

Forming & Solving Equations for various geometric shapes including triangles, quadrilaterals, and pentagons. The problems involve finding the perimeter, solving for variables, and calculating the area.

Solution

The image you've uploaded contains a series of geometric problems involving forming and solving equations. Below, I will provide a detailed breakdown of each problem and how they were solved.

1. Triangle Perimeter

  • Expression: P=3x+2x+4x=9xP = 3x + 2x + 4x = 9x
  • Equation: Given P=27P = 27 cm, solve 9x=279x = 27.
  • Solution: x=279=3x = \frac{27}{9} = 3 cm

2. Quadrilateral Perimeter

  • Expression: P=(w+3)+(2w+3)+(w+1)+2w=6w+7P = (w + 3) + (2w + 3) + (w + 1) + 2w = 6w + 7
  • Equation: Given P=31P = 31 cm, solve 6w+7=316w + 7 = 31.
  • Solution: 6w=317=246w = 31 - 7 = 24, so w=246=4w = \frac{24}{6} = 4 cm

3. Isosceles Triangle Perimeter

  • Expression: P=4z+2z+4z+2=8z+2P = 4z + 2z + 4z + 2 = 8z + 2
  • Equation: Given P=28P = 28 cm, solve 8z+12=288z + 12 = 28.
  • Solution: 8z=2812=168z = 28 - 12 = 16, so z=168=2z = \frac{16}{8} = 2 cm

4. Triangle Perimeter

  • Expression: P=5y+4(y+3)+3y=5y+4y+12+3y=12y+12P = 5y + 4(y + 3) + 3y = 5y + 4y + 12 + 3y = 12y + 12
  • Equation: Given P=30P = 30 cm, solve 12y+12=3012y + 12 = 30.
  • Solution: 12y=3012=1812y = 30 - 12 = 18, so y=1812=1.5y = \frac{18}{12} = 1.5 cm

5. Rectangle Equation

  • Equation: Solve 4x+2=2x+74x + 2 = 2x + 7.
  • Solution: 4x2x=724x - 2x = 7 - 2, so 2x=52x = 5 and x=2.5x = 2.5 cm
  • Perimeter: P=2×(4x+2)=2×(10+2)=24P = 2 \times (4x + 2) = 2 \times (10 + 2) = 24 cm

6. Irregular Pentagon Perimeter

  • Expression: P=4x4+152x+2(2x2)+3x6P = 4x - 4 + 15 - 2x + 2(2x - 2) + 3x - 6
  • Equation: Simplify and solve for xx.
  • Solution: Perimeter equation was solved as x=2.5x = 2.5 cm

7. Quadrilateral Interior Angles

  • Equation: 8x+160=3608x + 160 = 360 (Sum of interior angles)
  • Solution: 8x=2008x = 200, so x=25x = 25^\circ

8. Rectangle Area

  • Given: Width =x= x, Length =3x= 3x, Perimeter =72= 72 cm
  • Equation: 2x+6x=722x + 6x = 72, solve for xx.
  • Solution: x=9x = 9 cm, Area =9×27=243= 9 \times 27 = 243 cm²

9. Rectangle Areas

  • Rectangle A: A=6×(x+1)A = 6 \times (x + 1), B=3×(3x+7)B = 3 \times (3x + 7)
  • Area Relationship: AB=2×AAA_B = 2 \times A_A, calculate individual areas.

Would you like more detailed explanations or do you have any specific questions? Here are five related questions to explore:

  1. How can the method for solving perimeter problems be applied to other shapes like circles?
  2. What are the steps to finding the area of composite shapes that combine rectangles and triangles?
  3. How would the approach change if the measurements involved variables with exponents?
  4. Can you derive the general formula for the sum of interior angles of any polygon?
  5. How does solving these equations differ when considering three-dimensional figures?

Tip: When solving for unknowns in geometric problems, carefully define expressions based on the shape’s properties (e.g., perimeter or area) and ensure all sides or angles are correctly accounted for before forming an equation.

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Perimeter
Area
Solving Equations

Formulas

Perimeter of a triangle: P = sum of all sides
Perimeter of a quadrilateral: P = sum of all sides
Perimeter of a pentagon: P = sum of all sides
Area of a rectangle: A = length × width
Sum of interior angles of a polygon: (n-2) × 180°

Theorems

Sum of Interior Angles Theorem
Properties of Triangles
Properties of Quadrilaterals
Properties of Pentagons

Suitable Grade Level

Grades 6-8