Learn on PengiThe Art of Problem Solving: Prealgebra (AMC 8)Chapter 11: Perimeter and Area

Lesson 1: Measuring Segments

Property.

Section 1

Congruent Segments

Property

Two segments are congruent if they have equal lengths: ABCD\overline{AB} \cong \overline{CD} if and only if AB=CDAB = CD. Congruent segments are marked with identical tick marks in geometric diagrams.

Examples

Section 2

Perimeter

Property

In mathematics, the perimeter means the length of the boundary of a region. To find the perimeter of a straight-sided figure, measure the length of each side and add up the lengths of the sides.

Examples

  • A rectangular garden is 10 meters long and 7 meters wide. Its perimeter is the sum of all its sides: 10+7+10+7=3410 + 7 + 10 + 7 = 34 meters.
  • A triangular park has sides measuring 50 feet, 60 feet, and 75 feet. The perimeter is found by adding these lengths: 50+60+75=18550 + 60 + 75 = 185 feet.
  • For a square window with one side measuring 2 feet, the perimeter is 4×2=84 \times 2 = 8 feet, since all four sides are equal.

Explanation

Imagine you are an ant walking around the edge of a shape. The total distance you walk in one full circuit is its perimeter. It measures the length of the boundary, not the space inside.

Section 3

Geometry applications with perimeter

Property

Perimeter problems use the total distance around a shape to form an equation. For a rectangle with length LL and width WW, the perimeter is P=2L+2WP = 2L + 2W. Some problems may involve fencing only some sides, such as L+2WL + 2W for a yard against a house.

Examples

  • The perimeter of a rectangular park is 100 meters. The length is 10 meters more than the width. The system is {2L+2W=100L=W+10\begin{cases} 2L + 2W = 100 \\ L = W + 10 \end{cases}. The dimensions are L=30L=30 m and W=20W=20 m.
  • A fence is built around three sides of a rectangular yard adjacent to a house, using 140 feet of fencing. The length is 20 feet more than twice the width. The system is {L+2W=140L=2W+20\begin{cases} L + 2W = 140 \\ L = 2W + 20 \end{cases}. The dimensions are L=68L=68 ft and W=36W=36 ft.

Book overview

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Chapter 11: Perimeter and Area

  1. Lesson 1Current

    Lesson 1: Measuring Segments

  2. Lesson 2

    Lesson 2: Area

  3. Lesson 3

    Lesson 3: Circles

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Congruent Segments

Property

Two segments are congruent if they have equal lengths: ABCD\overline{AB} \cong \overline{CD} if and only if AB=CDAB = CD. Congruent segments are marked with identical tick marks in geometric diagrams.

Examples

Section 2

Perimeter

Property

In mathematics, the perimeter means the length of the boundary of a region. To find the perimeter of a straight-sided figure, measure the length of each side and add up the lengths of the sides.

Examples

  • A rectangular garden is 10 meters long and 7 meters wide. Its perimeter is the sum of all its sides: 10+7+10+7=3410 + 7 + 10 + 7 = 34 meters.
  • A triangular park has sides measuring 50 feet, 60 feet, and 75 feet. The perimeter is found by adding these lengths: 50+60+75=18550 + 60 + 75 = 185 feet.
  • For a square window with one side measuring 2 feet, the perimeter is 4×2=84 \times 2 = 8 feet, since all four sides are equal.

Explanation

Imagine you are an ant walking around the edge of a shape. The total distance you walk in one full circuit is its perimeter. It measures the length of the boundary, not the space inside.

Section 3

Geometry applications with perimeter

Property

Perimeter problems use the total distance around a shape to form an equation. For a rectangle with length LL and width WW, the perimeter is P=2L+2WP = 2L + 2W. Some problems may involve fencing only some sides, such as L+2WL + 2W for a yard against a house.

Examples

  • The perimeter of a rectangular park is 100 meters. The length is 10 meters more than the width. The system is {2L+2W=100L=W+10\begin{cases} 2L + 2W = 100 \\ L = W + 10 \end{cases}. The dimensions are L=30L=30 m and W=20W=20 m.
  • A fence is built around three sides of a rectangular yard adjacent to a house, using 140 feet of fencing. The length is 20 feet more than twice the width. The system is {L+2W=140L=2W+20\begin{cases} L + 2W = 140 \\ L = 2W + 20 \end{cases}. The dimensions are L=68L=68 ft and W=36W=36 ft.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 11: Perimeter and Area

  1. Lesson 1Current

    Lesson 1: Measuring Segments

  2. Lesson 2

    Lesson 2: Area

  3. Lesson 3

    Lesson 3: Circles