Math Problem Statement

Find the area and perimeter of the rectangle above.

Find the area and perimeter of the rectangle above.

Solution

To solve this problem for the rectangle shown:

1. Area of the Rectangle:

The formula for the area AA of a rectangle is:

A=length×widthA = \text{length} \times \text{width}

Given:

  • Length =7in.= 7 \, \text{in.}
  • Width =3in.= 3 \, \text{in.}

A=7in.×3in.=21square inchesA = 7 \, \text{in.} \times 3 \, \text{in.} = 21 \, \text{square inches}

2. Perimeter of the Rectangle:

The formula for the perimeter PP of a rectangle is:

P=2×(length+width)P = 2 \times (\text{length} + \text{width})

P=2×(7in.+3in.)=2×10in.=20in.P = 2 \times (7 \, \text{in.} + 3 \, \text{in.}) = 2 \times 10 \, \text{in.} = 20 \, \text{in.}

Final Results:

  • Area = 21square inches21 \, \text{square inches}
  • Perimeter = 20inches20 \, \text{inches}

Would you like further details or explanations on any part of this solution?

Here are 5 related questions to deepen your understanding:

  1. How would the area change if the dimensions of the rectangle were doubled?
  2. What is the diagonal of this rectangle, and how can we calculate it?
  3. How do the formulas for area and perimeter change for a square?
  4. If the length of the rectangle was reduced by 2 inches, what would the new perimeter be?
  5. Can you think of any situations in real life where calculating the area and perimeter of rectangles is useful?

Tip: The area formula can be applied to any rectangle, but remember the perimeter depends on both the length and width being counted twice.

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Math Problem Analysis

Mathematical Concepts

Geometry
Measurement

Formulas

Area = length × width
Perimeter = 2 × (length + width)

Theorems

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Suitable Grade Level

Grades 3-5