Area and perimeter word problem: table dimensions | Khan Academy | Summary and Q&A

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September 24, 2013
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Area and perimeter word problem: table dimensions | Khan Academy

TL;DR

Determine the length and width of a rectangular table with a perimeter of 20 feet and an area of 24 square feet.

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Questions & Answers

Q: How can we solve for the length and width of the rectangular table?

We can solve for the dimensions by trying out different whole number combinations that lead to an area of 24. We then check if the corresponding perimeter equals 20.

Q: Why did we create a table to try out different combinations?

The table allows us to organize our attempts to find the right dimensions by systematically testing all possible pairs of whole numbers as length and width values.

Q: How did we determine if a combination satisfied the perimeter constraint?

We added the width and length twice, representing the sides of the rectangular table, and checked if the sum equaled 20. If it did not, we crossed out that combination.

Q: How did we find the correct dimensions of the table?

By testing all possible combinations, we discovered that a width of 4 feet and a length of 6 feet satisfy both the area and perimeter constraints.

Summary & Key Takeaways

  • Given a rectangular table with a perimeter of 20 feet and an area of 24 square feet, determine its length and width.

  • The perimeter equation is width + width + length + length = 20.

  • The area equation is width × length = 24.

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