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Rectangular shape related questions in linear equations

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Rectangle Word Problems

A rectangle word problem relates a rectangle's length and width, often with its perimeter, and asks you to find the actual dimensions. Name the width as the unknown, express the length algebraically, and use the perimeter formula to build and solve a linear equation. See a worked example solved step by step.

Setting up a rectangle word problem

A rectangle word problem describes a relationship between a rectangle's length and width, often together with its perimeter or area, and asks you to find the actual dimensions. Naming the width as the unknown lets you express the length algebraically.

Worked example

A rectangle's length is 3 more than twice its width, and its perimeter is 36. Let w be the width, so the length is 2w + 3.

Setting up a rectangle word problem A rectangle's length is 3 more than twice its width, and its perimeter is 36. Using width w and length 2w plus 3, the perimeter equation 2 times the quantity w plus 2w plus 3 equals 36 simplifies to 6w plus 6 equals 36, so w equals 5 and the length is 13. length = 2w + 3 width = w Setting up the equation Perimeter = 2(w + length) = 36 2(w + 2w + 3) = 36 → 6w + 6 = 36 6w = 30 → w = 5, length = 13
Naming the width lets the length be written algebraically before building the perimeter equation.

The perimeter formula P = 2(w + length) gives 2(w + 2w + 3) = 36, which simplifies to 6w + 6 = 36, so 6w = 30 and w = 5. The length is 2(5) + 3 = 13.

Perimeter versus area setups

The same naming approach works for area word problems, though area gives a quadratic rather than a linear equation, since length × width multiplies two expressions in w together. This modeling process is the same one used for money word problems, unknown-number word problems, and distance-and-time word problems.

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