area word problems

JacquelinepdaguiaoTuguegarao East Central School

Area word problems

A. You will accounter these area word problems often in math. Many of them will require familiarity with basic math, algebra skills, or a combination of both to solve the problems

As I solve these area word problems, I will make an attempt to give you some problems solving skills

Word problem #1:

The area of a square is 4 inches. What is the length of a side?

Important concept: Square. It means 4 equal sides.

Area = s × s= 4 × 4 = 16 inches2

Word problem #2:

A small square is located inside a bigger square. The length of one side of the small square is 3 inches and the length of one side of the big square is 7 inches

What is the area of the region located outside the small square, but inside the big square?

Important concept: Draw a picture and see the problem with your eyes. This is done below:

The area that you are looking for is everything is red. So you need to remove the area of the small square from the area of the big square

Area of big square = s × s = 7 × 7 = 49 inches2

Area of small square = s × s = 3 × 3 = 9 inches2

Area of the region in red = 49 - 9 = 40 inches2

Word problem #3:


A classroom has a length of 20 feet and a width of 30 feet. The headmaster decided that tiles will look good in that class. If each tile has a length of 24 inches and a width of 36 inches, how many

tiles are needed to fill the classroom?

B. Question 480823: Solving Area and Perimeter Problems: Need to use the RSTUV

Method to solve The area of a rectangle is 96 square centimeters. If the width of the rectangle is 4 centimeters less than its length, what are the dimensions of the rectangle?

RSTUV1. Read the problem.2. Select the unknown.3. Translate into an equation.4. Use the rules to solve.5. Verify the result

1.area of a rectangle is 96 square cm.width is 4 cm less than its length.what are dimensions of rectangle.

2.let L = lengthlet W = widthlet A = area

3. A = L * W W = L - 4 cm Since W = L-4, then the equation for the area of a rectangle becomes: A = L * (L-4)

4. A = 96 so equation of: A = L * (L-4) becomes: 96 = L * (L-4)perform indicated operations to get: 96 = L^2 - 4Lsubtract 96 from both sides of the equation to get:0 = L^2 - 4L - 96commute to get:


L^2 - 4L - 96 = 0this is a quadratic equation.factor it to get:(L - 12) * (L + 8) = 0solve for L to get:L = 12 or L = -8L can't be negative so L = 12.back to original equation of:A = L * W which becomes:96 = 12 * Wdivide both sides of this equation by 12 to get:96/12 = W which becomes:W = 85. the dimensions of the rectangle are: L = 12 W = 8 the area of the rectangle is 36 square cm. A = L * W becomes A = 12 * 8 which becomes A = 96 this is the same area we started with so the dimensions of the rectangle look good. the dimensions are: L = 12 cm W = 8 cm

area word problems

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