You choose these problems of Word area often in mathematics. Many of them, it will require familiarity with basic math, algebra skills or a combination of the two for the problems

As I solve these problems of area Word, I will make an attempt to give a few problems to solve their problems

**Word problem # 1:**

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

**Important concept**: Square. This means 4 equal sides.

Area = s × s = 4 × 4 = 16 po2

**Word problem # 2:**

A small square is located within a larger square. The length of one side of the square is 3 inches and the length of one side of the large 7-inch

What is the area located outside the square, but in the great place?

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

The area that you are looking for, is that everything that is red. If you remove the area of the square of the area of the great place

Big square area = s × s = 7 × 7 = 49 po2

Small square area = s × s = 3 × 3 = 9 po2

Area in the region in red = 49-9 = 40 po2

**Word problem # 3:**

One classroom has a length of 20 feet and a width of 30 feet. The headmaster decided that the tiles will look good in this category. If each tile has a length of 24 inches and a width of 36 inches, how many tiles are needed to fill the classroom?

**Important concept**:

Find the area occupied by entire class

Find the zone for a tile

Decide what unit to use. In this case, we could use feet

Classroom area = length × width = 20 × 30 = 600 sq. ft.

Before we get the area of each tile, to convert dimensions to feet as he has already reported

Since 1 foot = 12 inches, 36 inches = 3 feet and 24 inches = 2 feet

Each tile length × width = 2 × 3 = area = 6 sq. ft. If a tile takes 6 FT2, it must be 100 tiles to cover the 600 sq. ft. (6 × 100 = 600)

**Word problem # 4:**

Sometimes word of zone issues may require skills in algebra, such as factoring and solving second degree equations

A room with area 24 sq. ft. has a length that is longer than the width of 2 feet. What are the dimensions of the room?

**Solution # 1** Use of basic mathematical skills and trial and error

Pretend wide = 1, length = 3 (1 + 2)

1 × 3 is not equal to 24 and 3 is not close at all to 24. Try as many

leave the width = 3, while the length = 5 (3 + 2)

3 × 5 = 15. It is still not equal to 24

leave the width = 4, while the length = 4 (4 + 2)

4 × 6 = 24. We are in fact!

**Solution # 2** Use of algebra

Leave the width = x, so length = x + 2 = length × width sector

24 = x x (x + 2)

24 = x 2 + 2 x

x 2 + 2 x = 24 (swap the left side with the right side)

x 2 + 2 x – 24 = 0.

(x + 6) × (x – 4) = 0

x =-6, and x = 4

So width = 4, length = 4 + 2 = 6

As you can see Word of zone problems can become very complicated as shows in the last problem. And the use of basic mathematical skills is perhaps not the best way to go to the problems of password area. Algebra is sometimes better because the trial and error can take a long time!

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