Word Problems on Area and Perimeter

We will solve different types of word problems on area and perimeter.

1. The perimeter of a rectangular field is 260 m. If its length is 90 m, find its breadth.

Solution:

Perimeter of a rectangle = 2(b + ℓ)

                      ⟹ 260m = 2(b + 90 m)

                      ⟹ 260 m = 2b + 180 m

                      ⟹ (260 - 180) m = 2b

                      ⟹ 80 m = 2b

                      ⟹ 2b = 80 m

                      ⟹ b = \(\frac{80}{2}\) m

                      ⟹ b = 40 m

Hence, the breadth of the rectangular field = 40 m

2. Find the cost of fencing a square park of side 250 m at the rate of $ 20 per metre.

Solution:

Side of the park = 250 m

To calculate the cost of fencing, we require perimeter.

Perimeter of the square = 4 × one side

                                   = 4 × 250 m

                                   = 1000 m

Now, the total cost of fencing the park = $ 20 × 1000 = $ 20000

3. The side of a square plot of land is 35 m. Find the cost of levelling the plot, if the rate is $ 4 per square metre.

Solution:

             Area of square = side × side

                                   = 35 m × 35 m

                                   = 1225 cm²

Levelling cost of 1 m² = $ 4

Levelling cost of 1225 m² = $ (4 × 1225)

                                      = $ 4900

4. How many tiles, each measuring 2 m by 1 m, are required to cover a rectangular hall 12 m long and 8 m broad? Find the cost of tiles at $ 25 per tile.

Solution:

Length of a tile = 2 m

Breadth of a tile = 1 m

Therefore, Area of each tile = 2 m × 1 m

                                        = 2 m²

Length of rectangular hall = 12 m

Breadth of rectangular hall = 8 m

Area of the hall = 12 m × 8 m

                       = 96 m²

Hence, number of tiles required

                        = \(\frac{\textrm{Area of the hall}}{\textrm{Area of one tile}}\)

                        = \(\frac{96}{2}\)

Therefore, Cost of tiles 48 × $ 25 = $ 1200

Worksheet on Word Problems on Area and Perimeter:

1. The sides of a triangular field are 20 cm, 15 cm, and 12 cm. Find the total distance travelled by a boy in taking 2 complete rounds of this field.

Answer:

1. 94 cm

2. The length and breadth of a rectangular park are 615 m and 550 m, respectively. Find the cost of fencing the park at the rate of $ 9.25 per metre.

Answer:

2. $ 21552.5

3. The length of a piece of wire is 78 m. If the wire is used to make a regular pentagon and a regular hexagon, find the difference in the lengths of its sides.

Answer:

3. 15.6 m - 13 m = 2.6 m

4. Find the length of a side of an equilateral triangle whose perimeter is 3 m 12 cm.

Answer:

4. 1.04 m

5. Stephen takes 2 rounds of a square park of side 135 m and Olivia takes 3 rounds of a rectangular park of length 70 m and breadth 45 m. Who covers more distance and by how much?

Answer:

5. Stephen; 390 m

6. Find the length of a rectangular park whose area is 204 sq m and breadth is 12 m.

Answer:

6. Length of a rectangular park = 17 m

7. The perimeter of a rectangular park is 270 m. If its length is 75 m, what will be its breadth?

Also calculate the area of this rectangular park.

Answer:

7. breadth = 60 m

Area of the rectangular park = 4500 m²

8. Find the area of a square if the length of its sides is doubled. Also, find the ratio of new area to its previous area.

Answer:

8. Area of the new square becomes 4 times when side is doubled.

Ratio of new area to its previous area = 4 : 1.

9. Two rectangles of length ℓ and breadth b are placed side by side along the length without any gap. Find the area of the new rectangle thus formed.

Answer:

9. Area of the new rectangle = 2ℓ × b = 2ℓb

10. The floor of a hall is completely covered by 25 carpets, each measuring 40 m × 2.5 m. What is the area of the floor?

Answer:

10. Area of the floor = 2500 m²

5th Grade Geometry

5th Grade Math Problems

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