krit.club logo

Fields and Fences - Fields and Fences

Grade 4CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The perimeter is the total length of the boundary of a closed shape. It is the distance around the outside of the shape.

•

To find the perimeter of any shape, we sum the lengths of all its sides.

•

For a square, all four sides are equal in length. Therefore, the perimeter is 44 times the length of one side.

•

For a rectangle, opposite sides are equal. The perimeter is calculated by adding the length and width and then multiplying the sum by 22.

•

Units of perimeter include centimeters (cmcm), meters (mm), and kilometers (kmkm).

•

If a shape is drawn on a grid of 1 cm1\text{ cm} squares, the perimeter can be found by counting the number of 1 cm1\text{ cm} segments along the boundary.

📐Formulae

Perimeter of a Square=4×side\text{Perimeter of a Square} = 4 \times \text{side}

Perimeter of a Rectangle=2×(Length+Breadth)\text{Perimeter of a Rectangle} = 2 \times (\text{Length} + \text{Breadth})

Perimeter of a Triangle=Side1+Side2+Side3\text{Perimeter of a Triangle} = \text{Side}_1 + \text{Side}_2 + \text{Side}_3

Perimeter of any polygon=Sum of all side lengths\text{Perimeter of any polygon} = \text{Sum of all side lengths}

💡Examples

Problem 1:

A rectangular field has a length of 25 m25\text{ m} and a width of 15 m15\text{ m}. What is the length of the fence required to cover the boundary of the field?

Solution:

Perimeter=2×(25+15)=2×40=80 m\text{Perimeter} = 2 \times (25 + 15) = 2 \times 40 = 80\text{ m}

Explanation:

Since the fence goes around the boundary, we need to find the perimeter of the rectangle. Using the formula 2×(L+B)2 \times (L + B), we add 2525 and 1515 to get 4040, then multiply by 22 to get 80 m80\text{ m}.

Problem 2:

Find the perimeter of a square garden where each side measures 12 m12\text{ m}.

Solution:

Perimeter=4×12=48 m\text{Perimeter} = 4 \times 12 = 48\text{ m}

Explanation:

A square has four equal sides. To find the boundary length, we multiply the length of one side (12 m12\text{ m}) by 44.

Problem 3:

Ganpat's field is shaped like a triangle with sides 15 m15\text{ m}, 15 m15\text{ m}, and 20 m20\text{ m}. Calculate the total boundary length using vertical addition.

Solution:

1515+2050\begin{array}{r} 15 \\ 15 \\ + 20 \\ \hline 50 \end{array} The perimeter is 50 m50\text{ m}.

Explanation:

The perimeter of a triangle is the sum of its three sides. Adding 1515, 1515, and 2020 gives a total boundary length of 50 m50\text{ m}.