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Perimeter and Area - Perimeter

Grade 6CBSE

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

🔑Concepts

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Perimeter is the total length of the boundary of a closed figure. It is the distance covered while going around the figure once.

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The perimeter of a polygon is the sum of the lengths of all its sides.

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For regular polygons (where all sides and angles are equal), the perimeter can be calculated by multiplying the length of one side by the total number of sides.

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The units of perimeter are the same as the units of length, such as mmmm, cmcm, mm, or kmkm.

📐Formulae

P=Sum of all sidesP = \text{Sum of all sides}

Prectangle=2×(l+b)P_{\text{rectangle}} = 2 \times (l + b) (where ll is length and bb is breadth)

Psquare=4×sP_{\text{square}} = 4 \times s (where ss is the side length)

Pequilateral triangle=3×sP_{\text{equilateral triangle}} = 3 \times s

Pregular pentagon=5×sP_{\text{regular pentagon}} = 5 \times s

Pregular hexagon=6×sP_{\text{regular hexagon}} = 6 \times s

💡Examples

Problem 1:

Find the perimeter of a rectangle whose length is 15 cm15\text{ cm} and breadth is 10 cm10\text{ cm}.

Solution:

P=2×(l+b)P = 2 \times (l + b) P=2×(15+10)P = 2 \times (15 + 10) P=2×25P = 2 \times 25 P=50 cmP = 50\text{ cm}

Explanation:

To find the perimeter of a rectangle, we add the length and breadth together and then multiply the result by 22 because a rectangle has two lengths and two breadths.

Problem 2:

The perimeter of a square park is 144 m144\text{ m}. Find the length of each side of the park.

Solution:

P=4×sP = 4 \times s 144=4×s144 = 4 \times s s=1444s = \frac{144}{4} s=36 ms = 36\text{ m}

Explanation:

Since a square has four equal sides, we divide the total perimeter by 44 to find the length of a single side.

Problem 3:

Calculate the perimeter of a regular hexagon with each side measuring 8.5 cm8.5\text{ cm}.

Solution:

P=6×sP = 6 \times s P=6×8.5P = 6 \times 8.5 P=51.0 cmP = 51.0\text{ cm}

Explanation:

A regular hexagon has 66 equal sides. Multiplying the length of one side (8.5 cm8.5\text{ cm}) by 66 gives the total perimeter.

Problem 4:

Find the cost of fencing a rectangular field of length 250 m250\text{ m} and breadth 150 m150\text{ m} at the rate of Rs 2020 per metre.

Solution:

First, find the perimeter: P=2×(250+150)P = 2 \times (250 + 150) P=2×400=800 mP = 2 \times 400 = 800\text{ m} Now, find the cost: Cost=800×20\text{Cost} = 800 \times 20 Cost=Rs 16000\text{Cost} = \text{Rs } 16000

Explanation:

Fencing is done along the boundary, so we first calculate the perimeter. Then, we multiply the total perimeter by the cost per unit length to find the total expenditure.