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Mensuration - Perimeter of Rectilinear Figures

Grade 6CBSE

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

🔑Concepts

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The perimeter is the total length of the boundary of a closed figure. For any rectilinear figure (made of straight line segments), it is the sum of the lengths of all its sides.

Rectangle diagram showing length and breadth with perimeter formula.
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A regular polygon is a figure where all sides and all angles are equal. The perimeter of a regular polygon with nn sides of length ss is calculated as n×sn \times s.

Regular pentagon with all sides labeled s.
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For irregular rectilinear figures, perimeter is found by identifying each individual segment length and adding them sequentially around the shape.

Irregular L-shaped figure with sides labeled A through F.
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To find the cost of fencing or bordering a field, first calculate the total perimeter and then multiply it by the rate per unit length: TotalCost=Perimeter×RateTotal Cost = Perimeter \times Rate.

📐Formulae

Perimeter of a Rectangle = 2×(length+breadth)2 \times (length + breadth)

Perimeter of a Square = 4×side4 \times side

Perimeter of an Equilateral Triangle = 3×side3 \times side

Perimeter of a Regular Pentagon = 5×side5 \times side

Perimeter of a Regular Hexagon = 6×side6 \times side

Perimeter of any Regular Polygon = n×siden \times side (where nn is the number of sides)

💡Examples

Problem 1:

Find the perimeter of a rectangular field whose length is 150 m150\ m and breadth is 100 m100\ m.

Solution:

Given: Length (ll) = 150 m150\ m, Breadth (bb) = 100 m100\ m Perimeter of rectangle = 2×(l+b)2 \times (l + b) Perimeter = 2×(150+100) m2 \times (150 + 100)\ m Perimeter = 2×250 m2 \times 250\ m Perimeter = 500 m500\ m

Explanation:

Since the field is rectangular, we apply the formula for the perimeter of a rectangle which accounts for two lengths and two breadths.

Problem 2:

A farmer wants to fence a square garden of side 120 m120\ m. If the cost of fencing is Rs 20Rs\ 20 per meter, find the total cost of fencing.

Solution:

Given: Side of square = 120 m120\ m Perimeter of square = 4×side=4×120=480 m4 \times side = 4 \times 120 = 480\ m Cost of fencing per meter = Rs 20Rs\ 20 Total cost = Perimeter×RatePerimeter \times Rate Total cost = 480×20=Rs 9,600480 \times 20 = Rs\ 9,600

Explanation:

Fencing is always done along the boundary, so we first calculate the perimeter of the square garden and then multiply that total length by the rate per meter.

Problem 3:

Find the perimeter of a regular hexagon with each side measuring 8.5 cm8.5\ cm.

Regular hexagon with one side labeled 8.5 cm.

Solution:

Perimeter=6×sidePerimeter = 6 \times side Perimeter=6×8.5 cmPerimeter = 6 \times 8.5\ cm Perimeter=51.0 cmPerimeter = 51.0\ cm

Explanation:

A regular hexagon has 6 equal sides. By multiplying the length of one side by 6, we obtain the total distance around the figure.

Problem 4:

Shikha runs around a square park of side 75 m75\ m. Priya runs around a rectangular park with length 60 m60\ m and breadth 45 m45\ m. Who covers less distance and by how much?

Comparison of a square park and a rectangular park with given dimensions.

Solution:

Distance covered by Shikha = Perimeter of square park =4×75 m=300 m= 4 \times 75\ m = 300\ m

Distance covered by Priya = Perimeter of rectangular park =2×(60 m+45 m)= 2 \times (60\ m + 45\ m) =2×105 m=210 m= 2 \times 105\ m = 210\ m

Difference in distance: 300−21090\begin{array}{r} 300 \\ - 210 \\ \hline 90 \end{array}

Priya covers less distance by 90 m90\ m.

Explanation:

We calculate the perimeters of both shapes. The perimeter represents the distance covered in one complete lap. Comparing the two values shows that Priya's rectangular path is shorter than Shikha's square path.