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Area and Perimeter - Relating Area and Boundary

Grade 5CBSE

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

🔑Concepts

Perimeter is the total length of the boundary of a closed figure. Imagine taking a piece of string and laying it exactly along the edges of a shape; the length of that string is the perimeter. It is measured in linear units like cmcm, mm, or kmkm.

Area represents the amount of surface or 'space' covered by a closed shape. Imagine a floor covered in identical square tiles; the number of tiles needed to cover the entire floor is its area. It is measured in square units such as sq cmsq\ cm or cm2cm^{2}.

Rectangles and squares are standard shapes where area and perimeter can be calculated using their dimensions. A rectangle has a length (ll) and a width (ww), while a square has four equal sides (ss). Visually, the area of a rectangle is the number of unit squares that fit into its grid of rows×columnsrows \times columns.

Different shapes can have the same perimeter but different areas. For example, a rectangle with sides 6 cm6\ cm and 4 cm4\ cm has a perimeter of 20 cm20\ cm and an area of 24 sq cm24\ sq\ cm. However, a very thin rectangle with sides 9 cm9\ cm and 1 cm1\ cm also has a perimeter of 20 cm20\ cm, but its area is only 9 sq cm9\ sq\ cm.

Conversely, shapes with the same area can have different perimeters. If you have 12 square tiles, you can arrange them in a 3×43 \times 4 block (perimeter 14 units14\ units) or a long 1×121 \times 12 line (perimeter 26 units26\ units). The more 'stretched out' a shape is, the larger its perimeter tends to be for the same area.

For irregular shapes drawn on a squared paper (grid), we estimate the area by counting the number of full squares and squares that are more than half-filled. We usually ignore squares that are less than half-filled. The perimeter is found by counting the number of unit segments along the outer edge of the shape.

The concept of 'Boundary' relates directly to fencing. If a problem asks how much wire is needed to go around a park, it is asking for the Perimeter. If it asks how much grass is needed to cover the park, it is asking for the Area.

📐Formulae

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

Area of a Rectangle=length×widthArea\ of\ a\ Rectangle = length \times width

Perimeter of a Square=4×sidePerimeter\ of\ a\ Square = 4 \times side

Area of a Square=side×sideArea\ of\ a\ Square = side \times side

Side of a Square=Perimeter4Side\ of\ a\ Square = \frac{Perimeter}{4}

Length of a Rectangle=AreawidthLength\ of\ a\ Rectangle = \frac{Area}{width}

💡Examples

Problem 1:

A rectangular garden has a length of 12 m12\ m and a width of 8 m8\ m. Find the cost of fencing it at the rate of Rs 50Rs\ 50 per meter and the cost of turfing (covering with grass) the garden at Rs 20Rs\ 20 per square meter.

Solution:

Step 1: Find the Perimeter for fencing. Perimeter=2×(length+width)Perimeter = 2 \times (length + width) Perimeter=2×(12+8)=2×20=40 mPerimeter = 2 \times (12 + 8) = 2 \times 20 = 40\ m Cost of fencing = 40×50=Rs 2,00040 \times 50 = Rs\ 2,000

Step 2: Find the Area for turfing. Area=length×widthArea = length \times width Area=12×8=96 sq mArea = 12 \times 8 = 96\ sq\ m Cost of turfing = 96×20=Rs 1,92096 \times 20 = Rs\ 1,920

Explanation:

Fencing is related to the boundary (Perimeter), while turfing covers the surface (Area). We calculate both using standard rectangle formulas and then multiply by their respective rates.

Problem 2:

A square shaped stamp has a side of 3 cm3\ cm. How many such stamps can be placed on a big rectangular sheet of paper that is 15 cm15\ cm long and 12 cm12\ cm wide?

Solution:

Step 1: Find the Area of one stamp. Area of stamp=side×side=3 cm×3 cm=9 sq cmArea\ of\ stamp = side \times side = 3\ cm \times 3\ cm = 9\ sq\ cm

Step 2: Find the Area of the rectangular sheet. Area of sheet=length×width=15 cm×12 cm=180 sq cmArea\ of\ sheet = length \times width = 15\ cm \times 12\ cm = 180\ sq\ cm

Step 3: Calculate the number of stamps. Number of stamps=Area of sheetArea of stamp=1809=20Number\ of\ stamps = \frac{Area\ of\ sheet}{Area\ of\ stamp} = \frac{180}{9} = 20 Total stamps = 2020

Explanation:

To find how many smaller shapes fit into a larger one, we divide the total area of the larger shape by the area of the smaller shape.