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Measurement - Perimeter and Area of Irregular Polygons

Grade 5IB

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

🔑Concepts

Irregular Polygons are closed shapes with straight sides where not all side lengths or interior angles are equal. Visualize a floor plan of a room with an 'L' or 'U' shape; unlike a perfect square or rectangle, these shapes have varying side lengths but are still made of straight lines.

The Perimeter is the total distance around the outside of the shape. Imagine placing a string along every edge of the polygon; the total length of that string when stretched out is the perimeter. To calculate it, you must add the lengths of every individual exterior side: P=s1+s2+s3++snP = s_1 + s_2 + s_3 + \dots + s_n.

Finding Missing Side Lengths is a required step when some side measurements are not labeled. In irregular polygons with right angles, the sum of shorter horizontal segments must equal the total length of the long horizontal side opposite them. For example, if the bottom side is 1010 cmcm and the top-left side is 66 cmcm, the remaining top-right horizontal segment must be 106=410 - 6 = 4 cmcm.

The Decomposition Method is the primary strategy for finding the Area of irregular polygons. This involves drawing 'imaginary' dotted lines to split the complex shape into smaller, simpler shapes like rectangles and squares. Imagine cutting an L-shaped piece of paper into two separate rectangles; you find the area of each piece and add them together.

The Subtraction Method is another strategy for calculating Area. Visualize the irregular shape as part of a larger, complete rectangle. You calculate the area of the large 'bounding' rectangle and then subtract the area of the 'empty' space that is missing. This is like calculating the area of a full piece of toast and then subtracting the area of a rectangular bite taken out of the corner.

Correct Units of Measurement are essential for differentiating between perimeter and area. Perimeter is measured in linear units like centimeters (cmcm) or meters (mm). Area measures the surface space inside the boundary and is always expressed in square units, such as square centimeters (cm2cm^2) or square meters (m2m^2). Imagine area as the number of small 1×11 \times 1 unit squares needed to perfectly cover the shape's interior.

📐Formulae

Perimeter(P)=Sum of all side lengthsPerimeter (P) = \text{Sum of all side lengths}

Area of Rectangle=l×wArea \text{ of Rectangle} = l \times w

Area of Square=s2Area \text{ of Square} = s^2

Total Area (Addition)=Area1+Area2Total \text{ Area (Addition)} = Area_1 + Area_2

Total Area (Subtraction)=ArealargeAreamissingTotal \text{ Area (Subtraction)} = Area_{large} - Area_{missing}

💡Examples

Problem 1:

An L-shaped polygon has a total height (left side) of 1212 cmcm, a total width (bottom side) of 1010 cmcm, a top horizontal side of 44 cmcm, and a far-right vertical side of 55 cmcm. Find the Perimeter and the total Area.

Solution:

  1. Find Missing Sides: The inner horizontal side is 104=610 - 4 = 6 cmcm. The inner vertical side is 125=712 - 5 = 7 cmcm. \n 2. Calculate Perimeter: P=12+10+5+6+7+4=44P = 12 + 10 + 5 + 6 + 7 + 4 = 44 cmcm. \n 3. Calculate Area: Split the shape into two rectangles. Rectangle A (left vertical) is 12×4=4812 \times 4 = 48 cm2cm^2. Rectangle B (bottom right) is 5×6=305 \times 6 = 30 cm2cm^2. \n 4. Total Area: 48+30=7848 + 30 = 78 cm2cm^2.

Explanation:

To solve this, we first use the properties of parallel sides to find the lengths of the two 'inner' sides. Then, we add all six sides for the perimeter. For the area, we decompose the shape into two non-overlapping rectangles and sum their individual areas.

Problem 2:

Find the area of a large rectangle measuring 1515 mm by 1010 mm that has a smaller 33 mm by 44 mm rectangular piece removed from one corner.

Solution:

  1. Area of Large Rectangle: Alarge=15×10=150A_{large} = 15 \times 10 = 150 m2m^2. \n 2. Area of Missing Piece: Amissing=3×4=12A_{missing} = 3 \times 4 = 12 m2m^2. \n 3. Total Area: 15012=138150 - 12 = 138 m2m^2.

Explanation:

This problem is best solved using the Subtraction Method. By treating the shape as a complete rectangle first and then removing the 'hole', we find the remaining surface area efficiently.