Introduction to Mensuration: Perimeter and Area Concepts
Mensuration: Definition and Scope
Mensuration is a branch of geometry within mathematics that deals with the measurement of geometric figures. It primarily focuses on calculating lengths, areas, and volumes of various shapes, both two-dimensional (plane figures) and three-dimensional (solid figures).
The term "Mensuration" originates from the Latin word 'mensura', which means 'measure'.
Scope of Mensuration
Mensuration covers a wide range of calculations related to the size and dimensions of shapes and objects. These include:
- Perimeter: The total length of the boundary of a closed two-dimensional figure. It's essentially the distance you would travel if you walked along the edges of the shape and returned to your starting point.
- Area: The measure of the surface or region enclosed by a closed two-dimensional figure. It quantifies how much space a flat shape covers on a plane.
- Surface Area: This applies to three-dimensional objects (solids) and is the total area of all the surfaces of the object. It can be further divided:
- Lateral Surface Area (LSA) or Curved Surface Area (CSA): The area of the side faces or curved surfaces of a solid, excluding the area of the top and bottom bases. For example, the area of the label on a cylindrical can is the LSA/CSA.
- Total Surface Area (TSA): The sum of the areas of all surfaces of a solid, including the bases (if any) and the lateral/curved surface. For example, the TSA of a closed box would be the area of all six faces.
- Volume: The measure of the space occupied by a three-dimensional object. It tells us how much a solid can hold or how much space it displaces. Volume is measured in cubic units.
Importance and Applications of Mensuration
Mensuration is not just an academic topic; it has immense practical importance across numerous fields and in everyday life:
- Architecture and Civil Engineering: Crucial for designing buildings, bridges, roads, and other structures. Calculations of area help determine floor space, land usage, and the amount of material like concrete, steel, and bricks required. Volume calculations are needed for concrete mixes, earthwork, etc.
- Construction Industry: Used extensively for estimating costs, determining the amount of paint needed for walls, tiles for floors, fencing for boundaries, and calculating the capacity of storage tanks or pits.
- Manufacturing: Essential for designing product dimensions, calculating the volume of packaging required, estimating the material needed to produce items, and determining the capacity of containers.
- Agriculture: Helps in measuring land area for cultivation, calculating irrigation needs (volume of water), and estimating yield per unit area.
- Interior Design: Used for calculating the area of rooms to determine carpeting, flooring, or painting requirements, as well as volume for space planning.
- Physics: Used in calculations related to density (mass per unit volume), pressure (force per unit area), and fluid dynamics.
- Everyday Life: Simple tasks like calculating the amount of cloth needed for a dress, the area of a room to buy a carpet, the volume of liquid in a container, or the distance around a sports field all involve mensuration.
In essence, mensuration provides the foundational quantitative tools to understand and interact with the physical space around us by quantifying the size, extent, and capacity of objects and regions.
Perimeter of Plane Figures: Definition and Basic Understanding
The perimeter of a closed two-dimensional (plane) figure is defined as the total length of its boundary. It represents the distance around the outside edge of the shape. The word "perimeter" comes from the Greek words 'peri' (around) and 'metron' (measure).
Think of it as the length you would measure if you were to trace the outline of the figure with a string and then measure the length of the string.
For example, if you have a picture frame, the perimeter is the total length of the wood or material used for the frame around the picture. If you are putting a decorative border around a tablecloth, the length of the border is the perimeter of the tablecloth.
Key Concepts
- Applicability: Perimeter is a concept applicable only to closed figures. Open figures do not enclose a region and therefore do not have a perimeter in the same sense. It is primarily calculated for polygons (shapes made of straight line segments like triangles, squares, rectangles, pentagons, hexagons, etc.) and circles. For a circle, the perimeter is given a special name: the circumference.
- Calculation:
- For any polygon, the perimeter is found by summing the lengths of all its sides.
- For a circle, the circumference is calculated using a specific formula involving its radius or diameter.
- Units: Perimeter is a measure of length. Therefore, it is expressed in linear units. The choice of unit depends on the size of the figure being measured. Common units include millimetres (mm), centimetres (cm), metres (m), kilometres (km), inches (in), feet (ft), etc. It is crucial to use consistent units throughout a calculation. For instance, if side lengths are given in cm, the perimeter will be in cm.
Examples
Example 1. A square garden has a side length of $8.5$ metres. A gardener wants to put a fence around it. What is the total length of the fence required?
Answer:
Given:
Side length of the square garden ($s$) $= 8.5$ m.
To Find:
The total length of the fence required, which is the perimeter of the square garden.
Solution:
A square has four equal sides. The perimeter of a square is the sum of the lengths of its four sides. The formula is:
$\text{Perimeter} = s + s + s + s = 4 \times s$
... (i)
Substitute the given side length into the formula:
$\text{Perimeter} = 4 \times 8.5 \$ \text{m}$
[Substitute $s=8.5$ m in (i)]
Now, perform the multiplication:
$\text{Perimeter} = 34.0 \$ \text{m}$
Thus, the total length of the fence required is $34$ metres.
Example 2. Find the perimeter of a rectangle whose length is $25$ cm and width is $18$ cm.
Answer:
Given:
Length of the rectangle ($l$) $= 25$ cm.
Width of the rectangle ($w$) $= 18$ cm.
To Find:
Perimeter of the rectangle.
Solution:
A rectangle has four sides, with opposite sides being equal in length. The perimeter is the sum of the lengths of its two lengths and two widths. The formula for the perimeter of a rectangle is:
$\text{Perimeter} = l + w + l + w = 2l + 2w = 2(l+w)$
... (i)
Substitute the given length and width into the formula:
$\text{Perimeter} = 2 \times (25 \$ \text{cm} + 18 \$ \text{cm})$
[Substitute $l=25$ cm and $w=18$ cm in (i)]
$\text{Perimeter} = 2 \times 43 \$ \text{cm}$
$\text{Perimeter} = 86 \$ \text{cm}$
The perimeter of the rectangle is $86$ cm.
Example 3. A triangular park has sides measuring $200$ m, $150$ m, and $220$ m. Find the perimeter of the park.
Answer:
Given:
Side lengths of the triangular park: $a = 200$ m, $b = 150$ m, $c = 220$ m.
To Find:
Perimeter of the triangular park.
Solution:
The perimeter of any polygon is the sum of the lengths of its sides. For a triangle with side lengths $a$, $b$, and $c$, the perimeter is:
$\text{Perimeter} = a + b + c$
... (i)
Substitute the given side lengths into the formula:
$\text{Perimeter} = 200 \$ \text{m} + 150 \$ \text{m} + 220 \$ \text{m}$
[Substitute values in (i)]
$\text{Perimeter} = (200 + 150 + 220) \$ \text{m}$
$\text{Perimeter} = 570 \$ \text{m}$
The perimeter of the triangular park is $570$ metres.
Example 4. A regular pentagon has a side length of $6$ cm. What is its perimeter?
Answer:
Given:
A regular pentagon with side length ($s$) $= 6$ cm.
To Find:
Perimeter of the regular pentagon.
Solution:
A pentagon is a polygon with $5$ sides. A regular pentagon has $5$ equal sides. The perimeter is the sum of the lengths of its sides.
$\text{Perimeter} = s + s + s + s + s = 5 \times s$
... (i)
Substitute the given side length:
$\text{Perimeter} = 5 \times 6 \$ \text{cm}$
[Substitute $s=6$ cm in (i)]
$\text{Perimeter} = 30 \$ \text{cm}$
The perimeter of the regular pentagon is $30$ cm.
In summary, finding the perimeter of a polygon involves simply adding up the lengths of all its sides. For curved shapes like circles, specific formulas are used, which will be covered in later sections.
Area of Plane Figures: Definition and Basic Understanding
The area of a closed two-dimensional (plane) figure is the measure of the surface or region enclosed within its boundary. It quantifies the amount of space the shape occupies on a flat plane. Area is a fundamental concept in geometry and has numerous practical applications.
Imagine the surface inside the boundary of a figure. The area is the measure of this surface. For example:
- If you are covering a floor with tiles, the area of the floor determines how many tiles are needed.
- If you are painting a wall, the area of the wall determines the amount of paint required.
- If you are buying land, the size of the plot is typically described by its area.
Key Concepts
- Applicability: Area is calculated for closed two-dimensional figures such as triangles, squares, rectangles, circles, parallelograms, trapeziums, kites, rhombuses, and other polygons, as well as combinations of these shapes.
- Measurement: Area measures the extent of a surface. It is a two-dimensional measurement, unlike perimeter, which is a one-dimensional measure of length.
- Units: Area is always measured in square units. This is because the standard way to quantify area is to determine how many unit squares (squares with side length 1 unit, e.g., 1 cm by 1 cm) can fit inside the figure without overlapping. Common units include square millimetres ($\text{mm}^2$), square centimetres ($\text{cm}^2$), square metres ($\text{m}^2$), square kilometres ($\text{km}^2$), hectares (ha), acres, square inches ($\text{in}^2$ or sq in), square feet ($\text{ft}^2$ or sq ft), etc. The superscript '2' indicates that it is a square unit derived from a linear unit.
Contrast with Perimeter
It is extremely important not to confuse area and perimeter. While both are measurements related to a closed 2D shape, they quantify different properties:
- Perimeter is the 1-dimensional measure of the length of the boundary (unit: length, e.g., m, cm).
- Area is the 2-dimensional measure of the size of the enclosed surface (unit: area, e.g., $\text{m}^2$, $\text{cm}^2$).
A key distinction is that two different shapes can have the same perimeter but vastly different areas, and vice versa.
Examples Illustrating Area and its Distinction from Perimeter
Example 1. Find the area of a rectangle with length $10$ cm and width $5$ cm. Also, find its perimeter.
Answer:
Given:
Length of the rectangle ($l$) $= 10$ cm.
Width of the rectangle ($w$) $= 5$ cm.
To Find:
Area and Perimeter of the rectangle.
Solution:
The area of a rectangle is calculated by multiplying its length by its width.
$\text{Area} = \text{Length} \times \text{Width} = l \times w$
... (i)
Substitute the given values:
$\text{Area} = 10 \$ \text{cm} \times 5 \$ \text{cm}$
[Substitute $l=10$ cm, $w=5$ cm in (i)]
$\text{Area} = 50 \$ \text{cm}^2$
The perimeter of a rectangle is calculated as twice the sum of its length and width.
$\text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2(l+w)$
... (ii)
Substitute the given values:
$\text{Perimeter} = 2 \times (10 \$ \text{cm} + 5 \$ \text{cm})$
[Substitute $l=10$ cm, $w=5$ cm in (ii)]
$\text{Perimeter} = 2 \times 15 \$ \text{cm}$
$\text{Perimeter} = 30 \$ \text{cm}$
The area of the rectangle is $50 \$ \text{cm}^2$ and its perimeter is $30 \$ \text{cm}$. Note the different units: $\text{cm}^2$ for area and cm for perimeter.
Example 2. Show that two rectangles can have the same perimeter but different areas.
Answer:
Consider two different rectangles:
Rectangle 1:
Length ($l_1$) $= 6$ m
Width ($w_1$) $= 4$ m
Perimeter ($P_1$) $= 2(l_1 + w_1) = 2(6 + 4) = 2(10) = 20$ m
Area ($A_1$) $= l_1 \times w_1 = 6 \times 4 = 24 \$ \text{m}^2$
Rectangle 2:
Length ($l_2$) $= 7$ m
Width ($w_2$) $= 3$ m
Perimeter ($P_2$) $= 2(l_2 + w_2) = 2(7 + 3) = 2(10) = 20$ m
Area ($A_2$) $= l_2 \times w_2 = 7 \times 3 = 21 \$ \text{m}^2$
Observation:
$\text{Perimeter of Rectangle 1} = 20 \$ \text{m}$
$\text{Perimeter of Rectangle 2} = 20 \$ \text{m}$
$\text{Area of Rectangle 1} = 24 \$ \text{m}^2$
$\text{Area of Rectangle 2} = 21 \$ \text{m}^2$
Conclusion:
Both rectangles have the same perimeter ($20$ m), but their areas are different ($24 \$ \text{m}^2$ and $21 \$ \text{m}^2$). This demonstrates that perimeter and area are independent properties of a shape.
Understanding the definition and concept of area, along with its distinction from perimeter and the use of square units, is fundamental before proceeding to the specific formulas for different geometric figures.
Units of Measurement for Perimeter and Area
Choosing the correct units and understanding conversions between them is essential for accurate calculations in mensuration. The unit used must be appropriate for the quantity being measured (length for perimeter, area for surface) and should be consistent within any given problem.
Units for Perimeter (Length)
Perimeter measures linear distance along a boundary. The standard SI unit is the metre (m).
Common units for length include:
- Metric System: Millimetre (mm), Centimetre (cm), Decimetre (dm), Metre (m), Decametre (dam), Hectometre (hm), Kilometre (km).
- Imperial System: Inch (in or "), Foot (ft or '), Yard (yd), Mile (mi).
Basic Metric Conversions (Length):
- $1 \$ \text{cm} = 10 \$ \text{mm}$
- $1 \$ \text{m} = 100 \$ \text{cm} = 1000 \$ \text{mm}$
- $1 \$ \text{km} = 1000 \$ \text{m}$
Conversions often follow powers of 10:
Unit | Symbol | Relationship to Metre |
---|---|---|
Millimetre | mm | $10^{-3}$ m ($0.001$ m) |
Centimetre | cm | $10^{-2}$ m ($0.01$ m) |
Decimetre | dm | $10^{-1}$ m ($0.1$ m) |
Metre | m | $1$ m |
Decametre | dam | $10^{1}$ m ($10$ m) |
Hectometre | hm | $10^{2}$ m ($100$ m) |
Kilometre | km | $10^{3}$ m ($1000$ m) |
Basic Imperial Conversions (Length):
- $1 \$ \text{foot (ft)} = 12 \$ \text{inches (in)}$
- $1 \$ \text{yard (yd)} = 3 \$ \text{feet (ft)} = 3 \times 12 = 36 \$ \text{inches (in)}$
- $1 \$ \text{mile (mi)} = 1760 \$ \text{yards (yd)} = 1760 \times 3 = 5280 \$ \text{feet (ft)}$
Metric-Imperial Conversions (Approximate):
- $1 \$ \text{inch} \approx 2.54 \$ \text{cm}$
- $1 \$ \text{foot} \approx 30.48 \$ \text{cm} \approx 0.3048 \$ \text{m}$
- $1 \$ \text{yard} \approx 0.9144 \$ \text{m}$
- $1 \$ \text{mile} \approx 1.60934 \$ \text{km}$
- $1 \$ \text{cm} \approx 0.3937 \$ \text{ inches}$
- $1 \$ \text{m} \approx 3.281 \$ \text{ feet} \approx 1.094 \$ \text{ yards}$
- $1 \$ \text{km} \approx 0.62137 \$ \text{ miles}$
Units for Area (Surface)
Area measures the extent of a surface and is expressed in square units. These units are derived by squaring the corresponding linear units.
Common units for area include:
- Metric System: Square millimetre ($\text{mm}^2$), Square centimetre ($\text{cm}^2$), Square metre ($\text{m}^2$), Are (a), Hectare (ha), Square kilometre ($\text{km}^2$).
- Imperial System: Square inch ($\text{in}^2$ or sq in), Square foot ($\text{ft}^2$ or sq ft), Square yard ($\text{yd}^2$ or sq yd), Acre, Square mile ($\text{mi}^2$ or sq mi).
Metric Conversions (Area):
Conversions for area are based on squaring the length conversion factors.
- Since $1 \$ \text{cm} = 10 \$ \text{mm}$, then:
$\text{1 cm}^2 = (10 \$ \text{mm}) \times (10 \$ \text{mm}) = 100 \$ \text{mm}^2$
[Squaring both sides]
- Since $1 \$ \text{m} = 100 \$ \text{cm}$, then:
$\text{1 m}^2 = (100 \$ \text{cm}) \times (100 \$ \text{cm}) = 10,000 \$ \text{cm}^2$
[Squaring both sides]
- Since $1 \$ \text{km} = 1000 \$ \text{m}$, then:
$\text{1 km}^2 = (1000 \$ \text{m}) \times (1000 \$ \text{m}) = 1,000,000 \$ \text{m}^2$
[Squaring both sides]
- Are (a): A unit often used in land measurement.
$\text{1 are} = 10 \$ \text{m} \times 10 \$ \text{m} = 100 \$ \text{m}^2$
- Hectare (ha): A very common unit for larger land areas.
$\text{1 hectare (ha)} = 100 \$ \text{m} \times 100 \$ \text{m} = 10,000 \$ \text{m}^2$
Note the relationship between hectare and $\text{km}^2$:
$\text{1 km}^2 = 100 \$ \text{ha}$
Imperial Conversions (Area):
- Since $1 \$ \text{ft} = 12 \$ \text{in}$, then:
$\text{1 ft}^2 = (12 \$ \text{in}) \times (12 \$ \text{in}) = 144 \$ \text{in}^2$
- Since $1 \$ \text{yd} = 3 \$ \text{ft}$, then:
$\text{1 yd}^2 = (3 \$ \text{ft}) \times (3 \$ \text{ft}) = 9 \$ \text{ft}^2$
- Acre: A common imperial unit for land area.
$\text{1 acre} = 4840 \$ \text{yd}^2$
Note the relationship between acre and $\text{mi}^2$:
$\text{1 sq mi} = 640 \$ \text{acres}$
Metric-Imperial Conversions (Area - Approximate):
- $1 \$ \text{in}^2 \approx 6.4516 \$ \text{cm}^2$
- $1 \$ \text{ft}^2 \approx 0.092903 \$ \text{m}^2$
- $1 \$ \text{yd}^2 \approx 0.836127 \$ \text{m}^2$
- $1 \$ \text{acre} \approx 0.404686 \$ \text{ha} \approx 4046.86 \$ \text{m}^2$
- $1 \$ \text{mi}^2 \approx 2.58999 \$ \text{km}^2$
- $1 \$ \text{cm}^2 \approx 0.155 \$ \text{ in}^2$
- $1 \$ \text{m}^2 \approx 10.7639 \$ \text{ ft}^2 \approx 1.19599 \$ \text{ yd}^2$
- $1 \$ \text{ha} \approx 2.47105 \$ \text{ acres}$
Consistency of Units
It is absolutely crucial to use consistent units throughout a calculation. If the dimensions of a figure are given in different units (e.g., length in metres and width in centimetres), you must convert them all to a single unit (either metres or centimetres) before applying any formula for perimeter or area.
Example 1. Find the area of a rectangular tabletop with length $1.5$ m and width $75$ cm.
Answer:
Given:
Length ($l$) $= 1.5$ m.
Width ($w$) $= 75$ cm.
To Find:
Area of the rectangular tabletop.
Solution:
The units for length and width are different (metres and centimetres). We need to convert one of them so that both are in the same unit.
Option 1: Convert to centimetres (cm)
Length ($l$) $= 1.5 \$ \text{m} = 1.5 \times 100 \$ \text{cm} = 150 \$ \text{cm}$
Width ($w$) $= 75 \$ \text{cm}$
Now, calculate the area using the formula $\text{Area} = l \times w$:
$\text{Area} = 150 \$ \text{cm} \times 75 \$ \text{cm}$
$\text{Area} = 11250 \$ \text{cm}^2$
Option 2: Convert to metres (m)
Length ($l$) $= 1.5 \$ \text{m}$
Width ($w$) $= 75 \$ \text{cm} = \frac{75}{100} \$ \text{m} = 0.75 \$ \text{m}$
Now, calculate the area using the formula $\text{Area} = l \times w$:
$\text{Area} = 1.5 \$ \text{m} \times 0.75 \$ \text{m}$
$\text{Area} = 1.125 \$ \text{m}^2$
Both results are correct and equivalent, as $1.125 \$ \text{m}^2 = 1.125 \times 10000 \$ \text{cm}^2 = 11250 \$ \text{cm}^2$. The key is to ensure units are consistent before calculation.
Understanding these units and conversions is vital for solving mensuration problems accurately. Always pay attention to the units given in the problem and the units required for the final answer, performing conversions as necessary.