Area (Mensuration)

Area of triangles, circles, polygons

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Mensuration is the branch of mathematics that deals with the measurement of geometric figures, including their areas, perimeters, and volumes. It involves calculating the space occupied by 2D shapes (like triangles, rectangles, circles) and understanding the relationships between different measurements. Mensuration problems commonly appear in aptitude tests and require knowledge of standard formulas and their applications.

Key Formulas

Rectangle: Area = length x breadth, Perimeter = 2(length + breadth)\text{Rectangle: Area = length x breadth, Perimeter = 2(length + breadth)}
Square: Area = side^2, Perimeter = 4 x side\text{Square: Area = side\textasciicircum{}2, Perimeter = 4 x side}
Triangle: Area = x base x height = sqrt[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2 (Heron’s formula)\text{Triangle: Area = x base x height = sqrt[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2 (Heron's formula)}
Circle: Area = pir^2, Circumference = 2pir = pid\text{Circle: Area = pir\textasciicircum{}2, Circumference = 2pir = pid}
Parallelogram: Area = base x height, Perimeter = 2(sum of adjacent sides)\text{Parallelogram: Area = base x height, Perimeter = 2(sum of adjacent sides)}
Rhombus: Area = x d x d (product of diagonals), Perimeter = 4 x side\text{Rhombus: Area = x d x d (product of diagonals), Perimeter = 4 x side}
Trapezium: Area = x (sum of parallel sides) x height\text{Trapezium: Area = x (sum of parallel sides) x height}
Sector of Circle: Area = ( /360 deg) x pir^2, Arc length = ( /360 deg) x 2pir\text{Sector of Circle: Area = ( /360 deg) x pir\textasciicircum{}2, Arc length = ( /360 deg) x 2pir}
Semicircle: Area = pir^2, Perimeter = pir + 2r = r(pi + 2)\text{Semicircle: Area = pir\textasciicircum{}2, Perimeter = pir + 2r = r(pi + 2)}
Ring (Annulus): Area = pi(R^2 - r^2) where R = outer radius, r = inner radius\text{Ring (Annulus): Area = pi(R\textasciicircum{}2 - r\textasciicircum{}2) where R = outer radius, r = inner radius}
Equilateral Triangle: Area = (sqrt3/4) x side^2, Height = (sqrt3/2) x side\text{Equilateral Triangle: Area = (sqrt3/4) x side\textasciicircum{}2, Height = (sqrt3/2) x side}
Isosceles Triangle: Area = (b/4) x sqrt(4a^2 - b^2) where a = equal sides, b = base\text{Isosceles Triangle: Area = (b/4) x sqrt(4a\textasciicircum{}2 - b\textasciicircum{}2) where a = equal sides, b = base}

Key Concepts

Area vs. Perimeter

Area measures the region enclosed within a shape (in square units), while perimeter measures the total boundary length (in linear units). For the same perimeter, a circle has the maximum area. Understanding when to use area versus perimeter formulas is crucial - area is needed for coverage problems (painting, carpeting) while perimeter is needed for boundary problems (fencing, framing).

Units and Conversions

Always ensure all measurements are in the same unit before calculating. Common conversions: 1 m = 100 cm = 1000 mm; 1 km = 1000 m; 1 hectare = 10,000 m^2 = 100 ares; 1 acre = 4046.86 m^2. Area conversions use square factors: 1 m^2 = 10,000 cm^2, not 100 cm^2. Volume conversions use cubic factors.

Composite Figures

Complex shapes can be broken down into simpler standard shapes. Calculate the required measurement (area/perimeter) for each component and combine appropriately. For area of composite figures: add areas of components that make up the shape. For shaded region problems: subtract the unshaded area from total area.

Pathway Problems

When a path runs around/inside a garden or field: Outer dimension minus path width gives inner dimension. For a path of uniform width 'w' around a rectangle of dimensions L x B: Inner rectangle dimensions are (L-2w) x (B-2w). Path area = Outer area - Inner area = 2w(L + B - 2w).

Cost Calculation Problems

To find cost of painting, fencing, or paving: First calculate the required area or perimeter, then multiply by the rate per unit. Example: Cost to paint a wall = (Area of wall) x (Rate per m^2). For cylindrical tanks or circular paths, be careful whether you need curved surface area or total surface area.

Ratio of Areas

When comparing similar figures or figures with related dimensions: If linear dimensions are in ratio a:b, then areas are in ratio a^2:b^2, and perimeters are in ratio a:b. Example: If two circles have radii in ratio 2:3, their areas are in ratio 4:9, and their circumferences are in ratio 2:3.

Solved Examples

Problem 1:

The length and breadth of a rectangle are in the ratio 5:3. If its perimeter is 128 m, find its area.

Solution:

Given: Length : Breadth = 5 : 3
Let length = 5x and breadth = 3x
Perimeter = 2(length + breadth) = 2(5x + 3x) = 2(8x) = 16x
Given perimeter = 128 m
So 16x = 128
x = 8
Length = 5x = 5 x 8 = 40 m
Breadth = 3x = 3 x 8 = 24 m
Area = length x breadth = 40 x 24 = 960 m^2

Problem 2:

Find the area of a triangle whose sides are 13 cm, 14 cm, and 15 cm.

Solution:

Using Heron's formula:
First calculate semi-perimeter s = (13 + 14 + 15)/2 = 42/2 = 21 cm
Area = sqrt[s(s-a)(s-b)(s-c)]
= sqrt[21 x (21-13) x (21-14) x (21-15)]
= sqrt[21 x 8 x 7 x 6]
= sqrt[7056]
= 84 cm^2

Problem 3:

The area of a rhombus is 216 cm^2. If one diagonal is 24 cm, find the other diagonal and the side of the rhombus.

Solution:

Given: Area = 216 cm^2, d = 24 cm
Area of rhombus = x d x d
216 = x 24 x d
216 = 12 x d
d = 216/12 = 18 cm

For a rhombus, diagonals bisect each other at right angles.
Each half of diagonals: 24/2 = 12 cm and 18/2 = 9 cm
Using Pythagoras theorem:
Side^2 = 12^2 + 9^2 = 144 + 81 = 225
Side = sqrt225 = 15 cm

Answer: Other diagonal = 18 cm, Side = 15 cm

Problem 4:

A circular park has a circumference of 440 m. A 7 m wide road runs around it. Find the area of the road.

Solution:

Given: Circumference of park = 440 m
2pir = 440
r = 440/(2pi) = 440/(2 x 22/7) = (440 x 7)/44 = 70 m

Inner radius (park) = 70 m
Outer radius (including road) = 70 + 7 = 77 m

Area of road = pi(R^2 - r^2) = pi(77^2 - 70^2)
= pi(77 + 70)(77 - 70) [using a^2 - b^2 = (a+b)(a-b)]
= pi x 147 x 7
= (22/7) x 147 x 7
= 22 x 147
= 3234 m^2

Tips & Tricks

  • Memorize all standard formulas for 2D shapes thoroughly. Create a formula sheet for quick revision.
  • Always verify units are consistent before calculating. Convert all measurements to the same unit first.
  • For composite figures, carefully identify which areas to add and which to subtract. Drawing a diagram helps visualize.
  • Remember that pi ~= 22/7 is a good approximation when calculations result in whole numbers, but use pi ~= 3.14 or calculator value for precision.
  • For pathway problems, be careful whether the path is inside (reducing dimensions) or outside (increasing dimensions) the main area.
  • When given ratios, use the technique of assuming actual values as ratio multiples (like 5x, 3x) and solve for x.
  • For cost-based problems, verify whether you're calculating cost based on area or perimeter before multiplying by the rate.
  • In right triangles, you can use x base x height, but for general triangles with three sides given, Heron's formula is essential.
  • For circle problems involving sectors and segments, always identify the central angle and use the appropriate fraction ( /360 deg) of the full circle values.
  • Double-check your final answer by estimating - if a rectangle has sides around 10 m each, the area should be around 100 m^2, not 1000 m^2.

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