Topic

Area

Area - Methods to calculate area of square, rectangle, triangle, circle and other shapes

Learning Content

📐 What is Area?

The surface of a two-dimensional figure or the space it occupiesSpread(Area) we say.

Spreadin square units(Square Units) are measured.

📏 Units
unit of length unit of area
meter (m) square meter (m²)
Centimeter (cm) Square cm. (cm²)
kilometer (km) sq. km. (km²)
feet (ft) square feet (sq.ft)
🔄 Unit Conversions
Change value
1 m² 10000 cm²
1 km² 1000000 m² (10⁶ m²)
1 hectare 10000 m²
1 acre 4046.86 m² ≈ 4047 m²
1 km² 100 hectares
1 sq.ft 929.03 cm² ≈ 0.0929 m²
📊 Basic shapes
1. Square

All four sides are equal in shape

If side = a, then area = a²

2. Rectangle

Opposite sides are congruent

If length = l, breadth = b, then area = l × b

3. Triangle

Three page format

If base = b, height = h, area = ½ × b × h

4. Circle

Points equidistant from the center

If radius = r, then area = πr²

📚 Key Codes
Code Meaning English
a side of the square Side of Square
l length Length
b Width / Bottom Breadth / Base
h height Height
r Radius Radius
d diameter Diameter
π Pi = 22/7 or 3.14 Pi
P Circumference Perimeter
A Spread Area
🎯 Perimeter vs Area
Perimeter Area
The length of the border of the shape The space inside the shape
One Dimensional Scale (1D) Two Dimensional Scale (2D)
Meter, cm units Square meter, square cm.
to fence Spread the floor
💡 Things to remember
  • π (pi)= 22/7 = 3.14159... (for round calculations)
  • √2= 1.414 (diagonal of square)
  • √3= 1.732 (equilateral triangle)
  • Diameter = 2 × radius (d = 2r)
  • Diagonal of square = a√2
  • Diagonal of rectangle = √(l² + b²)

📐 Area Formulas - All Shapes

All Area Formulas for TNPSC Exam!

◼️ Square
Spread A = a² a = page
Circumference P = 4a
Diagonal d = a√2
side (from surface) a = √A
side (from circumference) a = P/4
Area (diagonal) A = d²/2
▬ Rectangle
Spread A = l × b l = length, b = width
Circumference P = 2(l + b)
Diagonal d = √(l² + b²)
length l = A/b
width b = A/l
🔺 Triangle
General formula A = ½ × b × h b = base, h = height
Circumference P = a + b + c a, b, c = sides
Heron formula A = √[s(s-a)(s-b)(s-c)] s = (a+b+c)/2
Equilateral triangle A = (√3/4) × a² a = page
Isosceles triangle A = ¼ × b × √(4a² - b²) a = equilateral, b = foot
A right triangle A = ½ × feet × height
⭕ circle
Spread A = πr² r = radius
Circumference C = 2πr = πd d = diameter
Spread in diameter A = πd²/4
Radius (from surface) r = √(A/π)
Radius (from Circumference) r = C/2π
◗ Semicircle & Ring
form Spread Circumference
semicircle A = πr²/2 P = πr + 2r = r(π + 2)
foot circle A = πr²/4 P = πr/2 + 2r
Ring A = π(R² - r²) R = outer radius, r = inner radius
Sector A = (θ/360) × πr² θ = central angle
▰ Parallelogram
Spread A = b × h b = base, h = height
Circumference P = 2(a + b) a, b = adjacent sides
Diagonal area A = ½ × d₁ × d₂ × sinθ d₁, d₂ = diagonals
◇ Rhombus
Spread A = ½ × d₁ × d₂ d₁, d₂ = diagonals
Circumference P = 4a a = page
page a = ½ × √(d₁² + d₂²)
💡Note:The diagonals of a hypotenuse intersect perpendicularly
⏢ Trapezium
Spread A = ½ × (a + b) × h a, b = parallel sides, h = height
Circumference P = a + b + c + d A supplement of four pages
💡Memory:"Average of two parallel sides × height"
⬡ Regular Polygons
form Spread
Equilateral triangle A = (√3/4) × a²
A regular hexagon A = (3√3/2) × a²
n-sided polygon A = ¼ × n × a² × cot(π/n)
📐 Pathway Formulas
Path inside the rectangle:
Path area = 2w(l + b - 2w)
(w = track width, l = length, b = width)
Path outside the rectangle:
Path area = 2w(l + b + 2w)
Path around the square:
Path area = 4w(a + w) [out]
Path area = 4w(a - w) [in]
🔄 Key relationships
  • If the diagonal d of the square is:Area = d²/2, side = d/√2
  • In circle:Diameter d = 2r, perimeter = πd, area = πd²/4
  • In an equilateral triangle:Height h = (√3/2) × a
  • In the diagonal:a² = (d₁/2)² + (d₂/2)²

📝 Coverage - Solved Examples

10 Important Questions for TNPSC Exam

Question 1: Square

Question:The perimeter of a square is 64 cm. If so, what is its area?


Solution:

Perimeter P = 64 cm.

Side a = P/4 = 64/4 =16 cm.

Area A = a² = 16² =256 sq. cm.

Question 2: Rectangle

Question:Length of a rectangle is 15 m and width is 8 m. If , find its area and diagonal.


Solution:

Length l = 15 m, width b = 8 m.

Area A = l × b = 15 × 8 =120 sq m.

Diagonal d = √(l² + b²) = √(225 + 64) = √289 =17 m.

Question 3: Triangle

Question:The base of a triangle is 24 cm and the height is 15 cm. If so, find the area.


Solution:

Base b = 24 cm, height h = 15 cm.

Area A = ½ × b × h

A = ½ × 24 × 15 =180 sq. cm.

Question 4: Equilateral Triangle

Question:The side of an equilateral triangle is 14 cm. If so, find the area.


Solution:

Side a = 14 cm.

Area A = (√3/4) × a²

A = (√3/4) × 14² = (√3/4) × 196

A = 49√3 = 49 × 1.732 =84.87 sq. cm.

Question 5: Circle

Question:The radius of the circle is 14 cm. If so, find area and perimeter. (π = 22/7)


Solution:

Radius r = 14 cm.

Area A = πr² = (22/7) × 14 × 14

A = 22 × 2 × 14 =616 sq. cm.

Circumference C = 2πr = 2 × (22/7) × 14 =88 cm.

Question 6: Ring

Question:The outer radius of the ring is 21 cm and the inner radius is 14 cm. If so, what is the area of ​​the ring?


Solution:

Outer radius R = 21 cm, Inner radius r = 14 cm.

Area of ​​ring = π(R² - r²)

= (22/7) × (21² - 14²)

= (22/7) × (441 - 196)

= (22/7) × 245 = 22 × 35 =770 sq. cm.

Question 7: Rhombus

Question:The diagonals of a rhombus are 16 cm. and 12 cm. If so, see spread and page.


Solution:

d₁ = 16 cm, d₂ = 12 cm.

Area A = ½ × d₁ × d₂ = ½ × 16 × 12 =96 sq. cm.

Side a = ½ × √(d₁² + d₂²) = ½ × √(256 + 144)

a = ½ × √400 = ½ × 20 =10 cm

Question 8: Trapezium

Question:Parallel sides of a parallelogram are 20 cm. and 30 cm, height 15 cm. If so, find the area.


Solution:

Parallel sides a = 20 cm, b = 30 cm, height h = 15 cm.

Area A = ½ × (a + b) × h

A = ½ × (20 + 30) × 15

A = ½ × 50 × 15 =375 sq. cm.

Question 9: Heron's Formula

Question:The sides of a triangle are 13 cm, 14 cm, 15 cm. If so, find the area.


Solution:

Sides a = 13, b = 14, c = 15

Semicircumference s = (a + b + c)/2 = (13 + 14 + 15)/2 = 42/2 =21

s - a = 21 - 13 = 8

s - b = 21 - 14 = 7

s - c = 21 - 15 = 6

Area A = √[s(s-a)(s-b)(s-c)] = √(21 × 8 × 7 × 6)

A = √7056 =84 sq. cm.

Question 10: Pathway

Question:50 m. Length is 40 m. 2.5 m on the outside of a rectangular ground of width. There is a wide path. View the area of ​​the track.


Solution:

Length l = 50 m., Width b = 40 m., Track width w = 2.5 m.

Method 1: Direct formula

Outer track area = 2w(l + b + 2w)

= 2 × 2.5 × (50 + 40 + 5)

= 5 × 95 =475 sq m.

Method 2: Differentiation

Outer rectangle = (50 + 5) × (40 + 5) = 55 × 45 = 2475 sq m.

Inner rectangle = 50 × 40 = 2000 sq m.

Path = 2475 - 2000 =475 sq m.

📚 Additional practice questions
  1. Diagonal of a square is 10√2 cm. If, area =100 sq. cm.
  2. The circumference of a circle is 44 cm. If, area =154 sq. cm.
  3. The area of ​​the rectangle is 84 sq. m., width is 7 m. If, length =12 m.
  4. The base of the joint is 18 cm and the height is 12 cm. If, area =216 sq. cm.
  5. The radius of the semicircle is 7 cm. If, area =77 sq. cm.

⚡ Spread - cross paths

Super Tricks to Save Time in TNPSC Exam!

🚀 Trick 1: Square - Quick calculation
If the perimeter P is:Area = P²/16
If diagonal d is:Area = d²/2

Example:Circumference = 40 cm.

Area = 40²/16 = 1600/16 =100 sq. cm.

🚀 Trick 2: Rectangle - Diagonal Shortcut
3:4:5 aspect ratio rectangle:
If sides are 3k, 4k, diagonal = 5k

5:12:13 aspect ratio rectangle:
If sides are 5k, 12k, diagonal = 13k

Example:Length = 12 m., Width = 9 m.

This is a 3:4 ratio (9:12 = 3:4), k = 3

Diagonal = 5 × 3 =15 m.

🚀 Trick 3: Circle - Key Values
Radius (r) Perimeter (2πr) Area (πr²)
7 44 154
14 88 616
21 132 1386
28 176 2464
35 220 3850
💡Memory:The calculation for multiples of radius 7 (7, 14, 21...) is easy!
🚀 Trick 4: Page Change → Surface Change
square:x% increase in side → (2x + x²/100)% increase in area

Rectangle:Length x%, width y% increase → Area (x + y + xy/100)% increase

Example:10% increase on the side of the square

Increase in area = 2(10) + 100/100 = 20 + 1 =21%

🚀 Trick 5: Radius/Diameter conversion
x% increase in radius of circle:
Circumference → x% increase
Area → (2x + x²/100)% increase

Example:Doubling the radius (100% increase)

Circumference = 100% = double

Area = 2(100) + 10000/100 = 200 + 100 = 300% =Fourfold

🚀 Trick 6: Triangular memory
√3/4 = 0.433(equilateral triangle)
√3/2 = 0.866(equilateral height)
Equilateral side (a) Area = (√3/4)a²
2 √3 = 1.732
4 4√3 = 6.93
6 9√3 = 15.59
10 25√3 = 43.3
🚀 Trick 7: Pythagorean triples
Frequently used triads:
trinity application
3, 4, 5 Basic (folds: 6,8,10; 9,12,15; 12,16,20)
5, 12, 13 Multiples: 10,24,26; 15,36,39
8, 15, 17 Multiples: 16,30,34
7, 24, 25 Multiples: 14,48,50
🚀 Trick 8: Slider - Quick method
Horizontal area = average width × height
= [(a + b)/2] × h

Example:Parallel sides are 16 cm, 24 cm, height 10 cm.

Average width = (16 + 24)/2 = 20 cm.

Area = 20 × 10 =200 sq. cm.

🚀 Trick 9: Route Quick Formulas
Type formula
Path out of the rectangle 2w(l + b + 2w)
Path inside the rectangle 2w(l + b - 2w)
Path out of the square 4w(a + w)
Path inside the square 4w(a - w)
Cross paths (+) w(l + b - w)
🚀 Trick 10: Heron Formula Shortcut
After calculating s = (a+b+c)/2:
(s-a), (s-b), (s-c) Calculate → Multiply → √ Take

Example:Pages 5, 6, 7

s = (5+6+7)/2 = 9

s-a=4, s-b=3, s-c=2

Area = √(9×4×3×2) = √216 = 6√6 ≈14.7 square units

📊 TNPSC Frequently Asked Categories
Question type Crossroads
Spread from the perimeter Square: P²/16, Circle: C²/4π
Spread across the diagonal Square: d²/2, Diagonal: d₁d₂/2
% change x% → (2x + x²/100)% area
Circumference → Area A = C²/4π = Cr/2
Path width Outer: 2w(l+b+2w), Inner: 2w(l+b-2w)
💡 Important memory tips
  • π = 22/7- For circular calculations (if radius is 7 times)
  • π = 3.14- In other cases
  • √2 = 1.414- Square Diagonal
  • √3 = 1.732- Equilateral triangle
  • √3/4 = 0.433- Equilateral area coefficient
  • Circle:2 times r → 4 times the area
  • square:2 times a → 4 times the area
  • 1 hectare = 10000 m²
DISCLAIMER

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Please verify official sources before appearing for exams.