Comprehensive Study Notes: Mensuration 2D and Triangle Geometry

Overview of Mensuration and Geometry Dimensions

• Mensuration - 2D (क्षेत्रमिति - 2D): This field of study focuses on the measurement of two-dimensional geometric figures. In academic courses, this topic is typically covered in approximately 10 to 12 classes.
• Scope of Study: The primary shapes studied in 2D Mensuration include:
  • Triangles (त्रिभुज)
  • Quadrilaterals (चतुर्भुज)
  • Circles (वृत्त)
  • Polygons (बहुभुज)
• Classification of Dimensions (विमाये):
  • 0-Dimension Figure: A point (बिंदु) has no physical dimensions (length, width, or height).
  • 1-Dimension Figure: A line (रेखा) possesses only one dimension, which is length.
  • 2D Figures (Plane Figures): These figures have length and breadth but no thickness. Examples include:
    • Triangle, Rectangle, Circle, Octagon, Square, Pentagon, Hexagon, and Semicircle.
  • 3D Figures (Solid Figures): These figures possess three dimensions: length, breadth, and height/depth. Examples include:
    • Cube, Cuboid, Cone, Cylinder, Sphere, and Prism.

Fundamental Concepts of Perimeter and Area

• Perimeter (परिमाप):
  • Definition: The sum of all lengths of the outer boundary of any closed plane figure.
  • Nature: Essentially, perimeter is a measurement of length.
  • Units of Measurement: Typically measured in $cm$, $m$, or $km$.
• Area (क्षेत्रफल):
  • Definition: The amount of space occupied by a 2D surface or shape.
  • Units of Area: Usually measured in square units such as $cm^2$ or $m^2$.

Unit Conversions and Scale Measurements

• Length Conversions:
  • $1\,m = 100\,cm$
  • $1\,km = 1000\,m = 100000\,cm = 10^5\,cm$
• Area Conversions:
  • A square with sides of $1\,m$ has an area of $1\,m \times 1\,m = 1\,m^2$.
  • Converting to centimeters: $100\,cm \times 100\,cm = 10000\,cm^2 = 10^4\,cm^2$.
  • Therefore, $1\,m^2 = 10^4\,cm^2$.
• Land Measurement Units:
  • $1\text{ hectare} = 10000\,m^2$

Scaling Properties of 2D Figures

• Relationship Between Side, Perimeter, and Area:
  • When each side of a 2D figure is scaled by a factor of $k$:
    • Side Scaling: Each side changes by a factor of $k$.
    • Perimeter Scaling: The perimeter follows the linear scale and changes by a factor of $k$.
    • Area Scaling: The area follows the square of the scale and changes by a factor of $k^2$.
• Examples of Scaling:
  • If each side is doubled ($\times 2$):
    • Perimeter becomes $2 \times$ the original.
    • Area becomes $2^2 = 4 \times$ the original.
  • If each side is tripled ($\times 3$):
    • Area becomes $3^2 = 9 \times$ the original.
  • If each side is increased fivefold ($\times 5$):
    • Area becomes $5^2 = 25 \times$ the original.
  • If each side is increased by a factor of $x4$:
    • Perimeter increases by $x4$.
    • Area increases by $x16$.

Geometry and Area of a Triangle

• Standard Area Formula:
  • $\text{Area (A)} = \frac{1}{2} \times \text{Base} \times \text{Height}$
• Variable Sides and Altitudes:
  • For a triangle with sides $a, b, c$ and corresponding heights $h_1, h_2, h_3$:
    • $\text{Area (A)} = \frac{1}{2} \times a \times h_1 = \frac{1}{2} \times b \times h_2 = \frac{1}{2} \times c \times h_3$
  • This implies that the product of a side and its corresponding altitude is constant ($2A$):
    • $a \times h_1 = b \times h_2 = c \times h_3 = 2A$
• Inverse Relationship:
  • The height corresponding to a side is inversely proportional to the side length: $\text{height} \propto \frac{1}{\text{side}}$.
  • The ratio of the heights is the inverse of the ratio of the sides:
    • $h_1 : h_2 : h_3 = \frac{1}{a} : \frac{1}{b} : \frac{1}{c}$

Practical Problems and Solutions

• Problem 1: Map Scaling (CHSL PRE 2023)

  • Question: If a map of a city was drawn on a scale of $0.2\,cm$ to a $km$, what area would represent a country if it has an area of $22500\,km^2$?
  • Process:
    • Scale factor for side ($k$) = $0.2\,cm : 1\,km$.
    • Scale factor for area ($k^2$) = $(0.2)^2\,cm^2 : (1)^2\,km^2 = 0.04\,cm^2 / km^2$.
    • Given area = $22500\,km^2$.
    • Map area = $22500 \times 0.04 = 900\,cm^2$.
  • Answer: [d] $900\,cm^2$.

• Problem 2: Calculated Triangle Area

  • Question: What is the area of a triangle whose base is $21.4\,cm$ and height is $15.5\,cm$?
  • Process: $\text{Area} = \frac{1}{2} \times 21.4 \times 15.5 = 10.7 \times 15.5$.
  • Calculation: $10.7 \times 15.5 = 165.85\,cm^2$.
  • Answer: [a] $165.9\,cm^2$ (rounded to one decimal place).

• Problem 3: Area with Fractional Base (SSC GD 2024)

  • Question: Find the area of a triangle whose height is $35.7\,cm$ and the base is $1/7^{\text{th}}$ of the height.
  • Process:
    • $\text{Height} = 35.7\,cm$.
    • $\text{Base} = \frac{35.7}{7} = 5.1\,cm$.
    • $\text{Area} = \frac{1}{2} \times 5.1 \times 35.7 = 91.035\,cm^2$.
  • Answer: [b] $91.035\,cm^2$.

• Problem 4: Finding Height from Area Ratio

  • Question: Area of a triangle is $1470\,cm^2$. If the base is $3/5^{\text{th}}$ of the height, what is the height?
  • Process:
    • Let height = $h$, then base = $\frac{3}{5}h$.
    • $\text{Area} = \frac{1}{2} \times \frac{3}{5}h \times h = 1470$.
    • $\frac{3}{10}h^2 = 1470 \implies 3h^2 = 14700 \implies h^2 = 4900$.
    • $h = \sqrt{4900} = 70\,cm$.
  • Answer: [d] $70\,cm$.

• Problem 5: Perpendicular Length (DP CONSTABLE 2023)

  • Question: The area of a triangle is $182\,cm^2$ and one side is $28\,cm$. What is the length of the perpendicular drawn to this side from the opposite vertex?
  • Process:
    • $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$.
    • $182 = \frac{1}{2} \times 28 \times h$
    • $182 = 14h \implies h = \frac{182}{14} = 13\,cm$.
  • Answer: [b] $13\,cm$.

• Problem 6: Base and Altitude Relationship

  • Question: The base of a triangle is $3\,cm$ smaller than its altitude. Its area is $104\,cm^2$. What is the length of the base?
  • Process:
    • Let altitude (height) = $x$, then base = $x - 3$.
    • $\frac{1}{2} \times (x - 3) \times x = 104 \implies x(x - 3) = 208$.
    • Factors of $208$ that differ by 3 are $16$ and $13$ ($16 \times 13 = 208$).
    • Therefore, $x = 16$ (altitude) and $x - 3 = 13$ (base).
  • Answer: [a] $13\,cm$.

• Problem 7: Cultivation Cost and Dimensions

  • Question: The base of a triangular field is three times its altitude. If the cost of cultivating the field at $Rs.\,24.4/\text{hectare}$ is $Rs.\,448.35$, find its height (in meters).
  • Process:
    • $\text{Total area (hectares)} = \frac{448.35}{24.4}$.
    • Let height = $h$, base = $3h$.
    • $\text{Area in } m^2 = \text{Area in hectares} \times 10000$.
    • $\frac{1}{2} \times 3h \times h = \frac{448.35}{24.4} \times 10000$.
    • $1.5h^2 = 183750 \implies h^2 = 122500$.
    • $h = \sqrt{122500} = 350\,m$.
  • Answer: [b] $350\,m$.

• Problem 8: Corresponding Altitudes

  • Question: Two sides of a triangle are $12.8\,m$ and $9.6\,m$. If the height corresponding to $9.6\,m$ is $12\,m$, what is the height corresponding to $12.8\,m$?
  • Process:
    • Since area is constant: $side_1 \times h_1 = side_2 \times h_2$.
    • $12.8 \times h = 9.6 \times 12$
    • $h = \frac{9.6 \times 12}{12.8} = \frac{3 \times 12}{4} = 9\,m$.
  • Answer: [b] $9\,m$.

• Problem 9: Ratio of Altitudes (SSC CGL 2020 PRE)

  • Question: The ratio of the bases of two triangles is $x : y$ and the ratio of their areas is $a : b$. What is the ratio of their corresponding altitudes?
  • Process:
    • Area $\text{A} = \frac{1}{2} \times \text{Base} \times \text{Height} \implies \text{Height} \propto \frac{\text{Area}}{\text{Base}}$.
    • $\text{Ratio of Heights} = \frac{a}{x} : \frac{b}{y} = ay : bx$.
  • Answer: [c] $ay : bx$.