Chapter 6: Measuring Space: Perimeter and Area - Ultimate 100/100 Master Study Guide

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Topic 1: Foundations of 2D Measurement & Basic Triangles

Perimeter vs Area Fundamentals

Perimeter ($P$): The total linear length of the 1D boundary enclosing a 2D closed figure. Measured in linear units ($\text{cm}, \text{m}, \text{km}$).
Area ($A$): The measure of the 2D surface region enclosed inside the boundary. Measured in square units ($\text{cm}^2, \text{m}^2, \text{hectares}$).
Right-Angled Triangle Area: $$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times b \times h$$
Equilateral Triangle Area ($a$ = side): $$\text{Area} = \frac{\sqrt{3}}{4} a^2 \qquad \text{Height } h = \frac{\sqrt{3}}{2} a \qquad \text{Perimeter } P = 3a$$
Isosceles Triangle Area ($a$ = equal sides, $b$ = base): $$\text{Height } h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} = \frac{1}{2}\sqrt{4a^2 - b^2} \implies \text{Area} = \frac{b}{4}\sqrt{4a^2 - b^2}$$

Fig 6.1-6.2: Geometric Components of Right-Angled and Equilateral Triangles

Teacher's Exam Secret — Common Unit Conversion Pitfall

Always convert all given dimensions to the SAME unit before computing area!
Remember: $1\text{ m} = 100\text{ cm} \implies 1\text{ m}^2 = 10,000\text{ cm}^2$. Also, $1\text{ hectare} = 10,000\text{ m}^2$.

Immediate Solved Exam Applications

Example 1.1: Equilateral Triangle Area NCERT

Question: Find the area and height of an equilateral triangle of side $a = 12\text{ cm}$. (Take $\sqrt{3} \approx 1.732$).

Solution:

$$\text{Area} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3} \approx 36 \times 1.732 = \mathbf{62.352\text{ cm}^2}$$ $$\text{Height } h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3} \approx \mathbf{10.392\text{ cm}}$$

Topic 1: School Exam Practice Problem Kit

Problem 1.1 (Isosceles Triangle Height & Area) 2 MARKS

An isosceles triangle has equal sides of length $10\text{ cm}$ each and base of length $12\text{ cm}$. Find its height and area.

Step-by-Step Solution:

Perpendicular altitude bisects the base into two segments of $6\text{ cm}$.
In right $\Delta$: $h = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = \mathbf{8\text{ cm}}$. $$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 12 \times 8 = \mathbf{48\text{ cm}^2}$$

Fig 6.1A: Isosceles Triangle Height

Problem 1.2 (Equilateral Area from Perimeter) 1 MARK

The perimeter of an equilateral triangle is $36\text{ cm}$. Find its area.

Step-by-Step Solution:

Side $a = \frac{\text{Perimeter}}{3} = \frac{36}{3} = 12\text{ cm}$. $$\text{Area} = \frac{\sqrt{3}}{4} \times 12^2 = \mathbf{36\sqrt{3}\text{ cm}^2} \approx \mathbf{62.35\text{ cm}^2}$$

Topic 2: Heron's Formula for Arbitrary Triangles

Heron's Master Theorem

For any triangle with side lengths $a, b, c$ (whether acute, obtuse, or scalene) where height is not directly given:

$$\mathbf{\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}}$$

where $s$ is the semi-perimeter (half of the total perimeter):

$$\mathbf{s = \frac{a + b + c}{2}}$$

Fig 6.3: Scalene Triangle with Side Lengths $a, b, c$ for Heron's Formula

Teacher's Exam Secret — Prime Factorization Trick for Square Roots

Never multiply large numbers like $21 \times 8 \times 7 \times 6 = 7056$ and then try to find the square root manually!
Instead, break each term into prime factors:
$\sqrt{21 \times 8 \times 7 \times 6} = \sqrt{(7 \times 3) \times (2^3) \times 7 \times (2 \times 3)} = \sqrt{7^2 \times 3^2 \times 2^4} = 7 \times 3 \times 4 = \mathbf{84}$.

Immediate Solved Exam Applications

Example 2.1: Classic 13-14-15 Triangle NCERT FAVORITE

Question: Find the area of a triangle with sides $a = 13\text{ cm}$, $b = 14\text{ cm}$, $c = 15\text{ cm}$.

Solution:

1. Semi-perimeter $s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = \mathbf{21\text{ cm}}$.

2. Differences: $(s - a) = 8$, $(s - b) = 7$, $(s - c) = 6$.

$$\text{Area} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{(7 \times 3) \times (4 \times 2) \times 7 \times (3 \times 2)} = \sqrt{7^2 \times 3^2 \times 4^2} = 7 \times 3 \times 4 = \mathbf{84\text{ cm}^2}$$

Topic 2: School Exam Practice Problem Kit

Problem 2.1 (Triangular Park Area) 3 MARKS / NCERT

A triangular park has sides $120\text{ m}$, $80\text{ m}$, and $50\text{ m}$. A gardener has to put a fence all around it and plant grass inside. Find the area and fencing cost at ₹$20$ per metre leaving a $3\text{ m}$ gate.

Step-by-Step Solution:

$s = \frac{120 + 80 + 50}{2} = 125\text{ m}$. $$\text{Area} = \sqrt{125(125-120)(125-80)(125-50)} = \sqrt{125 \times 5 \times 45 \times 75} = \mathbf{375\sqrt{15}\text{ m}^2} \approx \mathbf{1452.37\text{ m}^2}$$ $$\text{Fencing Length} = (120+80+50) - 3 = 247\text{ m} \implies \text{Cost} = 247 \times 20 = \mathbf{\text{Rs. } 4940}$$

Fig 6.2A: Triangular Park

Problem 2.2 (Sides in Given Ratio) 3 MARKS

The sides of a triangle are in the ratio $3 : 5 : 7$ and its perimeter is $300\text{ m}$. Find its area.

Step-by-Step Solution:

Let sides be $3x, 5x, 7x \implies 3x + 5x + 7x = 300 \implies 15x = 300 \implies x = 20$.
Sides: $a = 60\text{ m}, b = 100\text{ m}, c = 140\text{ m}$. $s = 150\text{ m}$. $$\text{Area} = \sqrt{150(90)(50)(10)} = \sqrt{150 \times 90 \times 50 \times 10} = \mathbf{1500\sqrt{3}\text{ m}^2} \approx \mathbf{2598.08\text{ m}^2}$$

Topic 3: Quadrilaterals via Triangular Decomposition

General Quadrilateral Area

A general quadrilateral $ABCD$ can be divided by diagonal $AC = d$ into two triangles $\Delta ABC$ and $\Delta ADC$. If perpendicular heights from opposite vertices $B$ and $D$ onto diagonal $AC$ are $h_1$ and $h_2$:

$$\mathbf{\text{Area of Quadrilateral } ABCD = \frac{1}{2} \times d \times (h_1 + h_2)}$$

Fig 6.4: Quadrilateral $ABCD$ Decomposed by Diagonal $d$ with Offsets $h_1$ and $h_2$

Immediate Solved Exam Applications

Example 3.1: Quadrilateral Field Area NCERT

Question: A quadrilateral field $ABCD$ has diagonal $AC = 24\text{ m}$, and offsets from vertices $B$ and $D$ are $h_1 = 13\text{ m}$ and $h_2 = 8\text{ m}$. Find its area.

Solution:

$$\text{Area} = \frac{1}{2} \times d \times (h_1 + h_2) = \frac{1}{2} \times 24 \times (13 + 8) = 12 \times 21 = \mathbf{252\text{ m}^2}$$

Topic 3: School Exam Practice Problem Kit

Problem 3.1 (Right-Angled Composite Quadrilateral) 4 MARKS / NCERT

A park in the shape of a quadrilateral $ABCD$ has $\angle C = 90^\circ, AB = 9\text{ m}, BC = 12\text{ m}, CD = 5\text{ m}, AD = 8\text{ m}$. How much area does it occupy?

Step-by-Step Solution:

1. Join $BD$. In right $\Delta BCD$: $BD = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ m}$.
$\text{Area}(\Delta BCD) = \frac{1}{2} \times 12 \times 5 = \mathbf{30\text{ m}^2}$.
2. For $\Delta ABD$, sides are $9\text{ m}, 13\text{ m}, 8\text{ m} \implies s = 15\text{ m}$.
$\text{Area}(\Delta ABD) = \sqrt{15(6)(2)(7)} = \sqrt{1260} = \mathbf{6\sqrt{35}\text{ m}^2} \approx \mathbf{35.5\text{ m}^2}$. $$\text{Total Area} = 30 + 35.5 = \mathbf{65.5\text{ m}^2}$$

Fig 6.3A: Park $ABCD$

Topic 4: Special Quadrilaterals (Parallelogram, Rhombus, Trapezium)

Special Quadrilateral Area Formulas

Parallelogram: $\text{Area} = \text{Base} \times \text{Height} = b \times h$
Rhombus (Diagonals $d_1, d_2$): $$\text{Area} = \frac{1}{2} \times d_1 \times d_2 \qquad \text{Side } a = \frac{1}{2}\sqrt{d_1^2 + d_2^2}$$
Trapezium (Parallel sides $a, b$, Altitude $h$): $$\text{Area} = \frac{1}{2} \times (a + b) \times h$$

Fig 6.5-6.6: Diagonals of a Rhombus and Altitude of a Trapezium

Topic 4: School Exam Practice Problem Kit

Problem 4.1 (Rhombus Field Grazing Problem) 3 MARKS / NCERT

A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is $30\text{ m}$ and its longer diagonal is $48\text{ m}$, how much area of grass field will each cow be getting?

Step-by-Step Solution:

1. Diagonal divides rhombus into two congruent $\Delta s$ with sides $30\text{ m}, 30\text{ m}, 48\text{ m}$.
2. Semi-perimeter $s = \frac{30+30+48}{2} = 54\text{ m}$.
$\text{Area}(\Delta) = \sqrt{54(24)(24)(6)} = \sqrt{54 \times 6 \times 24^2} = 18 \times 24 = \mathbf{432\text{ m}^2}$.
3. Total Rhombus Area $= 2 \times 432 = \mathbf{864\text{ m}^2}$. $$\text{Area per cow} = \frac{864}{18} = \mathbf{48\text{ m}^2}$$

Fig 6.4A: Rhombus Field

Problem 4.2 (Trapezium Area with 4 Sides Given) 4 MARKS / NCERT

Find the area of a trapezium whose parallel sides are $25\text{ cm}$ and $10\text{ cm}$, and non-parallel sides are $14\text{ cm}$ and $13\text{ cm}$.

Step-by-Step Solution:

1. Draw line through top vertex parallel to non-parallel side, forming a parallelogram of base $10\text{ cm}$ and a triangle of sides $14\text{ cm}, 13\text{ cm}, (25-10)=15\text{ cm}$.
2. For $\Delta$ (13-14-15): $\text{Area} = \mathbf{84\text{ cm}^2}$.
3. Height $h$: $\frac{1}{2} \times 15 \times h = 84 \implies h = \frac{168}{15} = \mathbf{11.2\text{ cm}}$. $$\text{Trapezium Area} = \frac{1}{2} \times (25 + 10) \times 11.2 = 35 \times 5.6 = \mathbf{196\text{ cm}^2}$$

Topic 5: Circles, Annulus (Rings) & Track Measurements

Circular & Ring Geometry

Circumference of Circle: $C = 2\pi r = \pi d$
Area of Circle: $A = \pi r^2$
Area of Annulus / Circular Ring (Outer radius $R$, Inner radius $r$): $$\text{Area of Ring} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) = \mathbf{\pi (R - r)(R + r)}$$
Width of Circular Track: $w = R - r$

Fig 6.7: Annulus / Circular Ring with Outer Radius $R$ and Inner Radius $r$

Topic 5: School Exam Practice Problem Kit

Problem 5.1 (Circular Track Area) 3 MARKS

A circular garden of radius $14\text{ m}$ is surrounded by a circular running track of uniform width $7\text{ m}$. Find the area of the track. ($\pi = 22/7$).

Step-by-Step Solution:

Inner radius $r = 14\text{ m}$. Outer radius $R = 14 + 7 = 21\text{ m}$. $$\text{Track Area} = \pi (R^2 - r^2) = \frac{22}{7} (21 - 14)(21 + 14) = \frac{22}{7} \times 7 \times 35 = 22 \times 35 = \mathbf{770\text{ m}^2}$$

Fig 6.5A: Garden & Track

Topic 6: Sectors & Segments of Circles

Sector & Segment Formulas

Arc Length ($L$): $L = \frac{\theta}{360^\circ} \times 2\pi r$
Area of Sector ($A_{\text{sector}}$): $$A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{1}{2} \times L \times r$$
Area of Minor Segment ($A_{\text{segment}}$): $$A_{\text{segment}} = \text{Area of Sector } OAB - \text{Area of } \Delta OAB$$

Fig 6.8: Sector $OAB$ and Minor Segment Bounded by Chord $AB$

Topic 6: School Exam Practice Problem Kit

Problem 6.1 (Clock Minute Hand Swept Area) 3 MARKS / NCERT

The length of the minute hand of a clock is $14\text{ cm}$. Find the area swept by the minute hand in $5\text{ minutes}$.

Step-by-Step Solution:

In 60 mins, hand completes $360^\circ \implies \text{In 5 mins}, \theta = \frac{360^\circ}{60} \times 5 = \mathbf{30^\circ}$. $$\text{Area Swept} = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{12} \times \frac{22}{7} \times 196 = \frac{154}{3} = \mathbf{51.33\text{ cm}^2}$$

Fig 6.6A: Minute Hand 30°

Topic 7: Shaded Region Geometry & Composite Figures

Shaded Region Master Rule

For composite geometric shapes containing cutouts or overlapping figures:

$$\mathbf{\text{Area of Shaded Region} = \text{Area of Outer Container} - \text{Area of Inner Cutouts}}$$

Fig 6.9: Four Touching Circles Enclosed inside a Square

Topic 7: School Exam Practice Problem Kit

Problem 7.1 (Area Between 4 Touching Circles) 4 MARKS / NCERT

Find the area of the shaded region enclosed between 4 touching circles of radius $r = 7\text{ cm}$ each inside a square of side $14\text{ cm}$.

Step-by-Step Solution:

1. Area of Square $= 14^2 = \mathbf{196\text{ cm}^2}$.
2. Four quadrant cutouts form 1 full circle of radius $7\text{ cm}$: $$\text{Area of Circle} = \frac{22}{7} \times 7^2 = \mathbf{154\text{ cm}^2}$$ $$\text{Enclosed Shaded Area} = 196 - 154 = \mathbf{42\text{ cm}^2}$$

Fig 6.7A: Shaded Area = 42 cm²

Topic 8: Comprehensive Exam HOTS & Case Study Kit

Case Study 1: Triangular Roof Tiles Architecture CASE STUDY

A floral design on a floor is made up of 16 triangular tiles, each having sides $9\text{ cm}$, $28\text{ cm}$, and $35\text{ cm}$. Find the cost of polishing the tiles at the rate of $50\text{ paise}$ per $\text{cm}^2$.

Solution:

1. For 1 tile: $s = \frac{9 + 28 + 35}{2} = \frac{72}{2} = \mathbf{36\text{ cm}}$.

$$\text{Area of 1 tile} = \sqrt{36(36-9)(36-28)(36-35)} = \sqrt{36 \times 27 \times 8 \times 1} = \sqrt{36 \times (9 \times 3) \times (4 \times 2)} = 6 \times 3 \times 2 \sqrt{6} = \mathbf{36\sqrt{6}\text{ cm}^2}$$ $$\text{Since } \sqrt{6} \approx 2.45 \implies \text{Area of 1 tile} \approx 36 \times 2.45 = \mathbf{88.2\text{ cm}^2}$$

2. Total area of 16 tiles $= 16 \times 88.2 = \mathbf{1411.2\text{ cm}^2}$.

$$\text{Cost of polishing} = 1411.2 \times 0.50 = \mathbf{\text{Rs. } 705.60}$$

HOTS 1: Percentage Increase in Area when Sides Doubled HOTS FAVORITE

Question: Find the percentage increase in the area of a triangle if each of its sides is doubled.

Solution:

Let original sides be $a, b, c$ with semi-perimeter $s$. Original area $\Delta_1 = \sqrt{s(s-a)(s-b)(s-c)}$.

New sides $a' = 2a, b' = 2b, c' = 2c$. New semi-perimeter $s' = \frac{2a + 2b + 2c}{2} = 2s$.

$$\Delta_2 = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)} = \sqrt{16 s(s-a)(s-b)(s-c)} = 4 \Delta_1$$ $$\text{Increase in Area} = \Delta_2 - \Delta_1 = 4\Delta_1 - \Delta_1 = 3\Delta_1$$ $$\text{Percentage Increase} = \left(\frac{3\Delta_1}{\Delta_1}\right) \times 100\% = \mathbf{300\%}$$

9. 100/100 School Examination Master Checklist

Master Exam Checklist

Heron's Formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = (a+b+c)/2$.
Equilateral Triangle: $\text{Area} = \frac{\sqrt{3}}{4}a^2$, $\text{Height} = \frac{\sqrt{3}}{2}a$.
Rhombus Area: $\text{Area} = \frac{1}{2} d_1 d_2$.
Trapezium Area: $\text{Area} = \frac{1}{2} (a + b) h$.
Circular Ring: $\text{Area} = \pi (R^2 - r^2) = \pi (R-r)(R+r)$.
Sector Area: $\text{Area} = \frac{\theta}{360^\circ} \pi r^2$.
Sides Doubled Rule: Doubling each side of a triangle quadruples the area ($\text{Area}' = 4\text{Area}$) and gives a $300\%$ increase!