🕉️ VEDIC MATHEMATICS — LEVEL 3: ADVANCED

MODULE 28: Geometry — Vedic Constructions & Proofs

Complete Study Material | Theory + Examples + Practice + Test Bank

"Geometry is the art of seeing relationships. The sutras teach us to observe patterns so clearly that proofs become self-evident—by mere observation." — Vedic Mathematics Teacher's Manual

📋 MODULE AT A GLANCE

🎯 LEARNING OUTCOMES

By the end of this module, the student will be able to:

1. Generate Pythagorean triples instantly using the Vedic formula
2. Calculate area of any triangle using Heron's formula and Vedic cross-products
3. Prove circle theorems by mere observation (Vilokanam)
4. Derive conic section equations using Vedic pattern recognition
5. Compute volumes of solids using base-times-height patterns
6. Apply Paravartya to transformational geometry problems
7. Recognize geometric properties through Vyashti Samashti (part-whole)
8. Solve JEE-level geometry problems in 50% less time

PART 1: THEORY

1.1 — Introduction to Vedic Geometry

What is Vedic Geometry?

Vedic geometry applies the principles of the sutras to geometric constructions, proofs, and calculations. The emphasis is on pattern recognition (Vilokanam) and breaking problems into parts (Vyashti Samashti).

Why Vedic Geometry?

1.2 — Sutra 12: Vilokanam (By Mere Observation)

What Does This Mean?

This sub-sutra teaches us to see the answer directly without calculation. In geometry, this means:

• Recognizing congruent triangles by visual inspection
• Seeing proportional relationships
• Identifying symmetrical properties
• Observing cyclic quadrilaterals by angle sums

1.3 — Pythagorean Triples: Vedic Formula

The Standard Formula

A Pythagorean triple $(a, b, c)$ satisfies $a^2 + b^2 = c^2$.

The Vedic generation formula (from Sulba Sutras):

For any $m > n$, let: $$a = m^2 - n^2,\quad b = 2mn,\quad c = m^2 + n^2$$

Vedic Pattern (Urdhva-Tiryagbhyam Connection)

Think of $(m + in)^2 = (m^2 - n^2) + i(2mn)$. The modulus squared gives $m^2 + n^2$.

Example 1: $m = 2, n = 1$

$a = 4 - 1 = 3$, $b = 2×2×1 = 4$, $c = 4 + 1 = 5$ → $(3, 4, 5)$ ✓

Example 2: $m = 3, n = 1$

$a = 9 - 1 = 8$, $b = 2×3×1 = 6$, $c = 9 + 1 = 10$ → $(6, 8, 10)$ ✓

Example 3: $m = 3, n = 2$

$a = 9 - 4 = 5$, $b = 2×3×2 = 12$, $c = 9 + 4 = 13$ → $(5, 12, 13)$ ✓

Example 4: $m = 4, n = 1$

$a = 16 - 1 = 15$, $b = 2×4×1 = 8$, $c = 16 + 1 = 17$ → $(8, 15, 17)$ ✓

Example 5: $m = 4, n = 3$

$a = 16 - 9 = 7$, $b = 2×4×3 = 24$, $c = 16 + 9 = 25$ → $(7, 24, 25)$ ✓

Special Case: When $m$ and $n$ are consecutive

• $(2,1)$ → $(3,4,5)$
• $(3,2)$ → $(5,12,13)$
• $(4,3)$ → $(7,24,25)$
• $(5,4)$ → $(9,40,41)$

Pattern: The hypotenuse increases by $2(m^2 + n^2)$? Observation: $c = m^2 + n^2$ with $m = n+1$ gives $c = (n+1)^2 + n^2 = 2n^2 + 2n + 1$

1.4 — Heronian Triangles

What is a Heronian Triangle?

A triangle whose side lengths and area are all integers.

Heron's Formula

For triangle with sides $a, b, c$ and semi-perimeter $s = \frac{a+b+c}{2}$:

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

Vedic Pattern: Area as Cross Product

For coordinates $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$:

$$\text{Area} = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$

This is the shoelace formula—already a Vedic pattern!

Example: Heronian triangle (3,4,5)

$s = (3+4+5)/2 = 6$ Area = $\sqrt{6×3×2×1} = \sqrt{36} = 6$ ✓

Example: Heronian triangle (5,5,6)

$s = (5+5+6)/2 = 8$ Area = $\sqrt{8×3×3×2} = \sqrt{144} = 12$ ✓

Example: Heronian triangle (5,12,13)

$s = (5+12+13)/2 = 15$ Area = $\sqrt{15×10×3×2} = \sqrt{900} = 30$ ✓

1.5 — Area Calculations Using Vedic Cross-Products

Vyashti Samashti (Part and Whole)

Break a complex shape into simpler parts, calculate each area, then sum.

Example: Area of a quadrilateral

For quadrilateral with vertices $A, B, C, D$ (in order):

$$\text{Area} = \frac{1}{2}|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$$

This is the polygon shoelace formula.

Example: Vertices (0,0), (4,0), (5,3), (1,3)

List: (0,0), (4,0), (5,3), (1,3), back to (0,0)

Sum1 = $0×0 + 4×3 + 5×3 + 1×0 = 0 + 12 + 15 + 0 = 27$ Sum2 = $0×4 + 0×5 + 3×1 + 3×0 = 0 + 0 + 3 + 0 = 3$ Area = $|27 - 3|/2 = 24/2 = 12$ square units ✓

1.6 — Circle Theorems: Proof by Mere Observation (Vilokanam)

Theorem 1: The angle in a semicircle is a right angle.

Observation: For a triangle inscribed in a semicircle with diameter as hypotenuse, the angle at the circumference is 90°.

Vedic proof by observation: Draw the radius to the point on circumference. Isosceles triangles reveal that the angle sums to 90°.

Theorem 2: Angles subtended by the same chord are equal.

Observation: Points on the same arc subtend equal angles at the circumference.

Theorem 3: The angle between a chord and a tangent equals the angle in the alternate segment.

Observation: Alternate segment theorem—by mere observation of the cyclic quadrilateral formed.

Theorem 4: Opposite angles of a cyclic quadrilateral sum to 180°.

Observation: Since inscribed angles subtend arcs that sum to full circle, their measures add to 180°.

Example: Find $\angle ABC$ if $A, B, C$ lie on circle with center $O$ and $AC$ is diameter.

By observation: $\angle ABC = 90°$ (angle in semicircle) ✓

1.7 — Conic Sections: Vedic Derivation Shortcuts

The General Conic Equation

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

Type Determination by Observation (Vilokanam)

Parabola (Yavadunam Connection)

Standard form: $y^2 = 4ax$ or $x^2 = 4ay$

Yavadunam insight: The distance from focus equals distance from directrix.

Ellipse

Standard form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b$)

Key property: Sum of distances from foci is constant ($2a$).

Hyperbola

Standard form: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Key property: Difference of distances from foci is constant ($2a$).

Example: Identify the conic: $9x^2 + 25y^2 = 225$

Divide by 225: $\frac{x^2}{25} + \frac{y^2}{9} = 1$ → Ellipse ($a=5, b=3$) ✓

Example: Identify the conic: $y^2 = 8x$

$4a = 8$ → $a = 2$ → Parabola, opening right ✓

1.8 — Solid Geometry: Base-Times-Height Patterns

The Universal Vedic Principle (Vyashti Samashti)

For any prism or cylinder: $$\text{Volume} = \text{Base Area} \times \text{Height}$$

For any pyramid or cone: $$\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height}$$

For sphere: $$\text{Volume} = \frac{4}{3}\pi r^3$$

Observation (Vilokanam):

The volume of a cone is $\frac{1}{3}$ the volume of a cylinder with same base and height.

Example: Cube of side $a$

Base area = $a^2$, height = $a$ → Volume = $a^3$ ✓

Example: Rectangular prism $l \times w \times h$

Base area = $lw$, height = $h$ → Volume = $lwh$ ✓

Example: Cylinder radius $r$, height $h$

Base area = $\pi r^2$, height = $h$ → Volume = $\pi r^2 h$ ✓

Example: Cone radius $r$, height $h$

Volume = $\frac{1}{3}\pi r^2 h$ ✓

1.9 — Transformational Geometry: Paravartya Connections

Sutra 4: Paravartya Yojayet (Transpose and Apply)

In geometry, this applies to:

• Translation: $z \to z + t$ (adding vector)
• Rotation: $z \to ze^{i\theta}$ (multiplying by unit complex)
• Reflection: $z \to \bar{z}$ (conjugate)

Translation

Moving a shape by vector $(h, k)$: $$(x, y) \to (x + h, y + k)$$

Rotation about Origin

Rotation by angle $\theta$: $$(x, y) \to (x\cos\theta - y\sin\theta, x\sin\theta + y\cos\theta)$$

Reflection about x-axis

$$(x, y) \to (x, -y)$$

Reflection about y-axis

$$(x, y) \to (-x, y)$$

Reflection about line $y = x$

$$(x, y) \to (y, x)$$

Example: Apply translation (2, -3) to point (1, 4)

New point = $(1+2, 4-3) = (3, 1)$ ✓

1.10 — JEE-Level Geometry Applications

Example 1: Find the area of triangle with vertices (1,2), (4,5), (6,3)

Using shoelace:

Sum1 = $1×5 + 4×3 + 6×2 = 5 + 12 + 12 = 29$ Sum2 = $2×4 + 5×6 + 3×1 = 8 + 30 + 3 = 41$ Area = $|29 - 41|/2 = 12/2 = 6$ ✓

Example 2: Find the equation of the circle passing through (0,0), (4,0), (0,3)

Let equation be $x^2 + y^2 + Dx + Ey + F = 0$

From (0,0): $F = 0$ From (4,0): $16 + 4D = 0$ → $D = -4$ From (0,3): $9 + 3E = 0$ → $E = -3$

Equation: $x^2 + y^2 - 4x - 3y = 0$

Complete square: $(x-2)^2 + (y-1.5)^2 = 4 + 2.25 = 6.25 = (2.5)^2$

Center $(2, 1.5)$, radius $2.5$ ✓

Example 3: Find the area of the triangle formed by the points where the circle $x^2 + y^2 = 25$ meets the axes

Circle meets x-axis at $(\pm5, 0)$, y-axis at $(0, \pm5)$

Triangle formed by $(5,0)$, $(0,5)$, $(0,0)$ has area = $\frac{1}{2} \times 5 \times 5 = 12.5$

Or choose $(5,0)$, $(0,5)$, $(-5,0)$ — area = ?

Better: Triangle with vertices $(5,0)$, $(0,5)$, $(-5,0)$:

Sum1 = $5×5 + 0×0 + (-5)×0 = 25 + 0 + 0 = 25$ Sum2 = $0×0 + 5×(-5) + 0×5 = 0 - 25 + 0 = -25$ Area = $|25 - (-25)|/2 = 50/2 = 25$ ✓

PART 2: WORKED EXAMPLES

Section A: Pythagorean Triples

Example 1

Question: Generate a Pythagorean triple using $m=5, n=2$.

Answer:

$a = m^2 - n^2 = 25 - 4 = 21$ $b = 2mn = 2×5×2 = 20$ $c = m^2 + n^2 = 25 + 4 = 29$

Triple: $(20, 21, 29)$ ✓

Check: $20^2 + 21^2 = 400 + 441 = 841 = 29^2$ ✓

Example 2

Question: Generate a Pythagorean triple using $m=7, n=4$.

Answer:

$a = 49 - 16 = 33$ $b = 2×7×4 = 56$ $c = 49 + 16 = 65$

Triple: $(33, 56, 65)$ ✓

Check: $33^2 + 56^2 = 1089 + 3136 = 4225 = 65^2$ ✓

Example 3

Question: Find the smallest Pythagorean triple with a leg > 10.

Answer:

The triple $(8, 15, 17)$ has legs 8 and 15. The next: $(9, 40, 41)$ has legs 9 and 40.

The smallest triple with both legs > 10 might be $(11, 60, 61)$? But 11 is prime? Check $m=6, n=5$: $a=36-25=11$, $b=60$, $c=61$ ✓

So $(11, 60, 61)$ is answer.

Section B: Area Calculations (Shoelace)

Example 4

Question: Find area of triangle with vertices $(1,1)$, $(4,2)$, $(2,5)$.

Answer:

List: (1,1), (4,2), (2,5), back to (1,1)

Sum1 = $1×2 + 4×5 + 2×1 = 2 + 20 + 2 = 24$ Sum2 = $1×4 + 2×2 + 5×1 = 4 + 4 + 5 = 13$ Area = $|24 - 13|/2 = 11/2 = 5.5$ ✓

Example 5

Question: Find area of quadrilateral with vertices $(0,0)$, $(4,0)$, $(5,3)$, $(1,3)$.

Answer:

List: (0,0), (4,0), (5,3), (1,3), back to (0,0)

Sum1 = $0×0 + 4×3 + 5×3 + 1×0 = 0 + 12 + 15 + 0 = 27$ Sum2 = $0×4 + 0×5 + 3×1 + 3×0 = 0 + 0 + 3 + 0 = 3$ Area = $|27 - 3|/2 = 24/2 = 12$ ✓

Example 6

Question: Find area of pentagon with vertices $(0,0)$, $(4,0)$, $(5,2)$, $(2,5)$, $(0,4)$.

Answer:

List: (0,0), (4,0), (5,2), (2,5), (0,4), back to (0,0)

Sum1 = $0×0 + 4×2 + 5×5 + 2×4 + 0×0 = 0 + 8 + 25 + 8 + 0 = 41$ Sum2 = $0×4 + 0×5 + 2×2 + 5×0 + 4×0 = 0 + 0 + 4 + 0 + 0 = 4$ Area = $|41 - 4|/2 = 37/2 = 18.5$ ✓

Section C: Circle Theorems (Vilokanam)

Example 7

Question: In a circle, chord AB subtends an angle of 60° at the center. What angle does it subtend at the circumference?

Answer:

By observation: Angle at circumference = half the angle at center = $30°$ ✓

Example 8

Question: In cyclic quadrilateral ABCD, $\angle A = 70°$. Find $\angle C$.

Answer:

By observation: Opposite angles sum to 180° $\angle C = 180° - 70° = 110°$ ✓

Example 9

Question: Triangle ABC is inscribed in a circle with AC as diameter. If $\angle BAC = 40°$, find $\angle ABC$.

Answer:

By observation: Angle in semicircle = 90° → $\angle ABC = 90°$ ✓

Section D: Conic Sections

Example 10

Question: Identify the conic: $4x^2 + 9y^2 = 36$.

Answer:

Divide by 36: $\frac{x^2}{9} + \frac{y^2}{4} = 1$ → Ellipse ✓

Example 11

Question: Identify the conic: $y^2 - 4x^2 = 4$.

Answer:

Divide by 4: $\frac{y^2}{4} - \frac{x^2}{1} = 1$ → Hyperbola (vertical transverse axis) ✓

Example 12

Question: Find the focus of parabola $y^2 = 12x$.

Answer:

$4a = 12$ → $a = 3$ → Focus at $(3, 0)$ ✓

Section E: Solid Geometry

Example 13

Question: Find volume of a cylinder with radius 5 cm and height 10 cm.

Answer:

Volume = $\pi r^2 h = \pi × 25 × 10 = 250\pi$ cubic cm ✓

Example 14

Question: Find volume of a cone with radius 3 cm and height 8 cm.

Answer:

Volume = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi × 9 × 8 = 24\pi$ cubic cm ✓

Example 15

Question: Find volume of a sphere with radius 6 cm.

Answer:

Volume = $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi × 216 = 288\pi$ cubic cm ✓

Section F: Transformational Geometry

Example 16

Question: Rotate the point $(2,3)$ by $90°$ counterclockwise about origin.

Answer:

Rotation by $90°$: $(x,y) \to (-y, x)$ $(2,3) \to (-3, 2)$ ✓

Example 17

Question: Reflect the point $(3,5)$ about the line $y = x$.

Answer:

Reflection about $y = x$: $(x,y) \to (y, x)$ $(3,5) \to (5,3)$ ✓

Example 18

Question: Translate the triangle with vertices $(1,1)$, $(3,1)$, $(2,4)$ by vector $(2,-1)$.

Answer:

New vertices: $(3,0)$, $(5,0)$, $(4,3)$ ✓

PART 3: PRACTICE EXERCISES

Exercise Set A: Pythagorean Triples (15 Questions)

Generate Pythagorean triples for the given m,n.

A1. $m=2, n=1$
A2. $m=3, n=2$
A3. $m=4, n=3$
A4. $m=5, n=4$
A5. $m=5, n=2$
A6. $m=5, n=3$
A7. $m=6, n=5$
A8. $m=6, n=4$
A9. $m=7, n=6$
A10. $m=7, n=2$
A11. $m=8, n=3$
A12. $m=8, n=5$
A13. $m=9, n=2$
A14. $m=9, n=4$
A15. $m=10, n=7$

Exercise Set B: Area Calculations (Shoelace) (10 Questions)

Find the area of polygon with given vertices.

B1. Triangle: $(0,0), (5,0), (0,7)$
B2. Triangle: $(1,2), (4,6), (7,3)$
B3. Triangle: $(2,3), (5,8), (9,4)$
B4. Quadrilateral: $(0,0), (4,0), (5,3), (1,4)$
B5. Quadrilateral: $(0,0), (6,0), (5,4), (0,5)$
B6. Quadrilateral: $(2,1), (5,1), (6,4), (1,4)$
B7. Pentagon: $(0,0), (4,0), (5,2), (3,5), (0,4)$
B8. Pentagon: $(1,1), (4,1), (6,3), (3,5), (0,3)$
B9. Hexagon: $(0,0), (3,0), (4,2), (3,5), (0,5), (-1,2)$
B10. Hexagon: $(2,2), (5,2), (7,4), (5,7), (2,7), (0,4)$

Exercise Set C: Circle Theorems (10 Questions)

Answer by mere observation (Vilokanam).

C1. Angle subtended by a diameter at the circumference is ___
C2. In a cyclic quadrilateral, opposite angles sum to ___
C3. Angle between a chord and a tangent equals angle in the ___ segment
C4. In a circle, equal chords subtend ___ angles at the center
C5. The angle at the center is ___ the angle at the circumference subtended by the same chord
C6. A triangle inscribed in a semicircle is always ___-angled
C7. If a quadrilateral has opposite angles summing to 180°, it is ___
C8. The tangent at any point is ___ to the radius at that point
C9. Angles in the same segment of a circle are ___
C10. The angle between two tangents drawn from an external point is ___ to the difference of the intercepted arcs

Exercise Set D: Conic Sections (10 Questions)

Identify the conic and its key features.

D1. $x^2 + y^2 = 25$
D2. $\frac{x^2}{16} + \frac{y^2}{9} = 1$
D3. $\frac{x^2}{25} - \frac{y^2}{9} = 1$
D4. $y^2 = 16x$
D5. $x^2 = -8y$
D6. $9x^2 + 4y^2 = 36$
D7. $x^2 - y^2 = 1$
D8. $4x^2 + 4y^2 = 1$
D9. $y = x^2$
D10. $xy = 1$ (rectangular hyperbola)

Exercise Set E: Solid Geometry (10 Questions)

Find volume and/or surface area.

E1. Cube of side 5 cm (volume)
E2. Rectangular prism $l=6, w=4, h=3$ (volume)
E3. Cylinder radius 4 cm, height 7 cm (volume)
E4. Cone radius 3 cm, height 10 cm (volume)
E5. Sphere radius 5 cm (volume)
E6. Cylinder radius 2 cm, height 10 cm (surface area)
E7. Cone radius 6 cm, height 8 cm (slant height and total surface area)
E8. Sphere radius 7 cm (surface area)
E9. Hemisphere radius 4 cm (volume)
E10. Pyramid with square base side 6 cm and height 9 cm (volume)

Exercise Set F: Transformational Geometry (10 Questions)

Apply the given transformation.

F1. Translate $(3,4)$ by vector $(2,-5)$
F2. Rotate $(2,3)$ by $90°$ counterclockwise
F3. Rotate $(4,1)$ by $180°$ about origin
F4. Rotate $(5,2)$ by $270°$ counterclockwise ($-90°$)
F5. Reflect $(3,7)$ about x-axis
F6. Reflect $(4,5)$ about y-axis
F7. Reflect $(6,2)$ about line $y = x$
F8. Translate triangle with vertices $(1,1), (3,1), (2,4)$ by vector $(2,3)$
F9. Rotate point $(-2,5)$ by $90°$ clockwise
F10. Find the image of $(a,b)$ after reflection about line $y = -x$

Answer Key for Practice Exercises

Set A Answers (Pythagorean Triples):

A1. (3,4,5)
A2. (5,12,13)
A3. (7,24,25)
A4. (9,40,41)
A5. (21,20,29)
A6. (16,30,34) — divide by 2 → (8,15,17)
A7. (11,60,61)
A8. (20,48,52) — divide by 4 → (5,12,13)
A9. (13,84,85)
A10. (45,28,53)
A11. (55,48,73)
A12. (39,80,89)
A13. (77,36,85)
A14. (65,72,97)
A15. (51,140,149)

Set B Answers (Area):

B1. 17.5
B2. 9.5
B3. 13.5
B4. 19? Let me compute: (0,0),(4,0),(5,3),(1,4): Sum1=0+12+15+0=27, Sum2=0+0+3+4=7, Area=10
B5. 20.5? (0,0),(6,0),(5,4),(0,5): Sum1=0+24+20+0=44, Sum2=0+0+20+0=20, Area=12
B6. 12
B7. 17.5
B8. 16
B9. 19.5
B10. 27

Set C Answers (Circle Theorems):

C1. 90°
C2. 180°
C3. alternate
C4. equal
C5. twice
C6. right
C7. cyclic
C8. perpendicular
C9. equal
C10. supplementary

Set D Answers (Conics):

D1. Circle, center (0,0), radius 5
D2. Ellipse, a=4, b=3
D3. Hyperbola, a=5, b=3
D4. Parabola, a=4, focus (4,0)
D5. Parabola, a=2, focus (0,-2)
D6. Ellipse, a=3, b=2
D7. Hyperbola, a=1, b=1
D8. Circle, center (0,0), radius 1/2
D9. Parabola, vertex (0,0), opening up
D10. Rectangular hyperbola

Set E Answers (Solid Geometry):

E1. 125 cm³
E2. 72 cm³
E3. 112π cm³
E4. 30π cm³
E5. 500π/3 cm³
E6. 48π cm²
E7. l=10, TSA=96π cm²
E8. 196π cm²
E9. 128π/3 cm³
E10. 108 cm³

Set F Answers (Transformations):

F1. (5,-1)
F2. (-3,2)
F3. (-4,-1)
F4. (2,-5)
F5. (3,-7)
F6. (-4,5)
F7. (2,6)
F8. (3,4), (5,4), (4,7)
F9. (5,2)
F10. (-b,-a)

PART 5: TEACHER'S GUIDE

Common Mistakes & Corrections

Memory Aids

QUICK REFERENCE CARD

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║                     MODULE 28 — GEOMETRY: VEDIC APPROACH               ║
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║                                                                       ║
║  PYTHAGOREAN TRIPLES: (m²-n², 2mn, m²+n²)                            ║
║                                                                       ║
║  AREA (Shoelace): ½|Σ(xᵢyᵢ₊₁) - Σ(yᵢxᵢ₊₁)|                           ║
║                                                                       ║
║  CIRCLE THEOREMS (Vilokanam):                                         ║
║  • Angle in semicircle = 90°                                          ║
║  • Opposite angles cyclic quad = 180°                                 ║
║  • Angle at center = 2× angle at circumference                        ║
║                                                                       ║
║  CONIC SECTIONS:                                                      ║
║  Circle:   x² + y² = r²                                               ║
║  Ellipse:  x²/a² + y²/b² = 1                                          ║
║  Hyperbola: x²/a² - y²/b² = 1                                         ║
║  Parabola: y² = 4ax (right)                                           ║
║                                                                       ║
║  VOLUMES:                                                             ║
║  Prism/Cylinder: Base Area × Height                                   ║
║  Pyramid/Cone: ⅓ × Base Area × Height                                 ║
║  Sphere: ⁴⁄₃πr³                                                       ║
║                                                                       ║
║  TRANSFORMATIONS:                                                     ║
║  90° CCW: (-y, x) | 180°: (-x, -y)                                   ║
║  Reflect x-axis: (x, -y) | y-axis: (-x, y) | y=x: (y, x)             ║
║                                                                       ║
║  SUTRAS: Vilokanam (12), Vyashti Samashti (11), Urdhva (3)            ║
║                                                                       ║
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