Geometry – Study Notes for IIFT Quantitative Ability
Audience: IIFT‑PCM aspirants
Length: 4‑5 pages (≈1200 words)
Format: Markdown – self‑contained, easy to skim
1. Simple Explanation
Geometry is the part of mathematics that deals with the properties, measurement, and relationships of points, lines, angles, shapes, and solids. In IIFT’s Quant section, you’ll encounter problems on:
- Plane geometry – triangles, quadrilaterals, circles, polygons.
- Solid geometry – prisms, cylinders, pyramids, spheres.
- Coordinate geometry – points, slopes, distances, mid‑points, circles in the Cartesian plane.
- Trigonometry – (occasionally) basic ratios to compute lengths or angles.
The key to mastering geometry is recognizing patterns and using the right formula quickly.
2. Key Points & Formulas
| Category | Concept | Formula / Property | Quick Note |
|---|---|---|---|
| Angles | Adjacent/Vertical | Adjacent + Vertical = 180° | Use to find missing angles |
| Complementary | α + β = 90° | Often in right triangles | |
| Supplementary | α + β = 180° | For angles in straight line | |
| Triangles | Types | Equilateral – all sides equal. Isosceles – two equal sides. Scalene – none equal. | Recognise by sides/angles |
| Pythagoras | (a^2 + b^2 = c^2) (right triangle) | Identify hypotenuse first | |
| Area | (\frac{1}{2} \times \text{base} \times \text{height}) | Height may be alt. to base | |
| Heron’s Formula | (\sqrt{s(s-a)(s-b)(s-c)}) where (s=\frac{a+b+c}{2}) | For non‑right triangle | |
| Angle Bisector | (\frac{AB}{AC} = \frac{BD}{DC}) | Useful in ratio problems | |
| Similarity | (\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}) | Check proportional sides & angles | |
| Quadrilaterals | Parallelogram | Opposite sides =, Opposite angles = | Opposite sides parallel |
| Rectangle | All angles 90° | Diagonals equal | |
| Rhombus | All sides equal | Diagonals bisect at 90° | |
| Trapezium | One pair parallel | Only in British English; in Indian terms, trapezoid = only one pair | |
| Kite | Two distinct pairs of adjacent sides equal | Diagonal perpendicular to base | |
| Circles | Radius/diameter | (d = 2r) | Diameter = longest chord |
| Chord | (c = 2\sqrt{r^2 - d^2}) where d = distance from centre | Use when chord length needed | |
| Tangent | Tangent ⟂ radius | Length of tangent from external point = (\sqrt{PT^2 - r^2}) | |
| Coordinate Geometry | Distance | (d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}) | Basic |
| Mid‑point | (\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)) | ||
| Slope | (m = \frac{y_2-y_1}{x_2-x_1}) | Parallel lines have same (m); perpendicular lines have (m_1m_2 = -1) | |
| Equation of a line | (y-y_1 = m(x-x_1)) | ||
| Circle (Cartesian) | ((x-h)^2 + (y-k)^2 = r^2) | (h,k) = centre | |
| Solid Geometry | Cube | (V = a^3), (S = 6a^2) | a = side |
| Rectangular Prism | (V = lwh), (S = 2(lw+lh+wh)) | ||
| Cylinder | (V = \pi r^2h), (S = 2\pi r(h+r)) | ||
| Sphere | (V = \frac{4}{3}\pi r^3), (S = 4\pi r^2) | ||
| Pyramid | (V = \frac{1}{3}Bh) | B = base area | |
| Trigonometry (basic) | sin, cos, tan | (\sin\theta = \frac{\text{opp}}{\text{hyp}}) etc. | Needed only for “quick ratio” problems |
Tip: Write down “knowns” and “unknowns” on a separate sheet; geometry is all about matching the right formula to the right knowns.
3. Worked Example
Question (IIFT‑PCM style)
In a right‑angled triangle (ABC), ( \angle C = 90^\circ).
The altitude from (C) to the hypotenuse (AB) (meeting the hypotenuse at (D)) has length (CD = 6) cm.
The segment (BD = 10) cm.
Find the area of (\triangle ABC).
Step‑by‑Step Solution
- Identify what we know
- Right triangle → Pythagoras applicable.
- Altitude to hypotenuse splits it into two segments (AD) and (BD).
- (BD = 10) cm, (CD = 6) cm.
- Need area (= \frac{1}{2} \times \text{base} \times \text{height}).
- Here, base