Geometry – XAT Quantitative Ability Study Notes
(Designed for IIT‑JEE / competitive exam level)
Why Geometry?
Geometry questions test spatial reasoning, pattern identification, and algebraic manipulation. XAT frequently uses quick geometry – simple formulas, ratio tricks, and coordinate tricks – but does not shy away from challenging proofs or derivations. Mastery of core theorems, quick mental calculations, and the ability to spot “trick” angles or ratios will give you a big advantage.
Strategy
- Identify the shape (triangle, rectangle, circle, etc.)
- Recall the shortest formula for the requested quantity.
- Check for hidden symmetry or special ratios (30‑60‑90, 45‑45‑90, isosceles).
- Use coordinate tricks to eliminate algebra when possible.
- Cross‑verify with area/perimeter/volume relationships.
1. Basic Definitions & Quick Reference
| Symbol | Definition | Key Idea |
|---|---|---|
| (AB, ; BC,) etc. | Length of side | Use (AB = \sqrt{(x_B-x_A)^2+(y_B-y_A)^2}) in coordinate geometry |
| (m\angle XYZ) | Measure of angle | (m\angle XYZ = \arccos\frac{XZ^2+YZ^2-XY^2}{2XZ\cdot YZ}) |
| (r, R) | Inradius, circumradius | (r = \frac{2\Delta}{P}), (R = \frac{abc}{4\Delta}) |
| (s) | Semiperimeter | (s = \frac{a+b+c}{2}) |
| (\Delta) | Area of triangle | ( \Delta = rs = \sqrt{s(s-a)(s-b)(s-c)}) (Heron) |
| (d) | Distance | (d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}) |
| (m) | Slope | (m = \frac{y_2-y_1}{x_2-x_1}) |
| (r) | Radius | (r = \frac{1}{\sqrt{m^2+1}} |
2. Coordinate Geometry (CG)
2.1 Distance & Midpoint
- Distance between ((x_1,y_1)) and ((x_2,y_2)): [ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} ]
- Midpoint (M) of (\overline{AB}): [ M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) ]
2.2 Slope & Parallel/Perpendicular Lines
- Slope of (\overline{AB}): [ m_{AB} = \frac{y_B-y_A}{x_B-x_A} ]
- Parallelism: (m_1 = m_2).
- Perpendicularity: (m_1 m_2 = -1).
2.3 Equation of a Line
- Point–Slope: (y-y_1 = m(x-x_1)).
- General form: (Ax+By+C=0).
- Distance from point ((x_0,y_0)) to line: [ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2+B^2}} ]
2.4 Circle in the Plane
-
General Equation: [ x^2 + y^2 + Dx + Ey + F = 0 ] Center (\bigl(-\frac{D}{2}, -\frac{E}{2}\bigr)), radius (r = \sqrt{\left(\frac{D}{2}\right)^2 + \left(\frac{E}{2}\right)^2 - F}).
-
Intersecting a line: substitute (y = mx + c) → quadratic in (x).
- Discriminant > 0 → 2 intersection points.
- Discriminant = 0 → tangent.
- Discriminant < 0 → no intersection.
2.5 Quick Tricks
| Problem | Trick |
|---|---|
| Area of triangle with vertices ((x_1,y_1),(x_2,y_2),(x_3,y_3)) | Use shoelace formula: |
| (\displaystyle \Delta = \frac12 | x_1y_2+x_2y_3+x_3y_1 - x_2y_1 - x_3y_2 - x_1y_3 |
| Length of a diagonal in a rectangle | Use Pythagoras: (d = \sqrt{l^2 + w^2}). |
| Check if a point lies on a circle | Plug into circle equation. |
| Finding radius from 3 points on a circle | Compute pairwise perpendicular bisectors; intersection gives center; distance to any point is radius. |
3. Triangles
| Property | Formula / Statement |
|---|---|
| Pythagoras | (c^2 = a^2 + b^2) (right triangle |