In this guide
Geometry in SSC CHSL is lighter than in SSC CGL. Most questions test one property at a time: an exterior angle, a right triangle, an angle in a semicircle, a tangent. The difficulty is not the maths but recall under time pressure. If the right property comes to mind in five seconds, the question takes under a minute. If it doesn't, you can stare at it for three.
Tier 1 usually has a small number of geometry questions in the 25-question maths section. Tier 2 maths has some too, sometimes with two properties combined. Check recent papers for the current mix rather than trusting any fixed count.
The single best habit: draw a rough figure for every question and write the given values on it. Half the time, the answer is visible once the figure is marked.
Lines and angles
- Angles on a straight line add to 180°; angles around a point add to 360°.
- Vertically opposite angles are equal.
- When a transversal cuts parallel lines, corresponding angles are equal, alternate angles are equal, and co-interior angles (same side, between the lines) add to 180°.
Why co-interior angles add to 180°: each co-interior angle forms a straight line with an angle that is alternate to the other one. So the pair is really one straight angle split in two.
Triangles
- Angle sum = 180°.
- Exterior angle = sum of the two opposite interior angles. Why: the exterior angle and its adjacent interior angle make 180°, and so do all three interior angles. Remove the shared angle from both and what is left is equal.
- Triangle inequality: any two sides together are longer than the third. Use it to reject impossible options.
- The largest angle is opposite the longest side.
Pythagoras: in a right triangle, hypotenuse² = base² + height². Learn the common triplets and their multiples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (9, 40, 41). Spotting a triplet saves you the square root.
Congruence and similarity
Similar triangles have equal angles and sides in the same ratio. Two results carry most questions:
- Basic proportionality: a line parallel to one side of a triangle cuts the other two sides in the same ratio, and the small triangle it cuts off is similar to the whole.
- Areas of similar triangles are in the ratio of the squares of corresponding sides. Why: area is base × height ÷ 2, and both base and height scale by the same ratio k, so area scales by k².
Centres of a triangle
| Centre | Where lines meet | Key result |
|---|---|---|
| Centroid (G) | Medians | Divides each median 2 : 1 from the vertex; the three medians cut the triangle into six equal areas |
| Incentre (I) | Angle bisectors | ∠BIC = 90° + ∠A/2 |
| Circumcentre (O) | Perpendicular bisectors of sides | ∠BOC = 2∠A (acute triangle) |
| Orthocentre (H) | Altitudes | ∠BHC = 180° − ∠A |
Why ∠BIC = 90° + ∠A/2: in triangle BIC the angles at B and C are ∠B/2 and ∠C/2. So ∠BIC = 180° − (∠B + ∠C)/2 = 180° − (180° − ∠A)/2 = 90° + ∠A/2.
Why ∠BOC = 2∠A: it is the circle theorem below. B and C lie on the circumcircle, and O is its centre.
Circles
| Property | Statement |
|---|---|
| Angle at the centre | Twice the angle at the circumference on the same arc |
| Angle in a semicircle | 90° |
| Same segment | Angles on the same arc are equal |
| Cyclic quadrilateral | Opposite angles add to 180° |
| Perpendicular from centre to chord | Bisects the chord |
| Tangent | Perpendicular to the radius at the point of contact |
| Two tangents from a point | Equal in length |
Two formulas follow from Pythagoras. Tangent length from a point at distance d from the centre = √(d² − r²), because the radius meets the tangent at 90°. A chord at distance d from the centre has length 2√(r² − d²), because the perpendicular from the centre bisects it.
Polygons
- Sum of interior angles = (n − 2) × 180°. Why: diagonals from one vertex split an n-sided polygon into n − 2 triangles.
- Each exterior angle of a regular polygon = 360°/n. Why: walking once round any polygon turns you through one full turn.
- Number of diagonals = n(n − 3)/2.
For a regular polygon, find the exterior angle first (180° − interior). It is almost always the faster route to n.
Six worked questions
Q1. Two parallel lines are cut by a transversal. A pair of co-interior angles measure (3x + 10)° and (2x + 20)°. Find the larger angle.
Co-interior angles add to 180°: 5x + 30 = 180, so x = 30. The angles are 100° and 80°. Answer: 100°.
Q2. The angles of a triangle are in the ratio 2 : 3 : 4. Find the exterior angle at the vertex with the smallest angle.
Angles are 40°, 60° and 80°. The exterior angle = 180 − 40 = 140°, which also equals 60 + 80.
Q3. A chord of length 16 cm is 6 cm from the centre of a circle. Find the radius.
The perpendicular bisects the chord, giving half-chord 8 cm. Radius = √(8² + 6²) = 10 cm (a 6-8-10 triplet).
Q4. In triangle ABC, D is on AB and E is on AC with DE ∥ BC. AD = 3 cm, DB = 2 cm and the area of triangle ADE is 18 cm². Find the area of trapezium DBCE.
Triangle ADE is similar to ABC with ratio AD : AB = 3 : 5, so areas are 9 : 25. Area of ABC = 18 × 25/9 = 50 cm². Trapezium = 50 − 18 = 32 cm².
Q5. In triangle ABC, ∠A = 70°. I is the incentre and O is the circumcentre. Find ∠BIC and ∠BOC.
∠BIC = 90 + 35 = 125°. ∠BOC = 2 × 70 = 140°.
Q6. PA and PB are tangents from P to a circle with centre O, and ∠APB = 50°. Find ∠AOB.
In quadrilateral OAPB, the angles at A and B are 90° each. So ∠AOB = 360 − 90 − 90 − 50 = 130°.
Common mistakes
| Mistake | Fix |
|---|---|
| Taking area ratio = side ratio | Square the side ratio |
| Using the whole chord in Pythagoras | Use half the chord |
| Dividing 360 by the interior angle | Divide 360 by the exterior angle |
| Forgetting the right angle at the point of contact | Draw the radius to every tangent |
| Assuming a figure is drawn to scale | Use only what is given or proved |
Practice
- Two angles of a triangle are 40° and 75°. Find the third angle.
- A right triangle has legs of 9 cm and 12 cm. Find its hypotenuse.
- Two similar triangles have areas of 25 cm² and 64 cm². Find the ratio of their corresponding sides.
- An arc subtends 80° at the centre of a circle. What angle does it subtend at a point on the remaining part of the circle?
- Each interior angle of a regular polygon is 150°. Find the number of sides and the number of diagonals.
- In cyclic quadrilateral ABCD, ∠A = (2x + 10)° and ∠C = (3x − 5)°. Find ∠C.
- A point is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from the point.
- In triangle ABC, the incentre I gives ∠BIC = 116°. Find ∠A.
Answers:
- 65°. 180 − 40 − 75.
- 15 cm. A 3-4-5 triplet scaled by 3.
- 5 : 8. Side ratio = √25 : √64.
- 40°. Half the angle at the centre.
- 12 sides, 54 diagonals. Exterior angle = 30°, so n = 360/30 = 12; diagonals = 12 × 9/2.
- 100°. Opposite angles add to 180: 5x + 5 = 180, x = 35, so ∠C = 105 − 5.
- 12 cm. √(169 − 25) = √144.
- 52°. 90 + ∠A/2 = 116, so ∠A/2 = 26.
What to do next
- Write the centres table and the circle table from memory, then check them.
- Solve 20 mixed geometry questions, drawing a figure for each one.
- Learn the five triplets and test yourself on their multiples.
- Move on to mensuration, which uses these shapes, and trigonometry, which builds on right triangles.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Staff Selection Commission website .
Get the next SSC CHSL guide by email
New guides every week. No spam, unsubscribe any time.