Geometry Shortcuts for SSC CGL, CHSL and Other Exams

Geometry is the chapter students either love or avoid. If you avoid it, you are giving up a real chunk of the SSC quant section, because geometry, mensuration and trigonometry together tend to make up a noticeable share of questions in the papers. The exact split changes with each year and each exam, so check the latest notification and past papers for the number in your target exam. Either way, geometry is where students who decide to "skip it" often lose marks they could have taken.

Here is what makes it easier than it looks. SSC geometry is not proof-based. It is a set of about thirty facts that get used in different arrangements, and a diagram is almost always given or easy to draw. You do not need creativity. You need to recognise which fact fits the picture. That is a memory-and-practice problem, which means it can be fixed in a few weeks.

This post lists the facts that matter most, with the reasoning behind them, and works through numeric examples so you can see how they look in a question. Every number has been checked. If you are also preparing for trigonometry, the next post, trigonometry formulas for SSC, builds directly on the right-triangle work here.

Draw first, always

Before any formula, draw the figure on your rough sheet, even if a diagram is printed. Write the given values on it. Half the errors in geometry come from reading a value off the wrong side or mixing up two triangles that share a vertex. A neat, labelled sketch takes 10 seconds and saves 30.

Also use the answer options. In geometry, an option like "60 degrees" or "12 cm" lets you reject choices by rough proportion. If an angle in your drawing clearly looks less than 90, do not pick 110.

Triangles: the angle facts

The angles of a triangle sum to 180 degrees. An exterior angle equals the sum of the two opposite interior angles. These two facts alone solve most basic questions.

The special points of a triangle each come with an angle rule. Learn these four, because they are asked repeatedly.

Incentre (where the angle bisectors meet). If the vertex angle is A, the angle at the incentre, BIC, equals 90 + A/2. For A = 50 degrees, angle BIC = 90 + 25 = 115 degrees.

Circumcentre (perpendicular bisectors of the sides). Angle BOC = 2A. For A = 50, BOC = 100 degrees. This is the angle at the centre being twice the angle at the circumference.

Orthocentre (altitudes). Angle BHC = 180 - A. For A = 50, BHC = 130 degrees.

Excentre and the exterior angle bisectors. The angle between the bisectors of two exterior angles at B and C is 90 - A/2. For A = 50, that is 90 - 25 = 65 degrees.

A related fact: the angle between the altitude from A and the bisector of angle A is half the difference of the other two angles, |B - C|/2. If B = 70 and C = 40, the angle is 30/2 = 15 degrees.

Do not derive these in the exam. Memorise them as a small table and check them on a triangle you draw at the start of your study, so you trust them.

Point	Angle formed at the point
Incentre	BIC = 90 + A/2
Circumcentre	BOC = 2A
Orthocentre	BHC = 180 - A
Excentre (opposite A)	B(exc)C = 90 - A/2

Centroid and medians

The centroid is where the medians meet, and it divides each median in the ratio 2 : 1, the longer part being on the vertex side. So if a median is 9 cm long, the centroid is 6 cm from the vertex and 3 cm from the midpoint of the opposite side.

In a right-angled triangle, the median to the hypotenuse equals half the hypotenuse. For a 6-8-10 triangle, the median to the hypotenuse is 5. That also means the circumcentre of a right triangle is the midpoint of the hypotenuse, and the circumradius is half the hypotenuse.

The mid-point theorem: the line joining the midpoints of two sides is parallel to the third side and half as long. If D and E are midpoints of AB and AC and BC is 18, then DE = 9.

Pythagorean triplets

Memorise these until you can spot them in disguise:

3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29, 9-40-41, 12-35-37, and the multiples of each (6-8-10, 9-12-15, 12-16-20, 15-20-25, 10-24-26, and so on).

Why bother? Because half the time the question is designed so a triplet is hiding. If two sides of a right triangle are 15 and 20, you should hear "3-4-5 times 5" and answer 25 without squaring anything. If the hypotenuse is 26 and one side is 10, it is the 5-12-13 triplet times 2, so the other side is 24.

The same idea speeds up distance problems, tangent lengths, chord lengths and diagonal questions. A rectangle with sides 8 and 15 has a diagonal of 17. A rhombus with diagonals 16 and 30 has half-diagonals 8 and 15, so its side is 17, and its area is (16 x 30)/2 = 240.

Also useful: in a right triangle, the altitude to the hypotenuse equals (product of the legs)/hypotenuse. For a 6-8-10 triangle it is 48/10 = 4.8.

Heron's formula and the radii

For a triangle with sides a, b, c and semi-perimeter s = (a + b + c)/2:

Area = sqrt(s(s - a)(s - b)(s - c)).

Inradius r = Area / s. Circumradius R = abc / (4 x Area).

Take the 13-14-15 triangle, which SSC loves. s = 21. Area = sqrt(21 x 8 x 7 x 6) = sqrt(7,056) = 84. Inradius = 84/21 = 4. Circumradius = (13 x 14 x 15)/(4 x 84) = 2,730/336 = 8.125, or 65/8. The altitude to the side of 14 is (2 x 84)/14 = 12, which should not surprise you because 13-14-15 splits into a 5-12-13 triangle and a 9-12-15 triangle along that altitude. Remember that: 13-14-15 is two right triangles glued together, and it gives height 12.

For a right triangle, the inradius has a shortcut: r = (a + b - c)/2, where c is the hypotenuse. For 5-12-13, r = (5 + 12 - 13)/2 = 2. The circumradius is c/2 = 6.5.

For an equilateral triangle of side a, area = (sqrt(3)/4) a^2, height = (sqrt(3)/2) a, inradius = a/(2 sqrt(3)), circumradius = a/sqrt(3). For a = 12, area is (sqrt(3)/4) x 144 = 36 sqrt(3), about 62.35. Height is 6 sqrt(3), about 10.39. Inradius is 12/(2 sqrt(3)) = 2 sqrt(3), about 3.46. Circumradius is 12/sqrt(3) = 4 sqrt(3), about 6.93. Note that circumradius is exactly twice the inradius in an equilateral triangle, which you can use to check yourself.

Similar triangles and the basic proportionality theorem

If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio. And the small triangle it cuts off is similar to the whole one.

The key relationships for similar triangles: sides are in ratio k, areas are in ratio k squared, perimeters are in ratio k.

Here is a typical one. In triangle ABC, DE is parallel to BC with D on AB and E on AC. AD = 3 and DB = 5. The area of triangle ABC is 320. Find the area of the trapezium DBCE.

AD : AB = 3 : 8. So the area ratio of triangle ADE to triangle ABC is 9 : 64. Area of ADE = (9/64) x 320 = 45. Trapezium = 320 - 45 = 275. Do not compute the trapezium directly; find the small triangle and subtract.

Angle bisector theorem

The bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides. In triangle ABC with AB = 6, AC = 9 and BC = 10, the bisector from A meets BC at D with BD : DC = 6 : 9 = 2 : 3. So BD = 4 and DC = 6.

The external bisector divides the opposite side externally in the same ratio, and this rarely appears; if you see it, apply the same ratio with the external division.

Circles: the facts that repeat

The list of circle facts is short. Learn each with a small drawing.

• The angle subtended by an arc at the centre is twice the angle at any point on the remaining circle. If the angle at the centre is 80 degrees, the angle at the circumference is 40.
• The angle in a semicircle is 90 degrees.
• Angles in the same segment are equal.
• Opposite angles of a cyclic quadrilateral sum to 180 degrees. If one angle is 100, its opposite is 80. An exterior angle of a cyclic quadrilateral equals the interior opposite angle.
• A tangent is perpendicular to the radius at the point of contact.
• Two tangents from an external point are equal in length.
• The angle between a tangent and a chord equals the angle in the alternate segment.
• The perpendicular from the centre to a chord bisects the chord.

Now numerical examples.

Tangent length. A point is 13 cm from the centre of a circle of radius 5 cm. The tangent length is sqrt(13^2 - 5^2) = sqrt(144) = 12 cm. It is the 5-12-13 triplet, and the radius and tangent form the right angle.

Chord length. A circle has radius 13 cm and a chord lies 12 cm from the centre. Half the chord = sqrt(13^2 - 12^2) = sqrt(25) = 5, so the chord is 10 cm. Again a triplet.

Two chords intersecting inside. If chords AB and CD intersect at P inside the circle, PA x PB = PC x PD. If PA = 6 and PB = 4, and PC = 8, then PD = 24/8 = 3. Check: 6 x 4 = 24 and 8 x 3 = 24.

Secant and tangent from an outside point. If a tangent PT and a secant PAB are drawn from P, then PT^2 = PA x PB. If PT = 12 and the near intersection PA = 8, then PB = 144/8 = 18, and the chord AB = 18 - 8 = 10.

Common tangents

For two circles with radii r1 and r2 whose centres are d apart:

• Length of the direct common tangent = sqrt(d^2 - (r1 - r2)^2).
• Length of the transverse (cross) common tangent = sqrt(d^2 - (r1 + r2)^2).

For d = 17, r1 = 10, r2 = 2: direct tangent = sqrt(289 - 64) = sqrt(225) = 15. For d = 13, r1 = 3, r2 = 2: transverse tangent = sqrt(169 - 25) = sqrt(144) = 12.

The transverse tangent exists only when the circles do not overlap, meaning d is greater than r1 + r2. If d equals r1 + r2, the circles touch externally and the transverse tangent length is zero (they touch at a single point). If d equals r1 - r2, they touch internally.

The number of common tangents is 4 for separate circles, 3 for externally touching, 2 for overlapping, 1 for internally touching, and 0 for one circle inside the other. The count comes up as a direct question more often than you would expect.

Polygons

For a regular polygon with n sides:

• Each exterior angle = 360/n.
• Each interior angle = (n - 2) x 180/n.
• Number of diagonals = n(n - 3)/2.
• Sum of interior angles = (n - 2) x 180.

Quick values: a regular hexagon has interior angle 120, an octagon 135, and a decagon 144. If each exterior angle is 24 degrees, the polygon has 360/24 = 15 sides, and each interior angle is 156 degrees. An octagon has 8 x 5/2 = 20 diagonals.

The exterior angle route is nearly always faster. If a question says "each interior angle is 135 degrees", jump to the exterior angle of 45 and the polygon has 8 sides.

A regular hexagon of side a is made of six equilateral triangles of side a, so its area is 6 x (sqrt(3)/4) a^2 = (3 sqrt(3)/2) a^2. For a = 4, that is 24 sqrt(3), about 41.57. Its circumradius equals its side, and the inradius is (sqrt(3)/2) a.

Quadrilaterals and mensuration

Mensuration overlaps with geometry in most papers, and you should treat it as one block.

Rhombus. Diagonals bisect each other at right angles. Area = (d1 x d2)/2. Side = sqrt((d1/2)^2 + (d2/2)^2). For diagonals 16 and 30, side 17 and area 240.

Trapezium. Area = (1/2)(sum of parallel sides) x height. For parallel sides 10 and 14 and height 8, area = 12 x 8 = 96.

Triangle area with two sides and the included angle. Area = (1/2) ab sin C. For sides 10 and 12 with a 30-degree angle between them, area = (1/2)(10)(12)(1/2) = 30.

Parallelogram. Area = base x height. A rectangle is a parallelogram with a 90-degree angle, and a square is a rhombus with a 90-degree angle. The diagonal of a square of side a is a sqrt(2). The diagonal of a rectangle is sqrt(l^2 + b^2).

Circle. Circumference = 2 pi r, area = pi r^2. Sector area = (theta/360) x pi r^2. Arc length = (theta/360) x 2 pi r. Use pi = 22/7 when the radius is a multiple of 7, or 3.14 if the question says so.

Cylinder, cone, sphere, cube and cuboid. Volume formulas are their own list, and it is worth writing a one-page sheet: cuboid l x b x h, cube a^3, cylinder pi r^2 h, cone (1/3) pi r^2 h, sphere (4/3) pi r^3, hemisphere (2/3) pi r^3. Curved surface area of a cylinder is 2 pi r h, of a cone pi r l where l is slant height, and of a sphere 4 pi r^2.

The mensuration questions are usually the easiest in the quant section. If you have to choose between spending your last five minutes on geometry proofs or mensuration, take mensuration. It is arithmetic once you know the formula.

Parallel lines

The parallel line facts show up as an easy first question in many papers. When a transversal cuts two parallel lines: corresponding angles are equal, alternate angles are equal, and co-interior angles sum to 180 degrees. A common form asks you to find an angle when you are given two or three others and a pair of parallels. Draw an extra line parallel through the vertex and use alternate angles. Then it is addition.

How to read a geometry question quickly

Here is the process I would use for any question, in about 20 seconds.

1. Sketch, label, and mark equal items (equal sides with tick marks, equal angles with arcs).
2. Ask which object the question is really about: triangle, circle, quadrilateral. Most geometry questions are one of about ten types.
3. Look for a right angle. If there is one, look for a triplet. If you see a 3-4-5 scaled up, write it down.
4. Look for parallel lines or a circle. Those bring in the equal-angle and tangent facts.
5. Estimate. Check the options against the sketch.

Then apply the fact. If you cannot see which fact applies in 20 seconds, skip. That is what the mark-for-review button is for. Some geometry questions in the harder papers are built to take two facts in a row, and they are not worth two minutes.

A worked set: five questions in the SSC style

Here are five representative questions with full working, to show how the facts combine. All values were checked.

Question 1. In triangle ABC, angle A is 50 degrees. The bisectors of angles B and C meet at I. Find angle BIC. Answer: 90 + 50/2 = 115 degrees.

Question 2. The sides of a triangle are 13, 14 and 15. Find the inradius. Answer: s = 21, area = 84, r = 84/21 = 4.

Question 3. Two circles of radii 10 and 2 have their centres 17 apart. Find the length of the direct common tangent. Answer: sqrt(17^2 - 8^2) = sqrt(225) = 15.

Question 4. DE is parallel to BC in triangle ABC, with AD : DB = 3 : 5. If area of ABC is 320, find the area of the quadrilateral DBCE. Answer: 275, as computed above.

Question 5. Each exterior angle of a regular polygon is 24 degrees. Find the number of diagonals. Answer: n = 360/24 = 15. Diagonals = 15 x 12/2 = 90. Check: n(n - 3)/2 = 15 x 12/2 = 90.

Notice that none of them required more than three lines. The work was in recognising which fact to use.

Mistakes that keep costing marks

The first is using the wrong point of the triangle. The incentre, circumcentre and orthocentre have different angle rules. Students remember "90 plus A over 2" but apply it to the circumcentre. Write the four-line table on a card and check it on any question in this family.

The second is treating a figure as a scale drawing. SSC figures are often not to scale, and an angle that looks 60 might be 75. Trust the numbers, not the drawing.

The third is missing a shortcut triplet. A student computes sqrt(20^2 + 21^2) as sqrt(841) and stops to find the square root, when 20-21-29 is a known triple. Learn the seven or eight triplets and you will save minutes.

The fourth is confusing the direct and transverse tangent formulas. Direct uses the difference of the radii, transverse uses the sum. A quick memory aid: for a direct tangent the radii are on the same side, so the gap in radii matters; for a transverse tangent the circles are on opposite sides, so radii add.

The fifth is confusing area and perimeter ratios in similar figures. Perimeter goes with the ratio of sides (k). Area goes with k squared. Volume goes with k cubed.

The sixth is forgetting units, and mixing centimetres with metres in a mensuration problem. Convert at the start.

How to practise geometry

Geometry improves fastest when you handle it differently from arithmetic. The best approach I have seen is to build a one-page fact sheet and then do 10 to 15 questions per fact family, in this order.

Week one: triangles. Angle rules, centres, Pythagorean triplets, Heron's formula, inradius and circumradius. Around 40 questions.

Week two: circles. Angle facts, tangents, chords, common tangents. Around 40 questions.

Week three: similar triangles, polygons, quadrilaterals and mensuration. Around 40 questions.

Week four: mixed sets of 20, timed at about 25 minutes, plus review.

For each question you get wrong, write the fact you missed on the fact sheet in your own words. A short, personal sheet is much more useful than a long one from a book, because you will actually read it before a mock.

At the end of the fourth week, take a mock and treat geometry as its own block. See how many geometry and mensuration questions you attempted and how many you got right. The free mock tests on Pareeksha give you a section-wise breakdown, so you can see whether geometry is where you are giving away marks, and sectional practice is good for drilling one fact family in a short timed set. If you find geometry is taking more than a minute per question on average, the section-wise time management post is worth a read on how to allocate the paper.

Where geometry fits with the rest

Geometry pairs naturally with trigonometry, because heights and distances are just right-triangle geometry with angles attached. It also connects to the mensuration parts of the syllabus. The maths chapter weightage post explains how to count the geometry share in your target exam's previous papers, rather than trusting anyone's chart.

For RRB exams, geometry and mensuration questions tend to be simpler and more direct, closer to school level. For SSC CGL and CHSL, expect a wider range, including circles, tangents, and the triangle centres. For bank exams, geometry is minimal, mostly area and perimeter inside word problems, though it is worth checking each notification. If you are preparing for two exams at the same time, the SSC geometry set covers the RRB one as well.

What to skip

Skip proofs. SSC does not ask you to prove anything. Skip rare constructions. Do not spend a week on radical axes or Ptolemy's theorem unless you are running out of things to do; they are very rarely asked.

If you are short on time in the exam, skip a geometry question if the figure is complicated and no fact jumps out at you within 20 seconds. Mensuration and direct circle or triangle facts are better value.

Also skip trying to derive everything from scratch in the exam. The shortcuts exist so you can recall them quickly.

Your next step

Make a one-page fact sheet today with four blocks: the triangle centres table, the Pythagorean triplets, the circle facts, and the polygon formulas. Then do 20 triangle questions from any standard book, with a timer, and write down which fact you missed on each one you got wrong. After a week, re-solve the wrong ones without looking at the solutions.

FAQ

Do I need to learn all the proofs? No. SSC asks for values and applications. Knowing why a fact is true helps you remember it, but you will not be asked to prove it.

How many Pythagorean triplets should I know? The first seven or eight and their multiples are enough: 3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29, 9-40-41, 12-35-37.

Should I skip geometry if I am weak at it? Not entirely. Mensuration and direct triangle questions are usually quick and reward memory. You can skip the harder circle and multi-step ones without losing much.

Are geometry figures to scale in SSC? Often not, so rely on the given values and the facts, not on how the figure looks.

How is geometry different for RRB and bank exams? RRB questions are usually simpler, focused on area, perimeter, angles and mensuration. Bank papers rarely test geometry beyond basic area and perimeter. Check the notification for your exam.

How long does it take to get comfortable with geometry? Most students need around three to four weeks of steady practice, about 40 questions per fact family, plus a fortnightly revision.