◆ QA · Geometry
Trigonometry , formulas + CAT PYQs
Focused Geometry kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Trigonometry is here.
Geometry, formula sheet
Show the full Geometry formula sheet (explanations + basic examples)
1Lines & Angles
- The starting toolkit: most "find the angle" questions are solved just by knowing angles on a line make 180° and walking that around the figure.
- Angles on a straight line add to 180°; angles around a point add to 360°.
- Vertically opposite angles are equal.
- Parallel lines cut by a transversal: corresponding & alternate angles equal; co-interior angles sum to 180°.
- Exterior angle of a triangle = sum of the two remote interior angles.
- e.g. Two angles sit on a straight line and one is 110°. The other = 180° − 110° = 70°.
2Triangle, basics & area
- Pick the area formula that matches what you're given: base+height, all three sides (Heron), or two sides and the angle between them.
- Angle sum = 180°. Sum of any two sides > the third side.
- Area = ½ × base × height.
- Heron: Area = √[s(s−a)(s−b)(s−c)], s = (a+b+c)/2.
- Area = ½·a·b·sinθ (θ = included angle); Area = r × s (r = inradius); Area = abc/(4R) (R = circumradius).
- e.g. Sides 3, 4, 5: s = 6, Area = √[6·3·2·1] = √36 = 6 (matches ½·3·4).
3Cosine & Sine rules
- Cosine rule links three sides and one angle (use when you have two sides + included angle, or all three sides); sine rule links sides to opposite angles.
- Cosine rule: c² = a² + b² − 2ab·cosθ.
- cosθ = (a² + b² − c²)/(2ab).
- Sine rule: a/sinA = b/sinB = c/sinC = 2R.
- e.g. Sides 5 and 8 with a 60° angle between them: third side² = 25 + 64 − 2·5·8·½ = 89 − 40 = 49 ⇒ side = 7.
4Angle-bisector & medians
- A bisector splits the far side in the ratio of the two sides it sits between; a median goes to the midpoint, and the centroid cuts it 2:1.
- Angle bisector divides the opposite side in the ratio of the adjacent sides: BD/DC = AB/AC.
- Apollonius: b² + c² = 2m² + ½a² (m = median to side a).
- Median of isosceles (b = c): m² = b² − a²/4.
- Centroid divides each median in ratio 2 : 1 from the vertex.
- e.g. AB = 6, AC = 4, bisector meets BC (length 5) at D: BD:DC = 6:4 = 3:2 ⇒ BD = 3, DC = 2.
5Pythagoras & altitude relations
- In a right triangle the squares of the legs add to the square of the hypotenuse; memorising the triplets saves time in the exam.
- Right triangle: hypotenuse² = base² + height².
- Altitude to hypotenuse (AD⊥BC, right-angled at A): AD² = BD·DC, AB² = BD·BC, AC² = CD·BC.
- Acute: AC² = AB² + BC² − 2·BC·BD; Obtuse: AC² = AB² + BC² + 2·BC·BD.
- Triplets: 3-4-5, 5-12-13, 8-15-17, 7-24-25.
- e.g. Legs 6 and 8: hypotenuse = √(36 + 64) = √100 = 10 (a scaled 3-4-5).
6Congruence & similarity
- Congruent = identical; similar = same shape, scaled. The big CAT lever is that areas of similar figures scale as the square of the side ratio.
- Congruence: SSS, SAS, ASA, AAS, RHS.
- Similarity: AA, SSS, SAS. Corresponding sides are proportional.
- Basic Proportionality (Thales): a line ∥ to one side cuts the others in equal ratios.
- Ratio of areas of similar triangles = (ratio of sides)².
- e.g. Two similar triangles with sides in ratio 2:3 have areas in ratio 4:9. If the smaller has area 8, the larger = 18.
7Special triangles
- Equilateral and the two "set-square" triangles (30-60-90, 45-45-90) have fixed side ratios, recognise them and you can write down sides instantly.
- Equilateral side a: Area = (√3/4)a², height = (√3/2)a, R = a/√3, r = a/(2√3).
- 30-60-90 sides ratio 1 : √3 : 2.
- 45-45-90 sides ratio 1 : 1 : √2.
- From an interior point of an equilateral triangle, sum of ⊥s to the three sides = its height.
- e.g. Equilateral triangle of side 4: area = (√3/4)·16 = 4√3 and height = (√3/2)·4 = 2√3.
8Geometric centres
- Four "centres", each the meeting point of a different set of cevians; the incentre and circumcentre are the ones that show up most in area/radius questions.
- Centroid, intersection of medians (2:1).
- Incentre, intersection of angle bisectors, centre of inscribed circle.
- Circumcentre, intersection of ⊥ bisectors of sides, centre of circumscribed circle.
- Orthocentre, intersection of altitudes.
- e.g. In a right triangle the circumcentre is the midpoint of the hypotenuse, so a 6-8-10 triangle has circumradius = 10/2 = 5.
9Circle, basics
- Two workhorses: the angle at the centre is twice the angle at the rim on the same arc, and any angle drawn on a diameter is a right angle.
- Circumference = 2πr; Area = πr².
- Equal chords subtend equal angles at the centre & are equidistant from it.
- ⊥ from the centre bisects the chord.
- Angle at the centre = 2 × angle at the circumference on the same arc.
- Angle in a semicircle = 90°.
- e.g. An arc subtends 40° at the centre, so it subtends 40°/2 = 20° at any point on the major arc.
10Chords, tangents, secants
- "Power of a point": from any point, the products of the two distances to the circle along a line are equal, chords, secants and tangents all obey it.
- Two chords meeting at P: PA·PB = PC·PD.
- Tangent-secant from external P: PA·PB = PT².
- Tangent ⊥ radius at the point of contact; tangents from an external point are equal.
- Alternate segment theorem: tangent-chord angle = angle in the alternate segment.
- e.g. Two chords cross with parts 3 & 8 on one and 4 & x on the other: 3·8 = 4·x ⇒ x = 6.
11Cyclic quadrilateral & tangents to 2 circles
- If all four corners lie on a circle, opposite angles are supplementary; the tangent formulas give the straight-line distance between two circles' touch points.
- Cyclic quad: opposite angles sum to 180°; exterior angle = opposite interior angle.
- Ptolemy: AB·CD + BC·DA = AC·BD.
- A parallelogram inscribed in a circle is a rectangle.
- Direct common tangent = √[d² − (r₁−r₂)²]; Transverse = √[d² − (r₁+r₂)²].
- e.g. In a cyclic quad one angle is 70°, so its opposite angle = 180° − 70° = 110°.
12Quadrilaterals
- Each special quadrilateral has its own area shortcut, base×height for parallelograms, half-product of diagonals for a rhombus, average of parallel sides times height for a trapezium.
- Parallelogram: opposite sides & angles equal; diagonals bisect each other. Area = base × height.
- Rectangle: all angles 90°, diagonals equal. Square: all sides equal + 90°.
- Rhombus: all sides equal; diagonals ⊥ & bisect each other. Area = ½·d₁·d₂.
- Trapezium: one pair of parallel sides. Area = ½(sum of parallel sides) × height.
- e.g. Rhombus with diagonals 6 and 8: area = ½·6·8 = 24; trapezium with parallel sides 5 & 9, height 4: area = ½·(5+9)·4 = 28.
13Polygons
- Everything flows from "(n−2)·180° of total interior angle"; for a regular polygon the quick route is via the exterior angle, which is just 360°/n.
- Sum of interior angles = (n − 2)·180°.
- Each interior angle (regular) = 180° − 360°/n.
- Each exterior angle (regular) = 360°/n; all exterior angles sum to 360°.
- Number of diagonals = n(n − 3)/2.
- e.g. A regular hexagon (n = 6): each exterior angle = 360°/6 = 60°, so each interior angle = 120°; diagonals = 6·3/2 = 9.
14Regular hexagon
- Think of it as 6 equilateral triangles glued at the centre, that single picture gives its area, diagonals and angles.
- Side s: Area = (3√3/2)·s².
- It is 6 equilateral triangles of side s.
- Longer diagonal = 2s; shorter diagonal = √3·s.
- Interior angle = 120°.
- e.g. Hexagon of side 2: area = (3√3/2)·4 = 6√3; long diagonal = 4, short diagonal = 2√3.
152D mensuration, perimeters & areas
- The basic flat-shape formulas; a sector is just a fraction θ/360 of the whole circle for both its arc and its area.
- Square: P = 4a, Area = a², diagonal = a√2.
- Rectangle: P = 2(l+b), Area = l·b, diagonal = √(l²+b²).
- Circle: C = 2πr, Area = πr².
- Sector (angle θ): arc = (θ/360)·2πr, area = (θ/360)·πr².
- e.g. A 90° sector of a radius-6 circle is ¼ of it: area = ¼·π·36 = 9π, arc = ¼·2π·6 = 3π.
16Cube & cuboid
- A cube is just a cuboid with l = b = h; the space diagonal (corner-to-corner through the body) uses a 3-term Pythagoras.
- Cuboid: Volume = l·b·h; TSA = 2(lb + bh + hl); LSA (4 walls) = 2(l+b)h; diagonal = √(l²+b²+h²).
- Cube edge a: Volume = a³; TSA = 6a²; LSA = 4a²; diagonal = a√3.
- Sum of all 12 edges: cuboid 4(l+b+h), cube 12a.
- e.g. Cube of edge 3: volume = 27, TSA = 6·9 = 54, space diagonal = 3√3; a 2×3×6 cuboid has diagonal √(4+9+36) = √49 = 7.
17Cylinder & cone
- A cone holds exactly one-third of the cylinder with the same base and height; its slant height is the hypotenuse of the radius-and-height right triangle.
- Cylinder: Volume = πr²h; CSA = 2πrh; TSA = 2πr(r+h).
- Cone slant l = √(r²+h²); Volume = ⅓πr²h; CSA = πrl; TSA = πr(r+l).
- Frustum volume = ⅓πh(R² + r² + Rr).
- e.g. Cone with r = 3, h = 4: slant = √(9+16) = 5, CSA = π·3·5 = 15π, volume = ⅓·π·9·4 = 12π.
18Sphere & hemisphere & prism
- The key exam idea is "recasting": when one solid is melted into another, volume stays the same even though surface area changes.
- Sphere: Volume = (4/3)πr³; Surface area = 4πr².
- Hemisphere: Volume = (2/3)πr³; CSA = 2πr²; TSA = 3πr².
- Prism: Volume = base area × height; LSA = base perimeter × height.
- Recast objects keep volume constant.
- e.g. Sphere of radius 3: volume = (4/3)·π·27 = 36π, surface area = 4·π·9 = 36π.
19Coordinate geometry, distance & section
- Distance is just Pythagoras on the coordinate differences; the midpoint is the special case of the section formula with ratio 1:1.
- Distance = √[(x₂−x₁)² + (y₂−y₁)²].
- Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2).
- Section (ratio m:n internal) = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)).
- Centroid = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3).
- e.g. Distance from (1, 2) to (4, 6) = √(3² + 4²) = 5; their midpoint = (2.5, 4).
20Coordinate geometry, area & slope
- Slope = rise over run. Equal slopes mean parallel; slopes multiplying to −1 mean perpendicular. The area formula needs only the three vertices.
- Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|.
- Slope of line through two points m = (y₂−y₁)/(x₂−x₁).
- Parallel lines: m₁ = m₂. Perpendicular: m₁·m₂ = −1.
- e.g. Triangle (0,0), (4,0), (0,3): area = ½|0(0−3)+4(3−0)+0| = ½·12 = 6.
21Coordinate geometry, lines & circle
- For the general circle, halve the x- and y-coefficients (with a sign flip) to read off the centre, then back out the radius.
- Slope-intercept: y = mx + c. Point-slope: y − y₁ = m(x − x₁).
- ⊥ distance of (x₁,y₁) from ax+by+c=0 = |ax₁+by₁+c|/√(a²+b²).
- Distance between parallel lines = |c₂−c₁|/√(a²+b²).
- Circle: (x−h)² + (y−k)² = r²; general x²+y²+2gx+2fy+c=0, centre (−g,−f), r = √(g²+f²−c).
- e.g. Distance of (0,0) from 3x + 4y − 10 = 0 = |−10|/√(9+16) = 10/5 = 2.
Trigonometry, CAT PYQs
Trigonometry
Trigonometry. Heights-and-distances and angle-of-elevation problems solved with sine/cosine and right-triangle ratios.
ModerateCAT 2003
A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45° to 60°. After how much more time will this car reach the base of the tower?
- (1) 5(√3 + 1)
- (2) 6(√3 + √2)
- (3) 7(√3 − 1)
- (4) 8(√3 − 2)
Show solution
(1) 5(√3 + 1). Let x be the distance of the second position of the car (at 60°) from the tower, and h the tower height. In △ABD, ∠ADB = 60°: h = x·tan 60° = x√3 = AB. Also, from the first position (at 45°), AB = BC·sin 45°, so BC = x√3. Then DC = BC − BD = x√3 − x = x(√3 − 1). The car covers DC in 10 minutes, so speed S = x(√3 − 1)/10. Remaining time = x/S = 10/(√3 − 1) = 5(√3 + 1) minutes.