Mensuration, 3D , formulas + CAT PYQs
Focused Geometry kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Mensuration, 3D is here.
Geometry, formula sheet
Show the full Geometry formula sheet (explanations + basic examples)
- 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°.
- 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).
- 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.
- 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.
- 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).
- 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.
- 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.
- 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.
- 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.
- "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.
- 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°.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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).
- 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.
- 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.
Mensuration, 3D, CAT PYQs
Mensuration, 3D
A square tin sheet of side 12 inches is converted into a box with open top in the following steps: The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box?
- (1) 3
- (2) 4
- (3) 1
- (4) 2
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Directions (Q. 2 to 4): Answer the questions on the basis of the information given below.
Consider a cylinder of height h cm and radius r = 2/π cms as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of n turns (in other words, the string's length is the minimum length required to wind n turns.)
2. What is the vertical spacing in cms between two consecutive turns?
- (1) h/n
- (2) h/√n
- (3) h/n²
- (4) Cannot be determined with given information
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3. The same string, when wound on the exterior four walls of a cube of side n cms, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale). The length of the string, in cms, is:
- (1) 2n
- (2) √17 n
- (3) n
- (4) √13 n
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4. In the setup of the previous two questions, how is h related to n?
- (1) h = √2 n
- (2) h = √17 n
- (3) h = n
- (4) h = √13 n
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Consider a right circular cone of base radius 4 cm and height 10 cm. A cylinder is to be placed inside the cone with one of the flat surface resting on the base of the cone. Find the largest possible total surface area (in sq. cm) of the cylinder.
- (1) 100π/3
- (2) 80π/3
- (3) 120π/7
- (4) 130π/9
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The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is
- (1) 1300
- (2) 1340
- (3) 1480
- (4) 1520
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A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to
- (1) 10
- (2) 50
- (3) 60
- (4) 20
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A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9π cubic centimeters. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is
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A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is
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If the rectangular faces of a brick have their diagonals in the ratio 3 : 2√3 : √15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is:
- (1) √3 : 2
- (2) 1 : √3
- (3) 2 : √5
- (4) √2 : √3
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The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is:
- (1) 12
- (2) 10√2
- (3) 8√3
- (4) 5√5
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A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is:
- (1) 1026(1 + π)
- (2) 8464π
- (3) 928π
- (4) 1044(4 + π)
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A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is:
- (1) 264
- (2) 256
- (3) 232
- (4) 243
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CAT 2024 & 2025, recent
The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is
- (A) 1125π
- (B) 750π
- (C) 1125π√2
- (D) 750π√2