Quantitative Aptitude

Mensuration Guide & Practice

Master area, perimeter, surface area, and volume formulas for 2D and 3D shapes with solved examples and free SSC/Railways mock tests. Explore dynamic solver blueprints, master fundamental equations, examine step-by-step solved examples, and practice with real exam-grade mock test sets.

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Mensuration - Set 5 Practice Test

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Mensuration - Set 1 Practice Test

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Mensuration - Set 5 Practice Test

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Mensuration - Set 4 Practice Test

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22 min
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Mensuration - Set 3 Practice Test

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Mensuration - Set 1 Practice Test

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1. Fundamentals & Definitions

Key Definitions

  • Mensuration: The branch of mathematics that deals with the geometric measurement of various parameters of 1D, 2D, and 3D figures, such as length, area, perimeter, and volume.
  • 2D Shapes: Flat, two-dimensional geometric figures bounded by a closed path on a single plane. They possess length and width but have no depth or height. Key parameters measured are perimeter and area.
  • 3D Shapes (Solid Shapes): Three-dimensional figures that occupy space. They possess length, width, and depth/height. Key parameters measured are volume, curved/lateral surface area, and total surface area.
  • Perimeter (P) / Circumference (C): The total length of the continuous boundary enclosing a 2D shape. Measured in linear units (e.g., mm, cmcm).
  • Area (A): The region or space enclosed by a closed 2D shape. Measured in square units (e.g., m2m^2, cm2cm^2).
  • Volume (V): The total 3D space occupied or enclosed by a solid shape. Measured in cubic units (e.g., m3m^3, cm3cm^3).
  • Curved Surface Area (CSA): The area of only the curved surfaces of a 3D solid (such as cylinders, cones, and spheres), excluding the flat ends.
  • Lateral Surface Area (LSA): The total area of all the lateral (vertical side) faces of a solid (such as prisms, pyramids, cubes, and cuboids), excluding the areas of the bases (top and bottom).
  • Total Surface Area (TSA): The sum of the areas of all surfaces (both flat and curved) of a 3D solid.
  • Inradius (rr): The radius of the circle inscribed inside a polygon (such as a triangle or regular polygon) that touches all its sides.
  • Circumradius (RR): The radius of the circle circumscribed around a polygon that passes through all its vertices.
  • Apothem (rar_a): The line segment from the center of a regular polygon to the midpoint of one of its sides (corresponds to the inradius of the polygon).
  • Slant Height (ll): The distance from the apex of a cone or regular pyramid along the surface to the boundary of its base.

2. Core Concepts & Formulas

2.1 Key Geometric Rules and Properties

Triangles

  • Semi-perimeter (ss): Half of the perimeter of the triangle: s=a+b+c2s = \frac{a+b+c}{2}
  • Trigonometric Area: If two sides aa and bb and their included angle θ\theta are known: Area=12absin⁡θ\text{Area} = \frac{1}{2}ab\sin\theta
  • Medians Formula: If d,e,fd, e, f are the lengths of the medians of a triangle, and u=d+e+f2u = \frac{d+e+f}{2}: Area=43u(u−d)(u−e)(u−f)\text{Area} = \frac{4}{3}\sqrt{u(u-d)(u-e)(u-f)}
  • Inradius and Circumradius Relations: Area=r×s\text{Area} = r \times s Area=abc4R\text{Area} = \frac{abc}{4R}

Quadrilaterals

  • Brahmagupta's Formula: For a cyclic quadrilateral (vertices lie on a circle) with sides a,b,c,da, b, c, d and semi-perimeter s=a+b+c+d2s = \frac{a+b+c+d}{2}: Area=(s−a)(s−b)(s−c)(s−d)\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}
  • Bretschneider's Formula: For any general quadrilateral with sides a,b,c,da, b, c, d, semi-perimeter ss, and opposite angles α\alpha and β\beta: Area=(s−a)(s−b)(s−c)(s−d)−abcdcos⁡2(α+β2)\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d) - abcd\cos^2\left(\frac{\alpha+\beta}{2}\right)}
  • Ptolemy's Theorem: A quadrilateral is cyclic if and only if the product of its diagonals equals the sum of the products of its opposite sides: ac+bd=pq(where p,q are diagonals)ac + bd = pq \quad (\text{where } p, q \text{ are diagonals})
  • Trapezium Intersecting Diagonals Property: In a trapezium ABCDABCD where parallel sides are ABAB and CDCD, and diagonals intersect at OO: Area(ΔAOD)=Area(ΔBOC)\text{Area}(\Delta AOD) = \text{Area}(\Delta BOC) Area(ΔAOB)×Area(ΔCOD)=Area(ΔBOC)×Area(ΔAOD)\text{Area}(\Delta AOB) \times \text{Area}(\Delta COD) = \text{Area}(\Delta BOC) \times \text{Area}(\Delta AOD)

Regular Polygons

  • Interior Angles: Sum of interior angles=(n−2)×180∘\text{Sum of interior angles} = (n-2) \times 180^\circ Each interior angle (regular)=(n−2)×180∘n\text{Each interior angle (regular)} = \frac{(n-2) \times 180^\circ}{n}
  • Exterior Angles: Each exterior angle (regular)=360∘n\text{Each exterior angle (regular)} = \frac{360^\circ}{n}
  • Diagonals: Number of diagonals=n(n−3)2\text{Number of diagonals} = \frac{n(n-3)}{2}

Solids and 3D Principles

  • Scaling and Percentage Changes:
    • If all linear dimensions of a 2D shape are scaled by a factor kk, the perimeter changes by a factor of kk and the area changes by a factor of k2k^2.
    • If all linear dimensions of a 3D solid are scaled by a factor kk, the surface area changes by a factor of k2k^2 and the volume changes by a factor of k3k^3.
    • If only the radius rr of a cylinder/cone changes by a factor xx and height hh changes by a factor yy: Volume change ratio=x2y\text{Volume change ratio} = x^2y Curved surface area change ratio (for cylinder)=xy\text{Curved surface area change ratio (for cylinder)} = xy
  • Melting and Recasting Rule: When a solid is melted and recast into another shape (or multiple smaller shapes), the total volume remains constant (assuming no wastage). If there is a wastage percentage ww: Volume of newly formed solid(s)=(1−w100)×Volume of original solid(s)\text{Volume of newly formed solid(s)} = \left(1 - \frac{w}{100}\right) \times \text{Volume of original solid(s)}

2.2 Comprehensive Formula Reference Table

Dimension TypeShape NameKey ParametersPerimeter / LSA / CSAArea / TSAVolume / Diagonal / Special Properties
2DSquareSide aaP=4aP = 4aA=a2A = a^2 or A=12d2A = \frac{1}{2}d^2Diagonal d=a2d = a\sqrt{2}
2DRectangleLength ll, width wwP=2(l+w)P = 2(l+w)A=l×wA = l \times wDiagonal d=l2+w2d = \sqrt{l^2 + w^2}
2DGeneral TriangleSides a,b,ca, b, c; semi-perimeter s=a+b+c2s = \frac{a+b+c}{2}P=a+b+cP = a+b+cA=s(s−a)(s−b)(s−c)A = \sqrt{s(s-a)(s-b)(s-c)} (Heron's)
A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}		Inradius r=Asr = \frac{A}{s}
Circumradius R=abc4AR = \frac{abc}{4A}
2DEquilateral TriangleSide aaP=3aP = 3aA=34a2A = \frac{\sqrt{3}}{4}a^2Height h=32ah = \frac{\sqrt{3}}{2}a
Inradius r=a23r = \frac{a}{2\sqrt{3}}, Circumradius R=a3R = \frac{a}{\sqrt{3}}
2DIsosceles TriangleEqual sides aa, base bbP=2a+bP = 2a + bA=b44a2−b2A = \frac{b}{4}\sqrt{4a^2-b^2}Height h=124a2−b2h = \frac{1}{2}\sqrt{4a^2-b^2}
2DRight-Angled TriangleBase bb, height hh, hypotenuse ccP=b+h+cP = b + h + cA=12bhA = \frac{1}{2}bhHypotenuse c=b2+h2c = \sqrt{b^2 + h^2}
2DCircleRadius rrCircumference C=2πrC = 2\pi rA=πr2A = \pi r^2Diameter d=2rd = 2r
2DCircular SectorRadius rr, arc angle θ\thetaArc Length L=θ360×2πrL = \frac{\theta}{360} \times 2\pi r
P=L+2rP = L + 2r		A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2 or A=12LrA = \frac{1}{2}Lrθ\theta is in degrees
2DCircular SegmentRadius rr, angle θ\thetaBoundary length = L+2rsin⁡(θ2)L + 2r\sin\left(\frac{\theta}{2}\right)A=r2(πθ360−sin⁡θ2)A = r^2 \left( \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right)Area = Area of Sector −- Area of Triangle
2DParallelogramAdjacent sides a,ba, b, height hhP=2(a+b)P = 2(a+b)A=b×hA = b \times h or A=absin⁡θA = ab\sin\thetaOpposite angles are equal; diagonals bisect
2DRhombusSide aa, diagonals d1,d2d_1, d_2P=4aP = 4aA=12d1d2A = \frac{1}{2}d_1 d_2Diagonals bisect at 90∘90^\circ
Side relation: 4a2=d12+d224a^2 = d_1^2 + d_2^2
2DTrapeziumParallel sides a,ba, b, height hhP=a+b+c+dP = a+b+c+dA=12(a+b)hA = \frac{1}{2}(a+b)hhh is distance between parallel sides
2DRegular PolygonNumber of sides nn, side ssP=nsP = nsA=12×ra×PA = \frac{1}{2} \times r_a \times P (where rar_a is apothem)Interior angle = (n−2)180∘n\frac{(n-2)180^\circ}{n}
Exterior angle = 360∘n\frac{360^\circ}{n}
3DCubeEdge aaLSA=4a2\text{LSA} = 4a^2TSA=6a2\text{TSA} = 6a^2V=a3V = a^3
Diagonal d=a3d = a\sqrt{3}
3DCuboidLength ll, width ww, height hhLSA=2h(l+w)\text{LSA} = 2h(l+w)TSA=2(lw+wh+hl)\text{TSA} = 2(lw + wh + hl)V=lwhV = lwh
Diagonal d=l2+w2+h2d = \sqrt{l^2 + w^2 + h^2}
3DSolid CylinderRadius rr, height hhCSA=2πrh\text{CSA} = 2\pi rhTSA=2πr(r+h)\text{TSA} = 2\pi r(r+h)V=πr2hV = \pi r^2 h
3DHollow CylinderOuter radius RR, inner rr, height hhCSA=2πh(R+r)\text{CSA} = 2\pi h(R+r) (Total curved)TSA=2πh(R+r)+2π(R2−r2)\text{TSA} = 2\pi h(R+r) + 2\pi(R^2-r^2)Volume of material V=πh(R2−r2)V = \pi h(R^2 - r^2)
3DRight Circular ConeRadius rr, height hhCSA=πrl\text{CSA} = \pi rlTSA=πr(r+l)\text{TSA} = \pi r(r+l)V=13πr2hV = \frac{1}{3}\pi r^2 h
Slant height l=r2+h2l = \sqrt{r^2 + h^2}
3DFrustum of a ConeEnd radii R,rR, r, height hhCSA=πl(R+r)\text{CSA} = \pi l(R+r)TSA=πl(R+r)+πR2+πr2\text{TSA} = \pi l(R+r) + \pi R^2 + \pi r^2V=13πh(R2+Rr+r2)V = \frac{1}{3}\pi h(R^2 + Rr + r^2)
Slant height l=h2+(R−r)2l = \sqrt{h^2 + (R-r)^2}
3DSphereRadius rrCSA=4πr2\text{CSA} = 4\pi r^2TSA=4πr2\text{TSA} = 4\pi r^2V=43πr3V = \frac{4}{3}\pi r^3
3DHemisphereRadius rrCSA=2πr2\text{CSA} = 2\pi r^2TSA=3πr2\text{TSA} = 3\pi r^2V=23πr3V = \frac{2}{3}\pi r^3
3DQuarter SphereRadius rrCSA=πr2\text{CSA} = \pi r^2TSA=2πr2\text{TSA} = 2\pi r^2V=13πr3V = \frac{1}{3}\pi r^3
3DRight PrismBase Area AbA_b, Base Perimeter PbP_b, height hhLSA=Pb×h\text{LSA} = P_b \times hTSA=LSA+2Ab\text{TSA} = \text{LSA} + 2A_bV=Ab×hV = A_b \times h
3DRight PyramidBase Area AbA_b, Base Perimeter PbP_b, height hhLSA=12Pb×l\text{LSA} = \frac{1}{2} P_b \times l (where ll is slant height)TSA=LSA+Ab\text{TSA} = \text{LSA} + A_bV=13Ab×hV = \frac{1}{3} A_b \times h
3DRegular TetrahedronSide aaLSA=334a2\text{LSA} = \frac{3\sqrt{3}}{4}a^2TSA=a23\text{TSA} = a^2\sqrt{3}V=a362V = \frac{a^3}{6\sqrt{2}} or V=212a3V = \frac{\sqrt{2}}{12}a^3
Height h=a23h = a\sqrt{\frac{2}{3}}

Typical Exam Weightage

ExamTypical Questions
SSC (CGL / CHSL / MTS)2–4 questions
Railways (RRB)1–2 questions
Defense (NDA / CDS)1–2 questions

3D mensuration (cone, cylinder, sphere combinations) is a recurring SSC CGL Tier 2 favorite.

Figures are typical ranges based on recent-year patterns, not a guarantee for any specific upcoming paper — always cross-check against the latest official syllabus and previous-year papers for Mensuration.

Solved Examples

1Example 1 (Easy)

Topic: Area and Perimeter of Basic 2D Shapes (Right-Angled Triangle) Exams: SSC MTS, RRB Group D

Question: Find the perimeter and the area of a right-angled triangle whose base is 12 cm12\text{ cm} and hypotenuse is 13 cm13\text{ cm}.

2Example 2 (Moderate)

Topic: Volume of Composite 3D Solids (Cylinder, Cone, and Hemisphere) Exams: SSC CHSL, Railway NTPC

Question: A solid cylindrical container of radius 6 cm6\text{ cm} and height 15 cm15\text{ cm} is completely filled with ice cream. This ice cream is to be distributed among children in cones of height 12 cm12\text{ cm} and radius 3 cm3\text{ cm}, having a hemispherical shape on the top. Find the number of such cones that can be completely filled.

3Example 3 (Hard)

Topic: Frustum Volume and Melting/Recasting into Hollow Cylinder Exams: SSC CGL Tier-II, NDA

Question: A solid metallic right circular cone of height 20 cm20\text{ cm} with a vertical angle of 60∘60^\circ is cut into two parts at the middle of its height by a plane parallel to its base. The bottom frustum portion is melted and drawn into a long cylindrical wire of uniform diameter 116 cm\frac{1}{16}\text{ cm}. Find the length of the wire in meters.