Understanding Three-Dimensional Geometry:
Most tests often include questions based on the knowledge of the geometries of 3-D objects such as cylinder, cone, cuboid, cube & sphere. The purpose of the article is to help you learn basics of 3-D geometry and encapsulate some of the important formulae and tricks.
The questions on Volume and Surface Area appear in all the competitive exams. Most of the students tend to avoid this topic considering it to be quite complex and calculative. This article would help you not only in memorizing the formulas, but also in understanding direct or indirect applications of these formulas. We strongly advice you go through each and every definition and formula given below to solve questions on Surface Area and Volume.
Mensuration Formulas
Surface area and Volume Formulas
Solids: Solids are three–dimensional objects, bound by one or more surfaces. Plane surfaces of a solid are called its faces. The lines of intersection of adjacent faces are called edges.
For any regular solid, Number of faces + Number of vertices = Number of edges + 2. This formula is called Euler’s formula.
Volume: Volume of a solid figure is the amount of space enclosed by its bounding surfaces. Volume is measured in cubic units.
Cone: A cone has one circle on one of its ending & rest is the curved circle part with a corner on the other end.
Volume = 1/3 πr2h. Surface Area (curved) = π r l, where l = slant height.
As per the Pythagoras theorem, l2 = r2 + h2. formula for surface area (total) = π r l + πr2.
1. If the whole surface area of a cone is 217 cm
2 and radius of the base is 3.5 cm, find the slant height of the cone.
1. 16(5/22)cm
2. 18(2/3)cm
3. 11(1/3)cm
4. 17(1/7)cm
5. 12(2/5)cm
Explanation: The surface area of cone = π x r x r + π x r x l where l is the slant height and r is the radius of the base.
Therefore we have, π x r x (r + l) = 217, which is same as π x 3.5 x (3.5 + l) = 217.
Therefore, the slant height = 16 5 / 22 cm.
2. 2. A pile of gravel falls into the shape of a cone. If it is 3 (1/2) metres high and 8 m in diameter, how many cubic metres of gravel does it contain?
1. 15(1/18)cu m
2. 16(2/21)cu m
3. 42(1/2)cu m
4. 58(2/3)cu m
5. 16(1/6)cu m
Explanation: Volume of gravel = (1/3) x π x r x r x h
= (1/3) x π x 4 x 4 x 3.5
= 58 2/3 m3.
Frustum of a cone: A frustum is lower part of a cone, containing the base, when it is cut by a plane parallel to the base of the cone. Slant height, L = √h2 + (R-r) 2 Curved Surface area of cone = π(R + r) L. Total surface area of a frustum = Base area + Area of upper circle + Area of lateral surface = π (R2 + r2 + RL + rL). Volume of frustum = πh/3 (R2 + r2 + Rr)
1. The frustum of a right circular cone has base diameter 10 cm. The diameter of the top is 6 cm and the height is 5 cm. Calculate its volume.
1. 327.38 cm3
2. 256.67 cm3
3. 392.5 cm3
4. 102.6 cm3
Explanation: Volume of frustum = πh/3 (R2 + r2 + Rr) = 1/3 × 22/7 × 5(25 + 9 + 15)= 256.67 cm3
Must Read 3D Geometry Articles
A pyramid is a solid, whose lateral faces are triangular with a common vertex and whose base is a polygon. A pyramid is said to be tetrahedron (triangular base), square pyramid, hexagonal pyramid etc, according to the number of sides of the polygon that form the base.
In a pyramid with a base of n sides, number of vertices = n + 1.
Number of faces including the base = n + 1.
Surface area of lateral faces = 1/2 x perimeter of base x slant height
Total surface area of pyramid = Base area + 1/2 x perimeter of base x slant height
Volume of pyramid = 1/3 × Base area x height. A cone is also a pyramid.
1. A right pyramid, 10 m high, has a square base with diagonal 10 m. Find its lateral surface.
Lateral surface of the pyramid = 1/2 x Base perimeter x slant height. Diagonal = 10 m ∴ side = 10/√2.
∴Base perimeter
(4x10)/√2 = 40√2. Also slant ht = √h2 + (side/2)2 Slant Height = √100 + (25/2) = √225/2 = 15/√2
∴ Lateral Surface = (1/2) x (40/√2) x (15/√2) 150m2.
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