mensu

Contents

1. INTRODUCTION
2. RIGHT PRISM
   1. CUBOID
   2. CUBE
   3. CYLINDER
3. RIGHT PYRAMID
   1. CONE
   2. FRUSTUM OF A CONE
   3. CONSTRUCTING A CONE FROM THE SECTOR OF A CIRCLE
4. SPHERE
5. HEMISPHERE

Mensuration

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INTRODUCTION

Mensuration is a branch of geometry where we study measurements of the perimeter, area and volume of geometrical figures and solid objects.

We have covered measurements of perimeter and area of two-dimensional figures while learning the concepts of triangles, quadrilaterals and circles. In this lesson, we will focus on the measurements of solids or three-dimensional objects.

A solid is a portion of space bounded by surfaces, known as faces. The intersection of the surfaces gives lines known as edges. The intersection of the edges gives points known as vertices.

REMEMBER:

• In a solid with F faces, V vertices and E edges, according to Euler’s rule:

F + V = E + 2 (if V ≠ 0 and E ≠ 0)

For example,

For a cube, F = 6, V = 8. Using Euler’s rule, we can find the number of edges as follows.

6 + 8 = E + 2

E = 12 which is true.

We will discuss the relations between the surface area, volume and diagonal lengths. In general, we will discuss two types of surface areas as follows:

Lateral Surface Area (LSA): The sum of the areas of all the vertical/lateral surfaces (excluding the top and bottom) is known as the lateral surface area of the solid

Total Surface Area (TSA): The sum of the areas of all surfaces (including the top and bottom) is known as the total surface area of the solid

TSA = LSA + Area of top and bottom surfaces

In the case of a rectangular room, the area of the four walls is the LSA. If we add the areas of the roof and the floor to the LSA, we get the TSA.

RIGHT PRISM

A solid in which the top and bottom surfaces are identical polygons and the lateral (vertical) surfaces are rectangular in shape and perpendicular to the top and bottom surfaces is known as a right prism. In a right prism, the corresponding sides of the top and bottom polygons are parallel.

A right prism with a triangular base is shown in the diagram.

In general, the relations applicable to all kinds of right prism are given by:

LSA = Perimeter of base height

TSA = LSA + 2 Area of base

Volume (V) = Area of base height

CUBOID

A right prism in which all the six surfaces are rectangular in shape is known as a cuboid or a rectangular parallelepiped.

For a cuboid with base length l, base breadth b and height h:

LSA = 2(lh + bh)

TSA = 2(lh + bh + lb)

Volume (V) = lbh

Example 1:

Find the lateral surface area, total surface area, volume and body diagonals of a cuboid with dimensions of length, breadth and height of 5 m, 4 m and 3 m respectively. Solution: LSA = 2(lh + bh) = 2(5 3 + 4 3) = 54 m2 TSA = LSA + 2lb = 54 + 2 5 4 = 94 m2 Volume (V) = lbh = 5 4 3 = 60 m3

Example 2:

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? [CAT 2003 Re-Test] (1) 3 (2) 4 (3) 1 (4) 2 Solution: When the tin sheet is cut across its corners as shown in the figure, the box formed will have a height of x inches and its base will be a square of side (12 – 2x) inches. Let the volume of the box, V = (12 – 2x)2 x = 4x3 48x2 + 144x i.e. 12x2 96x + 144 = 0 12(x 6)(x 2) = 0 x = 2 or x = 6 However, x cannot be 6 as the length of the side is (12 – 2x). x = 2 Hence, option 4. Alternatively, Since V = (12 – 2x)2 x = [2(6 – x)]2 x = 4x(6 – x)2, Substituting values of x from 1 to 5, we get V maximum when x = 2 (i.e. V = 128) Hence, option 4.

Example 3:

The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will [CAT 2006] (1) remain the same (2) decrease by 13.64% (3) decrease by 15% (4) decrease by 18.75% (5) decrease by 30% Solution: Let the original length, breadth and height of the room be 3x, 2x and x respectively.  The new length, breadth and height are 6x, x and x/2 respectively.  Area of four walls = (2 length height) + (2 breadth height)  Original area of four walls = 6x2 + 4x2 = 10x2  New area of four walls = 6x2 + x2 = 7x2  Area of wall decreases by [(10x2  7x2)/10x2] 100 = 30%  Hence, option 5.

CUBE

A right prism in which all the six surfaces are square in shape is known as a cube. It is a special case of the cuboid.

For a cube with side a:

LSA = 4a²

TSA = 6a²

Volume (V) = a³

Example 4:

Three metal cubes of body diagonals are melted and recast into a bigger cube of body diagonal without any loss of metal. Find x. Solution: Let the sides of the three smaller cubes be a, b and c. Then the body diagonals of these cubes will be which are given as Hence a = 6, b = 8 and c = 10. Since the cubes are recast without any loss of metal, the volume of the bigger cube would be same as the total of the volumes of the individual cubes. Now, since the body diagonal of the bigger cube is , its side will be x and its volume = x3 x3 = a3 + b3 + c3 = 63 + 83 + 103 = 216 + 512 + 1000 = 1728 x = 12

Example 5:

If the lengths of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be _______. [CAT 2004] (1) equal to the side of the cube (2) times the side of the cube (3) (4) impossible to find from the given information Solution: Let the side of the cube be a units. DF, AG and CE are body diagonals each of length a units. The triangle is an equilateral triangle of side a units. Circumradius of this equilateral triangle Circumradius = side of cube. Hence, option 1.

CYLINDER

A right circular cylinder is a special case of the right prism. The top and bottom surfaces are identical polygons with infinite number of sides, or are circles. Its lateral surfaces are curved; hence the concept of LSA is replaced by the concept of CSA (curved surface area).

For a cylinder with height h and radius of base circle r:

CSA = 2rh

TSA = 2rh + 2r²

Volume (V) = r²h

Example 6:

A rectangular sheet 4 cm in length and 6 cm in width can be folded to make a cylinder in two different ways. Find the curved surface area, total surface area and volume of both the cylinders. Solution: Refer to the diagrams above. In case (1), height h1 = 6 cm and circumference of the base = 2(R1) = 4, For case (2), height h2 = 4 cm and circumference of the base = 2(R2) = 6, CSA of both the cylinders = area of the given rectangular sheet = 4 6 = 24 cm2 TSA for case (1) CSA + 2 (R1)2 TSA for case (2) CSA + 2 (R2)2 Volume for case (1) (R1)2 h1 Volume for case (2) (R2)2 h2

Example 7:

From a cylinder of height 5 cm, a cylindrical ring of thickness 2 cm is formed by removing the inner concentric cylindrical portion. If the radius of the removed portion is 3 cm, find the total surface area and volume of the ring. Solution: A cylindrical ring of thickness 2 cm is made by removing a cylinder of radius 3 cm. i.e. the radius of the original cylinder is (2 + 3) = 5 cm. Also height of the cylinder = 5 cm Volume of the ring = Voriginal cylinder – Vremoved cylinder = (52)5 – (32)5 = 80 cm3 Surface area of the ring = Lateral surface area of the original cylinder + Lateral surface area of the removed portion + 2(Top surface area of the original cylinder – Top surface area of the removed portion of the cylinder) = (2 5 5) + (2 3 5) + 2( 52 – 32) = 112 cm2

Example 8:

The water from a roof, 9 sq. metres in area, flows down to a cylindrical container of 900 cm2 base. To what height will the water rise in cylinder if there is a rainfall of 0.1 mm? [SNAP 2008] (1) 0.1 cm(2) 0.1 metre (3) 0.11 cm(4) 1 cm Solution: For convenience, we convert the area of the roof to sq. cm. and the rainfall to cm. Hence, option 4.

RIGHT PYRAMID

A right pyramid is a solid in which:

- The bottom surface (i.e. base) is a polygon, each vertex of which is joined with the help of slanted edges to a single vertex at the top

- The lateral (slanted-vertical) surfaces are triangular in shape

The line joining the top vertex to the centre of the polygon which forms the bottom surface (also known as the height) is perpendicular to the bottom surface

In the above figure, a right pyramid with a square base is shown.

In general, the relations applicable to all the right pyramids are given by:

TSA = LSA + Area of base

Example 9:

A regular pyramid has a square base with side 10 cm and a vertical height of 20 cm. If the height increases by 10% of its original value and the volume is constant, the percentage change in the side of the square base with respect to its original value is approximately: [JMET 2010] (1) +5% (2) +10% (3) –5% (4) –10% Solution: Volume of square pyramid Original base = 10 cm Original height = 20 cm New height = 22 cm New base = b As volume remains constant, b 9.5 Percentage change in side of base Hence, option 3.

CONE

A Right Circular Cone is a special case of the right pyramid, in which the bottom surface is a polygon with an infinite number of sides; a circle. Its lateral surfaces are curved; hence the LSA generalizes to the concept of a CSA (curved surface area).

The figure above shows a right circular cone, with some important parameters marked.

For a cone with height h, slant height l and radius of the base circle r:

CSA = rl

TSA = rl + r²

Example 10:

Find the area of canvas required to make a conical tent of height 3 m and base radius 4 m. Solution: Slant height l = = 5 m Area of canvas required = CSA of the above cone = rl = 4 5 = 20 m2

Example 11:

What is the area of the copper sheet required to prepare a cone of base radius 30 cm with the height 40 cm? [FMS 2009] (1) 7543 cm2(2) 5146 cm2 (3) 5432 cm2(4) 7246 cm2 Solution: The area of the copper sheet required to prepare a cone of base radius 30 cm and height 40 cm will be equal to the total surface area of the cone. Total Surface area of cone = r(r + l) Where l is the slant height of the cone and r is the radius of its base. Using r = 30 and h = 40, we get l = 50 cm Hence, option 1.

FRUSTUM OF A CONE

If the upper part of the cone (the part containing the top vertex) is cut off by a plane parallel to the base, the remaining part of the original cone is known as the frustum of the cone. The upper part so removed will also be a cone, similar to the original cone.

If the dimensions of the original cone are R, H and L and those of the removed cone are r, h and l, we can use the property of similarity (as studied earlier for triangles) to say that the ratio of the corresponding dimensions of the cones will be equal.

Applying this property of similarity to the cone in the figure above, we get

The ratio of the volumes of the two cones will be

Let L_f be the slant height of a frustum of a cone and H_f be the height of the frustum of the cone. {In the above case, L_f = (L – l) and H_f = (H – h)} Then,

REMEMBER:

• If the upper part of the Pyramid (the part containing the top vertex) is cut off by a plane parallel to the base, the remaining part of the original pyramid is known as the frustum of the pyramid.

If L_f - the slant height of a frustum of a pyramid

H_f - the height of the frustum of the pyramid

P - perimeter of the bottom surface of the frustum of the pyramid

p - perimeter of the top surface of the frustum of the pyramid

A_b – area of the bottom surface of the frustum of the pyramid

A_t – area of the top surface of the frustum of the pyramid

Then,

Example 12:

Find the height of a bucket of volume 168 square inches, if the radii of the top and bottom surfaces are 6 inches and 3 inches respectively. Solution: A bucket can be modelled as the frustum of a cone. We use the concept of similarity of the original cone and the cone removed to get a bucket-like shape. Since the ratio of the radii R/r = 6/3 = 2/1, the ratio of the heights H/h will also be equal to 2/1 and the ratio of the volumes V/v will equal (2/1)3 = 8. (Here, R, H and V correspond to the dimensions of the original cone, and r, h and v correspond to the dimensions of the smaller cone removed to get the bucket) V/v = 8 means that the volume of the bucket = (V – v) = (8v – v) = 7v Hence, 168 = Volume of the bucket = 7 (1/3) (3)2 h h = 8 inches The height of the bucket = (H – h) = (2h – h) = h = 8 inches

Example 13:

A child consumed an ice cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. If the height of the ice cream-cone was 8 cm, what was the height of the remaining ice cream-cone? [JMET 2010] (1) 2.5 cm (2) 3.0 cm (3) 3.5 cm (4) 4.0 cm Solution: Cone left for the mother after the child consumes = 12.5% = 1/8 Now, the change in the volume of the coin is proportional to the cube of change in height. The height of the cone that is left = original height (1/8) 1/3 = 8 1/2 = 4 Hence, option 4.

Example 14:

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. [CAT 2008] Solution: As shown in the figure, ABC is the cross section of a cone of height 10 cm and radius of base 4 cm. PQRS is the cross section of a cylinder which needs to be fitted inside the cone such that one of the flat faces of the cylinder (represented by RS) coincides with the base of the cone (represented by BC). ∆ABD and ∆PBR are similar triangles. Total surface area of the cylinder = S = Differentiating ‘S’ with respect to ‘r’, we get, to zero Maximum total surface area of the cylinder Hence, option 1.

CONSTRUCTING A CONE FROM THE SECTOR OF A CIRCLE

If we cut a sector of angle from a circular piece of paper with radius R and try to make a cone out of it, by joining the two radii:

- The centre of the circle will become the top vertex of the cone

- The radius of the original circle will become the slant height of the cone

- The length of the arc of the sector will become the circumference of the circle at the base of the cone

So, for the cone: slant height L = R, Circumference of the base circle 2r = 2R(/360).

Hence, r = R(/360)

Example 15:

Find the volume of the cone formed by joining the two radii of a semi circle which has radius 14 cm. Solution: Refer to the figure above. Slant height of the cone, l = 14 cm (= radius of the sector)

SPHERE

A sphere is a solid for which every point on the surface is at the same distance from its centre. The distance from the centre to any point on the surface is known as its radius.

For a sphere with radius r:

CSA = TSA = 4r²

Example 16:

Eight small solid spherical metal balls are melted, and this material is cast into a bigger spherical ball without any loss of metal. Find the percentage change in the total surface area. Solution: Let the radius and volume of the smaller ball be r and v and those of the bigger ball be R and V respectively. Since no metal is wasted, the total volume of the 8 smaller balls must equal the volume of the bigger ball. Thus, V = 8 v 4R3/3 = 8 4r3/3 R3 = 8 r3 i.e. R = 2r The TSA of the eight smaller balls = 8 4r2 and the TSA of the bigger ball is 4R2 = 4(2r)2 = 4 4r2. Hence, we can see that the TSA of the bigger ball is reduced to half that of the original 8 balls. The percentage change is equal to 50% (decrease).

Example 17:

Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be [CAT 2003 Leaked Test] (1) 85.5 (2) 92.5 (3) 90.5(4) 87.5 Solution: Ratio of areas of two spheres = s1 : s2 = 4 : 1 Ratio of their radii = r1 : r2 = 2 : 1 Ratio of their volumes = v1 : v2 = 8 : 1 Volume of A is 12.5% (1/8th) of the volume of B. But, volume of A is k% less than B. k = (100 12.5) = 87.5% Hence, option 4.

HEMISPHERE

If a sphere is cut into two halves symmetrically, either of the parts thus formed is called a hemisphere.

For a hemisphere with radius r:

CSA = 2r²

TSA = 3r²

Example 18:

A sphere is cut into two hemispheres; find the change in the total surface area. Solution: Total surface area of the original sphere = 4r2 Total surface area of the two hemispheres = 2 (3r2) = 6r2 The new TSA is 1.5 times that of the original, hence the percentage change is equal to 50% (increase).

Example 19:

A cylinder, a Hemi-sphere and a cone stand on the same base and have the same heights. The ratio of the areas of their curved surface is: [IIFT 2008] (1) 2 : 2 : 1 (4) None of the above Solution: Since in case of Hemi-sphere, radius is same as height, it implies that the height for all the three shapes is same, i.e. equal to r. Assuming the base radius to be r in all the three cases, we then have, In case of cylinder: The curved surface area = 2r2 In case of hemisphere: The curved surface area = 2r2 Hence, option 4.