﻿ Class X: Maths Mensuration,Mensuration,solids
Mensuration

 Mind Maps

Class X - Maths: Mensuration
Q) Definition of cone?

Q) Definition of sphere?

Q) 1000 cm3 = _________ L?

Q) The volume of a room is 80 m3. The height of the room is ?

Q) Volume of a cuboid = ?

Q) Find the area of a regular hexagon each of whose sides measures 6 cm?

Q) The surface of solid is called?

Q) Surface area of a cube = ?

Q) 1 cm3 = ______ mL ?

Q) Find the area of a regular octagon each of whose sides measures 5 cm?

Q) If the height of a cuboid becomes zero, it will take the shape of a ?

Q) Definition of a cube?

Q) Definition of cuboid?

Q) Definition of cylinder?

Q) The point A, B, C, D have respective co-ordinates (-4, -3), (10, -15) , (20, 9) and (0, 2). Find the area of quadrilateral ABC?

Q) The polar co-ordinates of the vertices of a triangle are (-a, π/8), (a, 3π/7) and (-2a, - 3π/5) find the area of the triangle?

Q) If the co-ordinates of the vertices of a △ ABC be (3, 5), (5, 8) and (8, 9) and if D and E divide AB and AC, respectively internally in the ratio 1 : 2, then show that the area of △ ABC = 9 . the area of △ADE?

Q) Find the altitude of the rhombus whose area is 345 cm2 and its perimeter is 210 cm?

Q) Find the area of the rhombus having each side equal to 20 cm and one of its diagonals equal to 18 cm?

Q) Find the area of a rhombus whose diagonals are of lengths 20 cm and 9.5 cm?

Q) The base radius and height of a right circular cylinder are 14 cm and 5 cm respectively.Its curved surface is ?

Q) The length of the parallel sides of a trapezium are in the rat: 5 : 2 and the distance between them is 20 cm. If the area of trapezium is 345 cm2, find the length of the parallel sides?

Q) Find the area of a trapezium whose parallel sides are AB = 12 cm, CD = 36 cm and the non-parallel sides are BC = 15 cm and AG = 15 cm? Q) Construct a quadrilateral ABCD in which AB = 3.6 cm, △ABC = 80°, BC = 4 cm, △BAD = 120° and AD = 5 cm. Q) The co-ordinates of the points A, B, C, D are (10, -7), (-7, 2), (15, 13) and (24, -15) respectively. Find the ratio in which AC divides BD?

Q) The height of two right circular cylinders are the same. Their volumes are respectively 16 and p; m3 and 81 and p; m3. The tation of their base radii is ?

Q) The height of two right circular cylinders are the same. Their volumes are respectively 16 ? m3, and 81? m3. The tation of their base radii is?

Q) The area of a trapezium is 40 cm2 . Its parallel sides are 12 cm and 8 cm. The distance between the parallel sides is ?

Q) Find the area of a regular pentagon each of whose sides measures 5 cm and the radius of the inscribed circle is 3.5 cm?

Q) The floor of a room is a square of side 6 m. Its height 4 m. The volume of the room?

Mensuration
Introduction

Mensuration is the skill of measuring the length of lines, areas of surfaces, and volumes of solids from simple data of lines and angles.

Mensuration in its literal meaning 'is to measure'.

It is generally used where geometrical figures are concerned, where one has to determine various physical quantities such as area, volume or length.

Measuring these quantities is called Mensuration. It can also be used where quantities like speed, acceleration and velocity are concerned.

We see so many things of different shapes (combination of two or more) around us. Houses stand on pillars, storage water tanks are cylindrical and are placed on cuboidal foundations, a cricket bat has a cylindrical handle and a flat main body, etc.

Think of different things around you.
Some of these are shown below: Of these objects like football have shapes where we know that the surface area and volume. We can however see that other objects can be seen as combinations of the solid shapes. So, their surface area and volume we now have to find. The table of the solid shapes, their areas and volumes are given below: Example 1:

A Sphere, a cylinder and a cone are of the same radius and same height. Find the ratio of their curved surface areas.

Solution: let r be the common radius of a sphere, a cone and cylinder.
Height of sphere = its diameter = 2r
Then, the height of cone = height of cylinder = height of sphere.
Then, the height of cone = 2r
Let l be the slant height of cone = sqrt(r^2 + h^2)
Let l be the slant height of cone = sqrt(r^2 + 2r^2)
:. s_1 = Curved surface area of sphere = 4πr^2
:. s_2 = Curved surface area of cylinder, 2πrh = 2πr × 2r = 4πr^2
:. s_3 = Curved surface area of cone =πrl = πr × sqrt(5r) = sqrt(5)πr^2
Ratio of curved surface area as
:. s_1 : s_2 : s_3 = 4πr^2 : 4πr^2 : sqrt(5)πr^2
:. s_1 : s_2 : s_3 = 4 : 4 : sqrt(5)

Example 2:

How many metres of cloth 1.1m wide will be required to make a conical tent whose vertical height is 12cm and base radius is 16m? Find also the cost of the cloth used at the rate of Rs. 14 per metre.

Solution:

Radius, r of base of conical tent = 16m.
Height, h of tent = 12m Slant height l = sqrt(h^2 + l^2) = sqrt(12^2 + 16^2) = sqrt(144 + 256) = 20m
Curved surface area of the tent = πrl = (22)/7 × 16 × 20 = (7040)/7m^2
Area of cloth = (7040)/7 m^2
Width of the cloth = 1.1m
Length of the cloth = (Area)/(width)
Length of the cloth = (7040)/7 -: 1.1
Length of the cloth = (7040)/7 × (10)/(11) = (6400)/7m = 914.29m.
Cost of (6400)/7m cloth = Rs. 14 × (6400)/7 = Rs. 12800.

Surface area of the combination of solids

In this unit we'll study three types of space figures that are not polyhedrons. These figures have curved surfaces, not flat faces.

Cylinder:

A cylinder is similar to a prism, but its two bases are circles, not polygons. Also, the sides of a cylinder are curved, not flat.

Cone:

A cone has one circular base and a vertex that is not on the base.

Sphere:

The sphere is a space figure having all its points an equal distance from the center point. Examples:

A right triangle, whose base and height are 15 cm. and 20 cm. respectively is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed (Use π=3.14).

Solution:

Let ABC be the right angled triangle such that AB = 15cm and AC = 20 cm
Using Pythagoras theorem in ΔABC we have: BC^2 = AB^2 + AC^2
BC^2 = 15^2 + 20^2
BC^2 = 225 + 400
BC = sqrt(625)
Let, OA = x and OB = y
In triangles ABO and ABC, we have /_BOA = /_BAC and /_ABO = /_ABC
so, by angle - angle criterian of similarity we have ΔBOA ~ ΔBAC
Therefore, (BO)/(BA) = (OA)/(AC) = (BA)/(BC)
rArr (y)/(15) = (x)/(20) = (15)/(25)
rArr (y)/(15) = (x)/(20) = (3)/(5)
rArr (y)/(15) = (3)/(5) and (x)/(20)= (3)/(5)
rArr y= (3)/(5) × 15 and x= (3)/(5) × 20
rArr y=9   x = 12
Thus, we have OA = 12 cm and OB = 9 cm
When the ABC is revolved about the hypotenuse .We get a double cone as shown in figure. Volume of the double cone = Volume of the cone CAA' + Volume of the cone BAA'
= 1/3π(OA)^2 × OC + 1/3π(OA)^2 × OB
= 1/3π ×12^2 × 16 + 1/3π×12^2 × 9
= 1/3π ×144 (16+9)
= 1/3× 3.14 ×144 ×25cm^3
= 3768cm^3
Surface area of the double cone = Curved surface area of cone CAA' + Curved surface area of cone BAA'
= (π ×OA ×AC) + (π ×OA ×AB)
= (π ×12 ×20) + (π ×12 ×15)cm^2
= 420 π cm^2
= 420 × 3.14 cm^2
= 1318.8 cm^2

2.A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. Solution: Height of Cylinder = 14 – 5 = 9 cm, radius = 2.5 cm Curved Surface Area of Cylinder
2πrh
= 2π × 2.5 × 9
= 45π cm^2
Curved Surface Area of two Hemispheres
= 4π r^2
= 4π × 2.5^2
= 25π cm^2
Total Surface Area
= 45π + 25π
= 70π = 220cm^2
Volume of combination of solids

Unlike the surface areas of combination of solids, while calculating the volume of combined solids we need to find the volume of each solid and add them toghether.

Examples:
1. A solid consisting of a right circular cone standing on a hemisphere, is placed upright in a right circular cylinder full of water and touches the bottom. Find the volume of water left in the cylinder, given that the radius of the cylinder is 3 cm. and its height is 6cm. The radius of the hemisphere is 2 cm. and the height of the cone is 4 cm.
Solution:

In the figure drawn here, ABCD is a cylinder and LMN is a Hemisphere OLM is a cone. We know that where a solid consisting of a cone and hemisphere is immersed in the cylinder full of water, then some water equal to the volume of the solid, is displaced. Volume of Cylinder = π r^2h = π × 3^2 × 6 = 54π cm^3
Volume of Hemisphere = 2/3 π r^3 = 2/3 × π × 2^3 = 16/3 π cm^3
Volume of Cone = 1/3 π r^2h = 1/3 π × 2^2 × 4 = 16/3 π cm^3
Volume of Cone and Hemisphere = 16/3 π +16/3 π
= 32/3π
Volume of water left in Cylinder = Volume of Cylinder - Volume of Cone and Hemisphere
= Volume of Cylinder - 32/3π
= 54π - 32/3π
= (162 π - 32 π)/3 = 130 π/3
= 130/3 × 22/7 = 2860/21 = 136.19 cm^3

2. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution:

Dimensions of cuboid = 15 cm × 10 cm × 3.5 cm, radius of cone = 0.5 cm, depth of cone = 1.4 cm
Volume of cuboid = length × width × height
= 15 × 10 × 3.5 = 525 cm^3
Volume of cone
= 1/3 π r^2h
= 1/3 × 22/7 × 0.5^2 × 1.4
= 11/30 cm^3
Volume of wood = Volume of cuboid – 6 x volume of cone
= 525 - 6 × 11/30
= 525 - 11/5
= 522.8 cm^3

Conversion of solid from one shape to another

Cut a watermelon into slices, they are converting a solid shape into other solid shapes. Regardless of the size and shape of the slices, there is one fact that holds true of the whole process. The volume of all the slices together exactly equals the volume of the original watermelon.

Convert a solid of a given shape to a solid of another shape, the surface area usually changes. However, the volume is preserved.      Examples

1.The diameter of a metallic sphere is 6cm. It is melted and drawn into a wire having diameter of the cross section as 0.2 cm. Find the length of the wire.

Solution:
We have, diameter of metallic sphere = 6cm
:. Radius of metallic sphere = 3cm
Also, we have, Diameter of cross - Section of Cylinder wire = 0.2 cm
Radius of cross section of cylinder wire = 0.1 cm
Let the length of wire be l cm
Since the metallic sphere is converted in to a cylinder sphere wire of length h cm
:. Volume of the metal used in wire = Volume of the sphere
π × (0.1)^2 × h = 4/3 × π × 3^3
π × (1/10)^2 × h = 4/3 × π × 27
π × 1/100 × h = 36 π
h = (36 π × 100)/π cm
=3600 cm = 36m
Therefore, the length of the wire is 36m

2.Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution:
Radii of spheres = 6 cm, 8 cm, 10 cm
Volume of sphere = 4/3 π r^3
Total volume of three spheres
= 4/3 π (6^3 + 8^3 + 10^3)
= 4/3 π (616 + 512 + 1000)
= 4/3 π × 1728
= ∛ (1728) = 12cm

3.How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution
Radius of coin = 0.875 cm, height = 0.2 cm
Dimensions of cuboid = 5.5 cm × 10 cm × 3.5 cm
Volume of coin:
= π × r^2h
= π × 0.875^2 × 0.2
= 0.48125 cm^3
Volume of cuboid = 5.5 × 10 × 3.5 = 192.5 cm^3
Number of coins = 192.5/0.48125 = 400
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