Mensuration
Category : 9th Class
MENSURATION
Learning Objectives
Introduction
Mensuration is a science of measurement of the lengths of lines, area of surfaces and volumes of solids.
Some Important Definitions and Formulae:
Area of Plane Geometrical Figures
(i) Right Triangle
(ii) Scalene Triangle (Heron's formula)
(iii) Isosceles Triangle
(iv) Equilateral Triangle
Right Triangle: Area of right triangle \[=\frac{1}{2}\times \left( perpendicular \right)\times Base=\frac{1}{2}\times AB\,\times BC\]
Scalene Triangle (Heron's formula): Let, a, b, c be the length of sides of a triangle
then area \[=\sqrt{s(s-a)(s-b)(s-c)}\] sq. unit, where \[s=\frac{1}{2}(a+b+c)\]
Isosceles Triangle: Area ot isosceies trsang Se
\[=\frac{1}{2}\times BC\times AD=\frac{1b}{4}\sqrt{4{{a}^{2}}-{{b}^{2}}}\]
Equilateral triangle:
Area \[=\frac{\sqrt{3}}{4}\,{{(side)}^{2}}=\frac{\sqrt{3}}{4}{{a}^{2}}\]
Circle
A circle is a geometrical figure consisting of all those points in a plane which are at a given distance from a fixed point in the same plane. The fixed point is called the centre and the constant distance is known as the radius.
A circle with centre O and radius r is generally denoted by C (0, r).
Circle Formulas
Quadrilateral
We know that a geometrical figure bounded by four lines segment is called quadrilateral In this section we will study about area and perimetre of different quadrilaterals.
Rectangle
(i) Area \[=(l\,\times b)\]square units |
(ii) Length \[=\frac{area}{breadth},\] breadth \[=\frac{area}{length},\] |
(iii) \[\operatorname{Diagonal}\,=\sqrt{{{l}^{2}}+{{b}^{2}}}\,units\] |
(iv) Perimetre \[=2(l+b)\]units |
Square
Let ABCD be a square with each side equal to 'a' units then we have
(i) Area \[={{a}^{2}}\] sq units |
(ii) Area \[=\left\{ \frac{1}{2}\times {{(diagonal)}^{2}} \right\}\] units |
(iii) Diagonal \[=a\sqrt{2}\] units |
(iv) Perimetre = 4a units |
Solids
The objects having definite shape and size are called solids. A solid occupies a definite space.
Cuboid
Solids like matchbox, chalk box, a tile, a book an almirah, a room etc. are in the shape of a cuboid.
.
For cuboid of length = I, breath = b and height = h, we have:
(i) Volume \[=\left( I\,\times b\,\times h \right)\] |
(ii) Total surface area \[=2 \left( Ib\,\times bh\,\times Ih \right)\] |
(iii) Lateral surface area \[=\left[ 2(l+b)\times h \right]\] |
Cube
Solids like ice cubes, sugar cubes, dice etc. are the shape of cube Formula for a cube having each edge = a units, we have:
(i) Volume \[={{a}^{3}}\] |
(ii) Total surface area \[=6{{a}^{2}}\] |
(iii) Lateral surface area \[=4{{a}^{2}}\] |
Cylinder
Solids like measuring jar, circular pencils, circular pipes, road rollers, gas cylinders, are said to have a cylindrical shape. Formula for a cylinder of base radius = r & height (or length) = h, we have
(i) Volume \[=\pi {{r}^{2}}h\] |
(ii) Curved surface area\[=2\pi rh\] |
(iii) Total surface area \[=(2\pi rh\,+\,2\pi {{r}^{2}})=2\pi r(h+r)\] |
Cone
Consider a cone in which base radius = r, height = h & slant height \[\operatorname{l}=\sqrt{{{h}^{2}}+{{r}^{2}}}\] then we have
(i) Volume of the cone \[=\frac{1}{3}\pi {{r}^{2}}h\] |
(ii) Curved surface area of the cone = (curved surface area) + (area of the base) |
\[=\pi rl+\pi {{r}^{2}}=\pi r(l+r)\]
Sphere
Objects like a football, a cricket ball, etc. are said to have the shape of a sphere. For a sphere of radius r, we have
(i) Volume of the sphere \[=\left( \frac{4}{3}\pi {{r}^{3}} \right)\] |
(ii) Surface area of the sphere \[=(4\pi {{r}^{2}})\] |
Hemisphere
A plane through the centre of a sphere cuts it into two equal parts, each part is called hemisphere.
For a hemisphere of radius r, we have:
(i) Volume of the hemisphere \[=\frac{2}{3}\pi {{r}^{2}}\] |
(ii) Total surface area of the hemisphere \[=(2\pi {{r}^{2}})\] |
(iii) Total surface area of the hemisphere \[=(3\pi {{r}^{2}})\] |
Commonly Asked Questions
(a) 7.1 sq. m. (b) 75 sq. m.
(c) 7.05 sq. m. (d) 7.9 sq.m.
(e) None of these
Answer (b)
Explanation: Height \[=\sqrt{{{(6.5)}^{2}}-{{6}^{2}}}=\sqrt{(42.25-36)}=\sqrt{6.25}=2.5m\]
\[\therefore \,\,\,\,Area\,=\frac{1}{2}\times 6\times 2.5=7.2\,sq\,.m.\]
(a) 24,000 sq. m. (b) 23,000 sq. m.
(c) 21,000 sq. m. (d) 22,000 sq. m.
(e) None of these
Answer (a)
Explanation: Let length and breadth (in metres) respectively be 5X and 3x
\[2\left( 5x+3x \right)=640\]
\[16x=640\,\Rightarrow 40\]
\[\therefore \,\,\,Area\,=\,5X\times 3X=15{{x}^{2}}=15{{(40)}^{2}}=24,000\,sq.\,m.\,\]
(a) 110. 5sq. m (b) 111. 4sq. m
(c) 112. 5sq. m (d) 110. 3sq. m
(e) None of these
Answer: (c)
Explanation: Area of sq. \[=\,\frac{1}{2}\times {{(15)}^{2}}=112.5\,sq.m\]
(a) 95 sq. m (b) 96 sq. m
(c) 92 sq. m (d) 94 sq. m
(e) None of these
Answer (b)
Explanation: Let ABCD be rhombus.
\[\angle AOB\] is a right angle
\[\operatorname{OB}=\frac{1}{2}(BD)\,=6cm,\,\,AB=10\,cm.\]
\[\Rightarrow AO 2 \left( AO \right)=16\,cm.\]
\[\therefore \,\,\,Area of rhombus =\frac{1}{2}\left( AC\times BD \right)=\frac{1}{2}\,\,\left( 16\times 12 \right)=96 sq. m.\]
Find the radius of the circle.
(a) 49 m (b) 47 m
(c) 46 m (d) 48 m
(e) None of these
Answer (a)
Explanation: If r be the radius,
\[2\pi r-2r=210\]
\[r=\frac{210}{2(\pi -1)}=\frac{105}{\frac{22}{7}-1}=\frac{7\times 105}{15}=49m.\]
(a) 124 cub. m (b) 123 cub. m
(c) 121 cub. m (d) 126 cub. m
(e) None of these
Answer (d)
Explanation: If r be the radius of the base, \[2\pi r=6\]
\[r=\frac{3}{\pi }\]
\[\therefore \,\,\,Area\,\,of\,base\,=\pi {{r}^{2}}=p{{\left( \frac{3}{p} \right)}^{2}}=\frac{9}{\pi }\,sq.m.\]
\[\therefore \,\,\,Volume=\frac{9}{\pi }\,\times 44=\frac{9}{22}\times 7\times 44=126\,cub.m.\]
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