Surface Area and Volume

Category : 9th Class

Surface Area and Volume

 

In this chapter, we will learn about some important formulas related to 2-D and 3-D geometrical shapes.

 

Area of a Triangle

  •           Area of a triangle \[=\frac{1}{2}\times (Perpendicular)\times Base\]
  •          Area of a triangle having lengths of the sides a, b and c is \[=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]sq. units, where \[s=\frac{1}{2}(a+b+c)\]

  •           Area of an equilateral triangle\[=\frac{\sqrt{3}}{4}{{a}^{2}}\], where a is the side of the equilateral triangle.

Circle                        

  •           Circumference of the circle = \[2\pi r\]
  •           Area of the circle =\[\pi {{r}^{2}}\]
  •           Area of the semicircle = \[\frac{1}{2}\pi {{r}^{2}}\]
  •           Perimeter of the semicircle \[=\pi r+2r\]

 

Length of Arc and Area of a Sector

Let an arc AB an angle 0 < 180° at the center (O) of a circle a4 radius; Then we have:

  •            Length of the arc \[AB=\frac{2\pi \theta }{360{}^\circ }\]
  •            Area of the sector \[OACB=\frac{\pi {{r}^{2}}\theta }{360{}^\circ }\]

  •            Area of the minor segments ACBA = area of sector OACB \[-\] of the corresponding triangle AOB
  •            Area of the major segment ADBA = area of the circle-area of the minor segment

 

Perimeter and Area of a Rectangle

Let ABCD be a rectangle in which length \[AB\text{ }=1\] units, breadth \[BC\text{ }=\text{ }b\] units then we have:

  •           \[Area\text{ }=\text{ }\left( l\text{ }\times \text{ }b \right)\text{ }square\text{ }units\]
  •           \[Length\text{ }\left( l \right)=\frac{area\text{ (A)}}{breadth\text{ (B)}}units\]

  •           \[breadth\text{ }\left( b \right)\text{ }=\text{ }\frac{area\text{ (A)}}{length\text{ (l)}}units\]
  •           Diagonal (d) \[=\sqrt{{{l}^{2}}+{{b}^{2}}}\text{ }units\]
  •           Perimeter \[\left( p \right)\text{ }=\text{ }2\left( l\text{ }+\text{ }b \right)\] units

 

Area of Four Walls of a Room

Let I, b and h are respectively the length/ breadth and height of a room, then area of four walls of the room =\[\left\{ 2\,\left( l+b \right)\times h \right\}\text{ }sq\text{ }units.\]

 

Perimeter and Area of Square

Let ABCD be a square with each side equal to ?a? units, then

  •            Area = a2 sq. units
  •           \[Area={{\left( \frac{1}{2}\times {{(diagonal)}^{2}} \right)}^{{}}}sq.\text{ }units\]

  •             \[Diagonal\text{ =a}\sqrt{2}\text{ units}\]
  •            Perimeter = 4a units

 

Area of Some Special Types of Quadrilateral

  •           Area of a parallelogram = (base \[\times \] height)

  •           Area of a rhombus \[=\frac{1}{2}\times (product\text{ of  diagonal})\]

  •           Area of a Trapezium \[=\frac{1}{2}\](Sum of length of parallel sides) \[\times \](distance between then)\[=\frac{1}{2}(a+b)\times h\]

 

Solids

The objects having definite shape and size are called solids. A solid occupies a definite space.

 

Cuboid

For a cuboid of length = l, breadth = b and height = h, we have:

  •            Volume = \[\left( l\times b\times h \right)\]cubic units
  •           Total surface area = \[2\left( lb + bh + lh \right)\] sq. units
  •            Lateral surface area \[=~\left[ 2\left( l\text{ }+\text{ }b \right)\text{ }x\text{ }h \right]\text{ }sq.\text{ }units\]
  •           Diagonal of a cuboid = \[\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}\]

        Cube

For a cube having each edge = a units, we have:

  •            Volume = a3 cubic units
  •           Total surface area = 6a2 sq. units
  •           Lateral surface area = 4a2 sq. units
  •           Diagonal of a cube = \[a\text{ }\sqrt{3}\]

 

Cylinder

Solids like jar, circular pencils, circular pipes, road rollers, gas cylinders are of cylindrical shape. For a cylinder of base radius = r units and height = h units, we have:

  •           Volume = \[\pi {{r}^{2}}h\] cubic units
  •           Curved surface area = \[2\pi rh\] square units
  •          Total surface area = \[(2\pi rh+2\pi {{r}^{2}})=\text{ }2\pi r(h+r)\text{ }sq.\text{ }units\]

 

Cone

Consider a cone in which base radius = r, height = h and slant height\[\left( l \right)=\sqrt{{{h}^{2}}+{{r}^{2}}}\], then we have:

  •            Volume of the cone \[\frac{1}{3}\pi {{r}^{2}}h\]
  •           Curved surface area of the cone = \[\pi rl\]
  •        Total surface area of the cone\[~=\left( curved\text{ }surface\text{ }area \right)+\left( area\text{ }of\text{ }the\text{ }base \right)=\pi rl+\pi {{r}^{2}}=\pi r\left( l+r \right)\]

 

Sphere

Objects like a football, a cricket ball, etc. are of spherical shapes. For a sphere of radius r, we have:

  •           Volume of the sphere \[=\frac{4}{3}\pi {{r}^{3}}\]
  •           Surface area of the sphere =\[4\pi {{r}^{3}}\]

Hemisphere

A plane through the centre of a sphere cuts it into two equal parts, each part is called a hemisphere. For a hemisphere of radius r, we have:

  •           Volume of the hemisphere =\[\frac{2}{3}\pi {{r}^{3}}\]
  •           Curved surface area of the hemisphere = \[2\pi {{r}^{2}}\]
  •          Total surface area of the hemisphere = \[~3\pi {{r}^{2}}\]

 

  •             Example:

Find the area of the triangle whose base is 25 cm and height is 10.8 cm.

            (a) 125 \[c{{m}^{2}}\]                                                             (b) 135 \[c{{m}^{2}}\]

            (c) 124 \[c{{m}^{2}}\]                                                              (d) 199 \[c{{m}^{2}}\]

            (e) None of these

Ans.     (b)

Explanation: Area of the given triangle \[=\left( \frac{1}{2}\times 25\times \text{10}\text{.8} \right)c{{m}^{2}}=135\text{ }c{{m}^{2}}\]

 

  •             Example:

A chord of a circle of radius 14 cm makes a right angle at the centre. Find the area of the major sector of the circle.

            (a) 590\[c{{m}^{2}}\]                                                 (b) 462\[c{{m}^{2}}\]

            (c) 595\[c{{m}^{2}}\]                                                               (d) 995\[c{{m}^{2}}\]

            (e) None of these

Ans.     (b)

Explanation: Area of the major sector \[\frac{270}{360{}^\circ }\times \pi {{r}^{2}}\]

            = \[\frac{270{}^\circ }{360{}^\circ }\times \frac{22}{7}\times 14\times 14=462\text{ c}{{\text{m}}^{\text{2}}}\]

 

  •            Example:

The length of a rectangular plot of land is twice its breadth. If the perimeter of the plot is 210 m, then find its area.

            (a) 2450\[{{m}^{2}}\]                                                              (b) 2251\[{{m}^{2}}\]

            (c) 5560\[{{m}^{2}}\]                                                              (d) 9060\[{{m}^{2}}\]

            (e) None of these

Ans.     (a)

Explanation: Let x metre be the breadth of the triangle, then its length will be 2x metre.

Now, \[2(x+2x)=210\Rightarrow \text{ }x=35\]

\[Area=35\times 70=2450\text{ }{{m}^{2}}\]

Other Topics

Notes - Surface Area and Volume
  15 10



LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

You need to login to perform this action.
You will be redirected in 3 sec spinner

Free
Videos