Mensuration

1.0Standard Units Of Volume

$$\begin{gathered} \text{ Value of }\pi =\frac{22}{7}\text{ or }3.14 \\ 1\mathrm{dm}=10\mathrm{cm} \end{gathered}$$ $1{\mathrm{dm}}^3$ or litre $=1\mathrm{dm}\times 1\mathrm{dm}\times 1\mathrm{dm}=(10\times 10\times 10){\mathrm{cm}}^3=1000{\mathrm{cm}}^3$ $1\mathrm{cm}=10\mathrm{mm}$ $1\mathrm{m}=100\mathrm{cm}$ $1\mathrm{m}=10\mathrm{dm}$ $1{\mathrm{m}}^3=(100\times 100\times 100){\mathrm{cm}}^3=1000000{\mathrm{cm}}^3$ $1{\mathrm{m}}^3=1000\times 1000{\mathrm{cm}}^3=1000$ litre 1 kilolitre $=1{\mathrm{m}}^3=1000$ litre $(\mathrm{mm})$ millimetre, $(\mathrm{cm})$ centimetre, $(\mathrm{dm})$ decimetre, $(\mathrm{m})$ metre

2.0Plane Figures

The geometrical figures which have only two dimensions are called as the plane figures i.e. Triangle, Square, Parallelogram, Trapezium etc.

Take a glass partly filled with water and pour some oil in it.

Observe : Do the surfaces of oil and water have any height? No, that is why that surface is called two-dimensional surface because it does not have height.

3.0Solid Figures

A figure which has three dimensions (as length, breadth and height) is not a plane figure and we cannot draw such figures on black board exactly. These three-dimensional figures are called solids. Three dimensions or 3D are length, breadth and height. E.g., Cube, Cuboid, Cylinder, Cone, Sphere, Prism, Pyramid etc.

Cube

Cylinder

Prism

Cone

4.0Triangle

(i) Area of triangle $=\frac{1}{2}\times$ Base $\times$ Altitude or $\frac{1}{2}(b\times h)$

(ii) Area of an equilateral triangle $=\frac{\sqrt{3}}{4}(\text{ side }{)}^2$ or $\frac{\sqrt{3}}{4}{\mathrm{a}}^2$

(iii) Area of an isosceles triangle (base $=b$, equal side $=a$ ) $=\frac{\text{ base }}{4}\times \sqrt{4(\text{ equal side }{)}^2-(\text{ base }{)}^2}\text{ or }\frac{\mathrm{b}}{4}\times \sqrt{4{\mathrm{a}}^2-{\mathrm{b}}^2}$

• Q. Find the area of each triangle below.

• (a)
  
  (b) Solution : Area of triangle $=\frac{1}{2}\times b\times h$ or $\frac{1}{2}\times$ base $\times$ height (a) Area $=\frac{1}{2}\times 5\times 4.2=10.5{\mathrm{cm}}^2$ (b) Area $=\frac{1}{2}\times 6\times 5.5=16.5{\mathrm{cm}}^2$
• Q. Calculate the area of an equilateral triangle whose perimeter is 24 m . Solution :
  
  Perimeter of an equilateral triangle $=3\mathrm{a}=24\mathrm{m}$ {" a " is the side of equilateral triangle $\}$ $\Rightarrow 3\mathrm{a}=24$ $\Rightarrow a=8$ Thus, the length of side is 8 m . Area of equilateral triangle $=\frac{\sqrt{3}}{4}\times (\text{ side }{)}^2$ $=\frac{\sqrt{3}}{4}\times 8^2$ $=16\sqrt{3}{\mathrm{m}}^2$
• Q. Calculate the area of an isosceles triangle whose sides are $13\mathrm{cm},13\mathrm{cm}$ and 24 cm . Explanation : Equal side (a) = 13 cm base (b) $=24\mathrm{cm}$ Area of isosceles triangle $=\frac{b}{4}\sqrt{4a^2-b^2}$ $$\begin{gathered} =\frac{24}{4}\sqrt{4(13{)}^2-(24{)}^2} \\ =6\sqrt{676-576} \\ =6\sqrt{100}=6\times 10=60{\mathrm{cm}}^2 \end{gathered}$$

5.0Quadrilateral

Area of quadrilateral $ABCD=\frac{1}{2}\times AC\times \left(h_1+h_2\right)$ $=\frac{1}{2}\times$ (length of a diagonal) $\times$ (sum of the lengths of perpendiculars from the remaining two vertices on it)

Here, $h_1$ and $h_2$ are called the offsets of the quadrilateral.

• Q. The diagonal of a quadrilateral is 20 m in length and the perpendiculars to it from the opposite vertices are 8.5 m and 11 m . Find the area of the quadrilateral.
  
  Solution : In quadrilateral $ABCD$, we have $AC=20\mathrm{m}$ Let $\mathrm{BL}\perp \mathrm{AC}$ and $\mathrm{DM}\perp \mathrm{AC}$ such that $\mathrm{BL}=8.5\mathrm{m}$ and $\mathrm{DM}=11\mathrm{m}$ $\therefore$ Area of quadrilateral $\mathrm{ABCD}=\frac{1}{2}\times \mathrm{AC}\times (\mathrm{BL}+\mathrm{DM})$ $=\frac{1}{2}\times 20\times (8.5+11)=10\times 19.5=195{\mathrm{m}}^2$

• Perimeter is the length of boundary of a closed figure. We can find perimeter by doing sum of all sides.

6.0Rectangle

Length $=\ell ,$ Breadth $=\mathrm{b}$ (i) Perimeter of rectangle $=2(\ell +b)$ (ii) Area of rectangle $=\ell \times \mathrm{b}$ (iii) Length $=\frac{\text{ Area }}{\text{ Breadth }}$ or Breadth $=\frac{\text{ Area }}{\text{ Length }}$ (iv) Diagonal $=\sqrt{{\ell }^2+{\mathrm{b}}^2}$

Area of path (shaded region) = Area of outer rectangle - Area of inner rectangle

Area of path (shaded region) $=$ Ar. $($ ABCD $)-$ Ar. $(PQRS)$

• We can find length of diagonal of rectangle by using Pythagoras theorem.

• Q. The perimeter of a rectangular sheet is 120 cm . If the length is $35cm$, find its breadth. Also find the area. Explanation : Perimeter of a rectangular sheet $=120\mathrm{cm}$ (given) Length $=35\mathrm{cm}$. Perimeter of rectangle $=2(\ell +b)$ $120=2(35+b)$ $\Rightarrow 60=35+b$ $\Rightarrow \mathrm{b}=25\mathrm{cm}$ $\therefore$ Breadth of rectangle $=25\mathrm{cm}$ Area of rectangle $=\ell \times \mathrm{b}=35\times 25=875{\mathrm{cm}}^2$

7.0Square

Length of each side $=\mathrm{a}$ (i) Perimeter of square $=4$ a or $4\times$ side (ii) Area of square $={\mathrm{a}}^2$ or (side $\times$ side) or $\frac{1}{2}(\text{ Diagonal }{)}^2$ (iii) Side of the square $=\sqrt{\text{ Area }}$ (iv) Diagonal of the square $=\sqrt{2}a$

• Same as rectangle, we can find the length of diagonal of a square by using Pythagoras theorem.

• Q. Find the area of the figure shaded in the diagram.
  
  Explanation : We can see from the diagram that ABCD is a rectangle and PQRS is a square. Area of rectangle $\mathrm{ABCD}=$ length $\times$ breadth $=8\mathrm{cm}\times 6\mathrm{cm}$ $=48{\mathrm{cm}}^2$ Area of square $PQRS=$ side $\times$ side $=2\mathrm{cm}\times 2\mathrm{cm}$ $=4{\mathrm{cm}}^2$ Area of shaded figure $=$ Area of rectangle $ABCD-$ Area of square $PQRS$ $=(48-4){\mathrm{cm}}^2$ $=44{\mathrm{cm}}^2$
• Q. Find the perimeter and area of each shape below. (a)
  
  (b)
  
  Solution : (a) The perimeter is found by adding the length of all the sides. Perimeter $=6+8+1+4+4+4+1+8=36\mathrm{cm}$ To find the area, consider the shape being split into a rectangle and a square
  
  Area $=$ Area of rectangle + Area of square $=(6\times 8)+4^2=48+16=64{\mathrm{cm}}^2.$ (b) Adding the length of the sides gives $P=10+7+8+2+2+5=34\mathrm{cm}$ The area can be found by considering the shape to be a rectangle with a square removed from it. Area of shape $=$ Area of rectangle - Area of square $=(7\times 10)-2^2=70-4=66{\mathrm{cm}}^2$
• Q. How many square tiles of side 10 cm will be required to tile a floor measuring $800\mathbf{cm}$ by 900 cm ? Solution: Number of square tiles $=\frac{\text{ Area of rectangular floor }}{\text{ Area of a square tile }}=\frac{800\times 900}{10\times 10}=7200$ tiles

• While solving questions, remember that the units must be same.
• Trapezium is a quadrilateral not a parallelogram as it has only one pair of parallel sides.

8.0Trapezium

A trapezium is a quadrilateral in which one pair of opposite sides are parallel.

Base

Each of the two parallel sides of a trapezium is called base of the trapezium.

Height or altitude

The distance between the two bases (parallel sides) is called the height or altitude of the trapezium.

• When we talk about the shortest distance, it is the perpendicular distance from a point to a line or between two lines.

• Q. The area of a trapezium shaped field is $480{\mathrm{m}}^2$, the height is $15\mathbf{m}$ and one of the parallel sides is $20\mathbf{m}$. Find the other side.
  
  Solution : We have, Area of trapezium ABCD $=480{\mathrm{m}}^2$ $\Rightarrow \frac{1}{2}(\mathrm{AB}+\mathrm{CD})\times \mathrm{AL}=480$ $\Rightarrow \frac{1}{2}(20+CD)\times 15=480$ $\Rightarrow 20+\mathrm{CD}=\frac{480\times 2}{15}$ $\Rightarrow 20+\mathrm{CD}=32\times 2$ $\Rightarrow 20+CD=64$ $\Rightarrow \mathrm{CD}=64-20=44$ Hence, the other side of the trapezium is 44 m .
• Q. Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to the given base is $9\mathbf{cm}$ long. Solution : Given Length of bases of trapezium $=15\mathrm{cm}$ and 9 cm . Length of height $=8\mathrm{cm}$ Area of trapezium $=\frac{1}{2}\times ($ sum of parallel sides $)\times$ height $$\begin{gathered} =\frac{1}{2}\times (15+9)\times 8 \\ =24\times 4=96{\mathrm{cm}}^2 \end{gathered}$$

9.0Parallelogram

(i) Area of parallelogram $=$ Base $\times$ Height or $\mathrm{AB}\times \mathrm{DM}$

• A parallelogram is a quadrilateral in which both pairs of sides are parallel.

• Q. The parallel sides of a trapezium are $20\mathbf{cm}$ and 10 cm . Its non-parallel sides are both equal, each being 13 cm . Find the area of the trapezium. Explanation :
  
  Let $ABCD$ be a trapezium such that $\mathrm{AB}=20\mathrm{cm},\mathrm{CD}=10\mathrm{cm}$ and $\mathrm{AD}=\mathrm{BC}=13\mathrm{cm}$ Draw $\mathrm{CL}|\mid \mathrm{AD}$ and $\mathrm{CM}\perp \mathrm{AB}$. Now, CL|AD and CD|AL. $\therefore$ ALCD is a parallelogram. $\Rightarrow \mathrm{AL}=\mathrm{CD}=10\mathrm{cm}$ and $\mathrm{CL}=\mathrm{AD}=13\mathrm{cm}$ In $△CLB$, we have $CL=CB=13\mathrm{cm}$ $\therefore \Delta$ CLB is an isosceles triangle. $\mathrm{LM}=\mathrm{MB}=\frac{1}{2}\mathrm{BL}=\frac{1}{2}\times 10\mathrm{cm}=5\mathrm{cm}[\because \mathrm{BL}=\mathrm{AB}-\mathrm{AL}=(20-10)\mathrm{cm}=10\mathrm{cm}]$ Applying Pythagoras theorem in $△$ CML, we have ${\mathrm{CL}}^2={\mathrm{CM}}^2+{\mathrm{LM}}^2$ $13^2={\mathrm{CM}}^2+5^2$ ${\mathrm{CM}}^2=169-25=144$ $\mathrm{CM}=\sqrt{144}=12\mathrm{cm}$ $\Rightarrow$ Area of $△\mathrm{CLB}=\frac{1}{2}\times \mathrm{BL}\times \mathrm{CM}=\frac{1}{2}\times 10\times 12=60{\mathrm{cm}}^2$ Area of parallelogram ALCD $=\mathrm{AL}\times \mathrm{CM}=(10\times 12)=120{\mathrm{cm}}^2$ Hence, Area of trapezium ABCD = Area of parallelogram ALCD + Area of $△$ CLB $=(120+60){\mathrm{cm}}^2=180{\mathrm{cm}}^2$
• Q. The base of a parallelogram is thrice its height. If the area is $867{\mathrm{cm}}^2$, find the base and the height of the parallelogram. Solution : Let the height of the parallelogram be xcm . Then, base $=3\mathrm{xcm}$ Area of the parallelogram $=(x\times 3x){\mathrm{cm}}^2=3{\mathrm{x}}^2{\mathrm{cm}}^2$ But area of the parallelogram is given as $867{\mathrm{cm}}^2$ $\therefore 3x^2=867$ $\Rightarrow x^2=289$ $\Rightarrow \mathrm{x}=\sqrt{289}=17$ Height $=17\mathrm{cm}$ and base $=(3\times 17)\mathrm{cm}=51\mathrm{cm}$.

10.0Rhombus

If $d_1$ and $d_2$ are the lengths of the diagonals of a rhombus, then

• A rhombus is a parallelogram in which all the sides are equal.
• Rhombus is the quadrilateral which is not a regular polygon but its all sides are equal.

• Q. If the area of a rhombus be $24{\mathrm{cm}}^2$ and one of its diagonals be $4\mathbf{cm}$, find the perimeter of the rhombus.
  
  Explanation : Let ABCD be a rhombus such that its one diagonal $\mathrm{AC}=4\mathrm{cm}$. Suppose the diagonals AC and BD intersect at 0 . Area of rhombus $\mathrm{ABCD}=24{\mathrm{cm}}^2$ $\Rightarrow \frac{1}{2}\times \mathrm{AC}\times \mathrm{BD}=24\Rightarrow \frac{1}{2}\times 4\times \mathrm{BD}=24$ $\Rightarrow 2\times \mathrm{BD}=24\Rightarrow \mathrm{BD}=12\mathrm{cm}$ Thus, we have $\mathrm{AC}=4\mathrm{cm}$ and $\mathrm{BD}=12\mathrm{cm}$ $\therefore \mathrm{OA}=\frac{1}{2}\mathrm{AC}=2\mathrm{cm}$ and $\mathrm{OB}=\frac{1}{2}\mathrm{BD}=6\mathrm{cm}$ Since the diagonals of a rhombus bisect each other at right angle. Therefore, $△\mathrm{OAB}$ is right triangle, right angled at 0. Using Pythagoras theorem in $△\mathrm{OAB}$, we have ${\mathrm{AB}}^2=0{\mathrm{A}}^2+0{\mathrm{B}}^2$ ${\mathrm{AB}}^2=2^2+6^2=40$ $\mathrm{AB}=\sqrt{40}=2\sqrt{10}\mathrm{cm}$ Hence, perimeter of rhombus $\mathrm{ABCD}=(4\times 2\sqrt{10})\mathrm{cm}=8\sqrt{10}\mathrm{cm}$
• Q. Find the area of rhombus (given).
  
  Solution : Using formula, $A=\frac{1}{2}{d}_1\times d_2$ $\mathrm{A}=\frac{1}{2}\times 5\times 8=20{\mathrm{cm}}^2$.

• The value of $\pi$ is $\frac{22}{7}$ or 3.14.

11.0Circle

(i) Circumference of circle $=2\pi r$ (ii) Radius of circle $=\frac{\text{ Circumference }}{2\pi }$ or $\sqrt{\frac{\text{ Area }}{\pi }}$ (iii) Area of circle $=\pi r^2$

• Draw a circle of any radius $r$. Divide it into 16 parts by paper folding and join them as shown in the figure. Exploring the concept
  
  The figure looks like a rectangle. Length of this rectangle = length of arc of a semicircle of radius $r$ $=\frac{1}{2}\times 2\pi r=\pi r$ and, Breadth of rectangle $=$ radius of circle $=r$. Drawing conclusion: $\therefore$ Area of circle $=$ area of rectangle $=\pi r\times r=\pi r^2$.
• Q. Find the circumference and area of this circle $(\pi =3.14)$.
  
  Solution : The circumference is found using $\mathrm{C}=2\pi \mathrm{r}$, which in this case gives $\mathrm{C}=2\pi \times 4=2\times 3.14\times 4=25.12\mathrm{cm}$ The area is found using $A=\pi r^2$, which gives $A=\pi \times 4^2=3.14\times 16=50.24{\mathrm{cm}}^2$

Q. Find the radius of a circle if $\left[\pi =\frac{22}{7}\right]$; (i) its circumference is $32\mathbf{cm}$, (ii) its area is $14.3{\mathbf{cm}}^2$. Solution : (i) Using $\mathrm{C}=2\pi \mathrm{r}$ gives $32=2\pi \mathrm{r}$ and dividing by $2\pi$ gives $\frac{32}{2\pi }=\mathrm{r}$ $\mathrm{r}=\frac{32\times 7}{2\times 22}=\frac{56}{11}=5.09\mathrm{cm}$ (ii) Using $\mathrm{A}=\pi {\mathrm{r}}^2$ gives $14.3=\pi {\mathrm{r}}^2$ $\Rightarrow \frac{14.3}{\pi }=r^2$ or $r=\sqrt{\frac{14.3}{\pi }}=\sqrt{\frac{14.3\times 7}{22}}=\sqrt{4.55}=2.13\mathrm{cm}$

12.0Area of a polygon

Polygon

Regular polygon

Types of polygon

• Q. Find the area of the pentagon $ABCDE$ shown in figure if $AD=8\mathrm{cm},AH=6\mathrm{cm},\mathrm{AG}=4cm$, $\mathrm{AF}=3\mathrm{cm},\mathrm{BF}=2\mathrm{cm},\mathrm{CH}=3\mathrm{cm}$ and $\mathrm{EG}=2.5\mathrm{cm}$. Solution : We have
  
  Area of pentagon $\mathrm{ABCDE}=$ Area of $△\mathrm{AFB}+$ Area of trapezium $\mathrm{FBCH}+$ Area of $△\mathrm{CHD}+$ Area of $△\mathrm{ADE}$ $=\frac{1}{2}(\mathrm{AF}\times \mathrm{BF})+\frac{1}{2}(\mathrm{BF}+\mathrm{CH})\times \mathrm{FH}+\frac{1}{2}(\mathrm{DH}\times \mathrm{CH})+\frac{1}{2}(\mathrm{AD}\times \mathrm{GE})$ $\Rightarrow \frac{1}{2}(\mathrm{AF}\times \mathrm{BF})+\frac{1}{2}(\mathrm{BF}+\mathrm{CH})\times (\mathrm{AH}-\mathrm{AF})+\frac{1}{2}\{(\mathrm{AD}-\mathrm{AH})\times \mathrm{CH}\}+\frac{1}{2}(\mathrm{AD}\times \mathrm{GE})$ $\Rightarrow \frac{(3\times 2)+(2+3)\times (6-3)+(8-6)\times 3+(8\times 2.5)}{2}$ $\Rightarrow \frac{6+15+6+20}{2}$ $\Rightarrow \frac{47}{2}=23.5{\mathrm{cm}}^2$

13.0Cuboid

A solid bounded by six rectangular plane regions is called a cuboid.

A cuboid has six faces OAQB, CMPN, OAMC, BQPN, PQAM and OCNB, eight vertices $A,B,C,O,P,Q$, $M$ and $N$, twelve edges : $OA,BQ,NP,CM,PM,CN,OB,AQ,BN,PQ,MA$ and $OC$ and four diagonals OP, BM, CQ and AN.

Surface area of a cuboid

Consider a cuboid whose length is $\ell \mathrm{cm}$, breadth is bcm and height is h cm as shown in figure. Area of face BNPQ $=$ Area of face AOCM $=\ell \times b$ Area of face BNCO $=$ Area of face AQPM $=b\times h$ Area of face CMPN $=$ Area of face $\mathrm{AQBO}=\ell \times \mathrm{h}$ $\therefore$ Total surface area of the cuboid $=$ sum of the areas of all its six faces $$\begin{gathered} =2(\ell \times \mathrm{b})+2(\mathrm{b}\times \mathrm{h})+2(\ell \times \mathrm{h}) \\ =2[\ell \mathrm{b}+\mathrm{bh}+\ell \mathrm{h}] \end{gathered}$$

Lateral surface area of a cuboid (Area of 4 walls) $=$ Perimeter of the base $\times$ Height $$\begin{gathered} =2\times (\text{ Length }+\text{ breadth })\times \text{ Height } \\ =2(\ell +b)\times h \end{gathered}$$

Volume of cuboid = Area of a rectangular sheet $\times h$ $=\text{ Length }\times \text{ Breadth }\times \text{ Height }$ $=(\ell \times \mathrm{b})\times \mathrm{h}=\ell \mathrm{bh}$

Diagonal of a cuboid $=\sqrt{{\ell }^2+b^2+h^2}$

• Q. The diagram shows a lorry.
  
  Find the volume of the load-carrying part of the lorry. Explanation : The load-carrying part of the lorry is represented by a cuboid, so its volume is given by $\mathrm{V}=2\times 2.5\times 4=20{\mathrm{m}}^3$.
• Q. There are two cuboidal boxes whose dimensions are given below. Which box requires the higher amount of material to make? Cuboid A : $\ell =23,\mathrm{b}=30,\mathrm{h}=40$ Cuboid B : $\ell =30,\mathrm{b}=12,\mathrm{h}=44$ Solution : Required amount of material will be measured by the total surface area. Total surface area of cuboidal box $\mathrm{A}=2(\ell \mathrm{b}+\mathrm{bh}+\ell \mathrm{h})$ $$\begin{gathered} =2(23\times 30+30\times 40+23\times 40) \\ =2\times 2810{\mathrm{cm}}^2 \\ =5620{\mathrm{cm}}^2 \end{gathered}$$ Total surface area of cuboidal box $B=2(\ell b+bh+\ell h)$ $$\begin{gathered} =2(30\times 12+12\times 44+30\times 44) \\ =2\times 2208{\mathrm{cm}}^2 \\ =4416{\mathrm{cm}}^2 \end{gathered}$$ So, cuboidal box A will require the higher amount of material to make.

14.0Cube

A cuboid whose length, breadth and height are all equal is called a cube. Since all the faces of a cube are squares of the same size i.e. for a cube we have $\ell =\mathrm{b}=\mathrm{h}$. Thus, if $\ell \mathrm{cm}$ is the length of the edge of side of a cube, then Total Surface area of the cube $=2(\ell \times \ell +\ell \times \ell +\ell \times \ell )$ $=2\times 3{\ell }^2=6{\ell }^2=6(\text{ Edge }{)}^2$

Lateral surface area of the cube $=2(\ell \times \ell +\ell \times \ell )$ $=2\left({\ell }^2+{\ell }^2\right)=4{\ell }^2=4(\text{ Edge }{)}^2$

Volume of a cube $=\ell \times \ell \times \ell ={\ell }^3$

• When the length of the edge of a cube is doubled, its surface area becomes four times and volume becomes eight times.

• Q. If the volume of a cube is $729{\mathrm{cm}}^3$, find its side. Solution Volume of cube $=(\text{ side }{)}^3$ $729=(\text{ side }{)}^3\Rightarrow \sqrt[3]{729}=$ side $\Rightarrow \sqrt[3]{9^3}=$ side $\Rightarrow 9=$ side Hence, side of cube $=9\mathrm{cm}$
• Q. Find the side of a cube whose surface area is $2400{\mathrm{m}}^2$. Explanation : Total surface area of cube $=6$ (side) ${}^2$ $\Rightarrow 6$ (side) ${}^2=2400$ $\Rightarrow (\text{ side }{)}^2=400$ $\Rightarrow$ side $=\sqrt{400}=20\mathrm{m}$ Right circular cylinder
  
  A solid generated by the revolution of a rectangle about one of its side is called a right circular cylinder. Radius of right circular cylinder is $\mathbf{r}$ and height is $\mathbf{h}$ Area of the lateral surface of the cylinder = Area of the paper strip ABCD $=2\pi r\times h$ Lateral (curved) surface area of cylinder $=2\pi \mathrm{rh}=($ Base perimeter $\times \mathrm{h})$ Area of each base of cylinder $=\pi r^2$ Total surface area of cylinder $=$ Area of the curved surface $+2\times$ Area of base $=2\pi rh+2\pi r^2=2\pi r(h+r)$ Thus, volume of right circular cylinder $=$ Area of the base $\times$ height $=\pi r^{2}h$
• Q. A milk tank is in the form of cylinder whose radius is 1.5 m and length $7m$. Find the quantity of milk in litres that can be stored in the tank.
  
  Solution : Radius ( $r$ ) $=1.5\mathrm{m}$, Height ( h ) $=7\mathrm{m}$, Volume of cylinder $=\pi r^2\mathrm{h}$ $=\left(\frac{22}{7}\times 1.5\times 1.5\times 7\right){\mathrm{m}}^3=49.5{\mathrm{m}}^3$ $\because 1{\mathrm{m}}^3=1000$ litres $\therefore 49.5{\mathrm{m}}^3=1000\times 49.5$ litres $=49500$ litres Hence, the required quantity of milk in litres in tank $=49500$ litres
• Q. The earth taken out while digging a pit, is evenly spread over a rectangular field of length 90 m , width 60 m . If the volume of the earth dug is $3078{\mathrm{m}}^3$, find the height of the field raised. Explanation : $3078{\mathrm{m}}^3=90\mathrm{m}\times 60\mathrm{m}\times \mathrm{h}$ $\mathrm{h}=\frac{3078}{90\times 60}=0.57\mathrm{m}$ $\therefore$ Height of field raised $=0.57\mathrm{m}=0.57\times 100=57\mathrm{cm}(\because 1\mathrm{m}=100\mathrm{cm})$

15.0Mind Map