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13.1 MENSURATION:

Figure lying in a plane is called a plane figure. A plane figure made up of lines or curve or both, is said to be a closed figure if it has on free ends. Closed figure in a plane covers some part of the plane, then magnitude $o$ that part of the plane is called the area of that closed figure. the unit of measurement of that part of the plane is called the area of that closed figure. the unit o measurement of area is square unit (i.e. square centimeter, square metre etc.)

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13.1 (a)Mensuration of a Triangle:

perimeter $=a+b+c$

$\begin{array}{c}\text{ Area }=\frac{1}{2}\times \text{ Base }\times \text{ Height }\\ =\frac{1}{2}ah\end{array}$

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Heron’s formula:

Area $=\sqrt{s(s-a)(s-b)(s-c)}$

Where’s $=$ semi - perimeter

$=\frac{a+b+c}{2}$

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13.1(b) Menstruation of a Rectangle:

Perimeter $=2(\ell +b)$

Area $=\ell \times b$

Length of diagonal $=\sqrt{{\ell }^2+b^2}$

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13.1(c) Menstruation of a Square:

Perimeter $=4a$

Area $=a^2$

Length of diagonal $=a\sqrt{2}$

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13.1(d) Menstruation of a parallelogram:

Perimeter $=2(a+b)$

Area $=a{h}_1=b{h}_2$

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13.1(e)Mensuration of a Rhombus:

Perimeter $=4a=2\sqrt{d_1^2+d_2^2}$

Area $=\frac{1}{2}{d}_1{d}_2$

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13.1 (f) Mensuration of a Quadrilateral:

Let $AC=d$

$\text{ Area }=\frac{1}{2}d(h_1+h_2)$

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13.1(g)Menstruation of a Trapezium:

Area $=\frac{1}{2}h(a+b)$

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13.2 AREA RELTED TO CIRCLE:

Circle: Circle is a point, which moves so such a manner that its distance from a fixed point id always equal. The fixed point is called center of the circle of the circle and the fixed distance is called radius of the circle.

Area of circle $(A)=\pi r^2$

Circumference $(\mathbf{C})=2\pi r$

Diameter $(D)=2r$

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RESULTS:

(i) If two circles touch internally. then the distance between their centers is equal to the difference of their radii,

(ii) If two circles touch externally, then the distance between their centers is equal to the sum of their radii.

(iii) Distance moved by a rotating wheel in one revolution is the circumference of the wheel.

(iv) Number of revolutions completed by a rotating wheel in one minute

$=\frac{\text{ Distance moved in one minute }}{\text{ Circumference }}$

(v) Angle described by minute hand is one minute $=6^{\circ }$.

(vi) Angle described by hour hand in one hour $={30}^{\circ }$.

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13.5 (a) Semicircle:

Perimeter $=\pi r+2r=(\pi +2)r$

Area (A) $=\frac{\pi r^2}{2}$

Semi-Circle

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13.2 (b) Sector:

Area (A) $=\frac{\pi r^2\theta }{{360}^0}$

Length of arc $(\ell )=\frac{\pi r\theta }{{180}^{\circ }}$

Area $(A)=\frac{1}{2}\times \ell \times r$

Perimeter $=\ell +2r$

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13.2(c)Segment :

Shaded portion in the figure id called segment of a circle.

Minor segment

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Major segment

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Minor Segment

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Area of minor segment $=$ Area of the sector -Area of triangle $OAB$

$A=\frac{\pi r^2\theta }{{360}^0}-r^2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$ OR $A=\frac{\pi r^2\theta }{{360}^0}-\frac{r^2}{2}\sin \theta$

Here, segment $ACB$ is called manor segment while ADB is called major segment.

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13.3 MENSURATION (SOLID FIGURES) :

If any figure such as cuboids, which has three dimensions length, width and height are height are known as three dimensional figures. Where as rectangle has only two dimensional i.e., length and width. Three dimensional figures have volume in addition to areas of surface from which these soils figures are formed.

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Some of the main solid figures are:

13.3 (a) Cuboid:

Total Surface Area (T.S.A.) : The area of surface from which cuboid is formed. There are six faces (rectangular), eight vertices and twelve edges $n$ a cuboid.

(i) Total Surface Area (T.S.A.) $=2[\ell \times b+b\times h+h\times \ell ]$

(ii) Lateral Surface Area (L.S.A.) $=2[b\times h+h\times \ell ]$

(or Area of 4 walls)

$=2h[\ell +b]$

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(iii) Volume of Cuboid $=($ Area of base $)\times$ height

(iv) Length of diagonal $=\sqrt{{\ell }^2+b^2+h^2}$

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13.3 (b) Cube :

Cube has six faces. Each face is a square.

(i) T.S.A $=2[.x+x.x+x.x]$

$=2$ $2x^2+x^2+x^2=2(3x^2)=6x^2$

(ii) L.S.A. $=2[x^2+x^2]=4x^2$

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(iii) Volume $=($ Area of base $)\times$ Height

$=(x^2)\cdot x=x^3$

(iv) Length of altitude $=x\sqrt{3}$

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13.3 (c) Cylinder :

Curved surface area of cylinder (C.S.A.) : It is the area of surface from which the cylinder is formed. When we cut this cylinder, we will find a rectangle with length $2\pi r$ and height $h$ units.

(i) C.S.A. of cylinder $=(2\pi r)\times h=2\pi rh$.

(ii) Total Surface Area (T.S.A.) :

T.S.A. = C.S.A. + circular top & bottom

$\begin{array}{rl}& =2\pi rh+2\pi r^2\\ & =2\pi r(h+r)\text{ sq. units. }\end{array}$

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(iii) Volume of cylinder :

Volume $=$ Area of base $\times$ height

$\begin{array}{rl}& =(\pi r^2)\times h\\ & =\pi r^{2}h\text{ cubic units }\end{array}$

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13.3 (d) Cone :

(i) C.S.A. $=\pi r\ell$

(ii) T.S.A. $=$ C.S.A. + Other area

$=\pi r\ell$

$=\pi r(\ell +r)$

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(iii) Volume $=\frac{1}{3}\pi r^{2}h$

Where, $h=$ height

$r=$ radius of base

$\ell =$ slant height

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13.3 (e) Sphere :

T.S.A. $=$ S.A. $=4\pi r^2$

Volume $=\frac{4}{3}\pi r^3$

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13.3 (f) Hemisphere :

C.S.A $=2\pi r^2$

T.S.A = C.S.A. + other area

$=2\pi r^2+\pi r^2$

$=3\pi r^2$

Volume $=\frac{2}{3}\pi r^3$

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13.3 (g) Frustum of a Cone :

When a cone is cut by a plane parallel to base, a small cone is obtained at top and other part is obtained at bottom. This is known as ‘Frustum of Cone’. $\mathrm{△}ABC\sim \mathrm{△}ADE$

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$\begin{array}{cc}\therefore & \frac{AC}{AE}=\frac{AB}{AD}=\frac{BC}{DE}\\ & \\ & \frac{h_1}{h_1-h}=\frac{\ell }{{\ell }_1-\ell }=\frac{r_1}{r_2}\\ & \\ \text{ Or }& \frac{h_1}{h}=\frac{{\ell }_1}{\ell }=\frac{r_1}{r_1-r_2}\end{array}$

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Volume of Frustum $=\frac{1}{3}\pi r_1^2{h}_1-\frac{1}{3}\pi r_2^2(h_1-h)$

$\begin{array}{rl}& =\frac{1}{3}\pi [r_1^2{h}_1-r_2^2(h_1-h)]\\ & =\frac{1}{3}\pi [r_1^2(\frac{r_{1}h}{r_1-r_2})-r_2^2(\frac{r_{1}h}{r_1-r_2}-h)]=\frac{1}{3}\pi h[\frac{r_1^3-r_2^3}{r_1-r_2}]\\ & =\frac{1}{3}\pi h[r_1^2+r_2^2+r_1{r}_2]\end{array}$

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Curved Surface Area of Frustum $=\pi r_1{\ell }_1-\pi r_2({\ell }_1-\ell )$

$\begin{array}{rl}& =\pi [r_1(\frac{r_1\ell }{r_1-r_2})-r_2(\frac{r_1\ell }{r_{1-r_2}}-\ell )]=\pi \ell [\frac{r_1^2}{r_1-r_2}-\frac{r_2^2}{r_1-r_2}]\\ & =\pi \ell (r_1+r_2)\end{array}$

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Total Surface Area of Frustum $=$ CSA of frustum $+\pi r_1^2+\pi r_1^2+\pi r_2^2$

$=\pi \ell (r_1-r_2)+\pi r_1^2+\ell r_2^2$

Slant height of a Frustum $=\sqrt{h^2+(r_1-r_2{)}^2}$

where,

$h$ - height of the frustum

$r_1=$ radius of larger circular end

$r_2=$ radius of smaller circular end

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ILLUSTRATION :

Ex. 1 A chord of circle $14cm$ makes an angle of ${60}^{\circ }$ at the center of the circle. Find :

(i) area of minor sector

(iii) area of the major sector

(ii) area of the minor segment

(iv) area of the major segment

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Sol. Given, $r=14cm,\theta ={60}^{\circ }$

(i) Area of minor sector OAPB $=\frac{\theta }{{360}^0}\pi r^2$

$\begin{array}{rl}& =\frac{{60}^0}{{360}^0}\times 3.14\times 14\times 14\\ & \\ & =102.57c{m}^2\end{array}$

(ii) Area of minor segment APB $=\frac{\pi r^2\theta }{{360}^0}-\frac{r^2}{2}\sin \theta$

$\begin{array}{rl}& =102.57-\frac{14\times 14}{2}\sin {60}^0\\ & =102.57-98\times \frac{\sqrt{3}}{2}\\ & =17.80c{m}^2\end{array}$

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(iii) Area of major sector $=$ Area of circle - Area of minor sector OAPB $=\pi (14{)}^2-102.57$

$=615.44-102.57=512.87c{m}^2$

(iv) Area of major segment AQB

$\begin{array}{rl}& =\text{ Area of circle }-\text{ Area of minor segment APB }\\ & =615.44-17.80\\ & =597.64c{m}^2.\end{array}$

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Ex. 2 $ABCP$ is a quadrant of a circle of radius $14cm$. With $AC$ as diameter, a semicircle is drawn. Find the area of the shaded portion (figure).

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Sol. In right angled triangle $ABC$, we have.

$\begin{array}{rl}& A{C}^2=A{B}^2+B{C}^2\\ & A{C}^2={14}^2+{14}^2\\ & AC=\sqrt{2\times {14}^2}=14\sqrt{2}cm\end{array}$

Now required Area $=$ Area $APCQA$

= Area ACQA - Area ACPA

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$=$ Area ACQA - (Area ABCPA - Area of $\mathrm{△}ABC$ )

$=\frac{1}{2}\times \pi \times (\frac{14\sqrt{2}}{2}{)}^2-[\frac{1}{4}\times \pi (14{)}^2+\frac{1}{2}\times 14\times 14]$

$=\frac{1}{2}\times \frac{22}{7}\times 7\sqrt{2}\times 7\sqrt{2}-\frac{1}{4}\times \frac{22}{7}\times 14\times 14+7\times 14$

$=154-154÷98=98c{m}^2$

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Ex. 3 The diameter of cycle wheel is $28cm$. How many revolution will it make in moving $13.2km$ ?

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Sol. Distance traveled by the wheel is one revolution $=2\pi r$

$=2\times \frac{22}{7}\times \frac{28}{2}=88cm$

$\begin{array}{rl}\text{ and the total distance covered by the wheel }& =13.2\times 1000\times 100cm\\ & =1320000cm\\ \therefore \text{ Number of revolution made by the wheel }& =\frac{1320000}{88}=15000.\end{array}$

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Ex. 4 How many balls, each of radius $1cm$, can be made from a solid sphere of lead of radius $8cm$ ?

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Sol. Volume of the spherical ball of radius $8cm=\frac{4}{3}\pi \times 8^{3}c{m}^3$

Also, volume of each smaller spherical ball of radius $1cm=\frac{4}{3}\pi \times 1^{3}c{m}^3$.

Let $n$ be the number of smaller balls that can be made. Then, the volume of the larger ball is equal to the sum of all the volumes of $n$ smaller balls.

Hence, $\frac{4}{3}\pi \times n=\frac{4}{3}\pi \times 8^3\Rightarrow n=8^3=512$

Hence, the required number of balls $=512$.

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Ex. 5 An iron of length $1m$ and diameter $4cm$ is melted and cast into thin wires of length $20cm$ each. If the number of such wires be 2000 , find the radius of each thin wire.

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Sol. Let the radius of each thin wire be $rcm$. The, the sum of the volumes of 2000 thin wire will be equal to the volume of the iron rod. Now, the shape of the iron rod and each thin wire is cylindrical.

Hence, the volume of the iron rod of radius $\frac{4}{2}cm=2cm$ is $\pi \times 2^2\times 100c{m}^3$

Again, the volume of each thin wire $=\pi r^2\times 20$

Hence, we have $\pi \times 2^2\times 100=2000\times \pi r^2\times 20$

$\Rightarrow 40r^2=4\Rightarrow r^2=\frac{1}{100}\Rightarrow r=\frac{1}{10}$

[Taking positive square root only]

Hence, the required radius of each thin wire is $\frac{1}{10}cm$. of $0.1cm$.

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Ex. 6 By melting a solid cylindrical metal, a few conical materials are to be made. If three times the radius of the cone is equal to twice the radius of the cylinder and the ratio of the height of the cylinder and the height of the cone is $4:3$ find the number of cones which can be made.

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Sol. Let $R$ be the radius and $H$ be the height of the cylinder and let $r$ and $h$ be the radius and height of the cone respectively. Then.

$3r=2R$

and $H:h=4:3$ …..(i)

$\Rightarrow \frac{H}{h}=\frac{4}{3}$

$\Rightarrow 3H=4h$ …..(ii)

Let be the required number of cones which can be made from the material of the cylinder. The, the volume of the cylinder will be equal to the sum of the volumes of $n$ cones. Hence, we have

$\begin{array}{rl}& \pi R^{2}H=\frac{n}{3}\pi r^{2}h\Rightarrow 3R^{2}H=n{r}^{2}h\\ & \Rightarrow n=\frac{3R^{2}H}{r^{2}h}=\frac{3\times \frac{9r^2}{4}\times \frac{4h}{3}}{r^{2}h}[\therefore \text{ From (i) and (ii), }R=\frac{3r}{2}\text{ and }H=\frac{4h}{3}]\\ & \Rightarrow n=\frac{3\times 9\times 4}{3\times 4}\\ & \Rightarrow n=9\end{array}$

Hence, the required number of cones is 9 .

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Ex. 7 The base diameter of solid in the form of a cone is $6cm$ and the height of the cone is $10cm$. It is melted and recast into spherical balls of diameter $1cm$. Find the number of balls, thus obtained.

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Sol. Let the number of spherical balls be $n$. Then, the volume of the cone will be equal to the sum of the volumes

of the spherical balls. The radius of the base of the cone $=\frac{6}{2}cm=3cm$

and the radius of the sphere $=\frac{1}{2}cm$

Now, the volume of the cone $=\frac{1}{3}\pi \times 3^2\times 10c{m}^3=30\pi c{m}^3$

and, the volume of each sphere $=\frac{4}{3}\pi (\frac{1}{2}{)}^{3}c{m}^3=\frac{\pi }{6}c{m}^3$

Hence, we have

$n\frac{\pi }{6}=30\pi \Rightarrow n=6\times 30=180$

Hence, the required number of balls $=180$.

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Ex. 8 A conical empty vessel is to be filled up completely by pouring water into it successively with the help of a cylindrical can of diameter $6cm$ and height $12cm$. The radius of the conical vessel if $9cm$ and its height is $72cm$. How many times will it required to pour water into the conical vessel to fill it completely, if, in each time, the cylindrical can is filled with water completely?

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Sol. Let $n$ be the required number of times. Then, the volume of the conical vessel will be equal to $n$ times the volume of the cylindrical can.

Now, the volume of the conical vessel $=\frac{1}{3}\pi \times 9^2\times 72c{m}^3=24\times 81\pi c{m}^3$

Add the volume of the cylindrical can $=\pi \times 3^2\times 12c{m}^3=9\times 12\pi c{m}^3$

Hence , $24\times 81\pi =9\times 12\pi \times n$

$\Rightarrow n=\frac{24\times 81}{9\times 12}=18$ Hence, the required number of times $=18$.

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Ex. 9 The height of a right circular cylinder is equal to its diameter. It is melted and recast into a sphere of radius equal to the radius of the cylinder, find the part of the material that remained unused.

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Sol. Let $n$ be height of the cylinder. Then, its diameter is $h$ and so its radius is $\frac{h}{2}$. Hence, its volume is

$V_1=\pi (\frac{h}{2}{)}^{2}h=\frac{\pi h^3}{4}$

Again, the radius of the sphere $=\frac{h}{2}$

Hence, the volume of the sphere is $V_2=\frac{4}{3}\pi (\frac{h}{2}{)}^3=\frac{\pi h^3}{6}$

$\therefore$ The volume of the unused material $=V_1-V_2=\frac{\pi h^3}{4}-\frac{\pi h^3}{6}=\frac{\pi h^3(3-2)}{12}=\frac{\pi h^3}{12}=\frac{1}{3}=\times \frac{\pi h^3}{4}=\frac{1}{3}{V}_1$

Hence, the required volume of the unused material is equal to $\frac{1}{3}$ of the volume of the cylinder.

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Ex. 10 Water flows at the rate of $10m$ per minute through a cylindrical pipe having its diameter as $5mm$. How much time till it take to fill a conical vessel whose diameter of the base is $40cm$ and depth $24cm$ ?

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Sol. Diameter of the pipe $=5mm\frac{5}{10}cm=\frac{1}{2}cm$.

$\therefore$ Radius of the pipe $=\frac{1}{2}\times \frac{1}{2}cm=\frac{1}{4}cm$.

In 1 minute, the length of the water column in the cylindrical pipe $=10m=1000cm$.

$\therefore$ Volume, of water that flows out of the pipe in 1 minute $=\pi \times \frac{1}{4}\times \frac{1}{4}\times 1000c{m}^3$.

Also, volume of the cone $=\frac{1}{3}\times \pi \times 20\times 20\times 24c{m}^3$.

Hence, the time needed to fill up this conical vessel $=(\frac{1}{3}\pi \times 20\times 20\times 24÷\pi \times \frac{1}{4}\times \frac{1}{4}\times 1000)$ minutes

$=(\frac{20\times 20\times 24}{3}\times \frac{4\times 4}{1000})=\frac{4\times 24\times 16}{30}$ minutes

$=\frac{256}{5}$ minutes $=51.2$ minutes.

Hence, the required time of 51.2 minutes.

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Ex. 11 A hemispherical tank of radius $1\frac{3}{4}$ is full of water. It is connected with a pipe which empties it at the rate of 7 liters per second. How much time will it take to empty the tank completely?

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Sol. Radius of the hemisphere $=\frac{7}{4}m=\frac{7}{4}\times 100cm=175cm$

$\therefore$ Volume of the hemisphere $=\frac{2}{3}\times \pi \times 175\times 175\times 175c{m}^3$

The cylindrical pipe empties it at the rate of 7 liters i.e., $7000c{m}^3$ of water per second.

Hence, the required time to empty the tank $=(\frac{2}{3}\times \frac{22}{7}\times 175\times 175\times 175÷7000)s$

$=\frac{2}{3}\times \frac{22}{7}\times \frac{175\times 175\times 175}{7000\times 60}min=\frac{11\times 25\times 7}{3\times 2\times 12}min=\frac{1925}{72}min$

$\cong 26.75$ $min$, nearly.

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Ex. 12 A well of diameter $2m$ is dug $14m$ deep. The earth taken out of its is spread evenly all around it to a width of $5m$ to from an embankment. Find the height of the embankment.

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Sol. Let $n$ be the required height of the embankment.

The shape of the embankment will be like the shape of a cylinder of internal radius $1m$ and external radius $(5+1)m=6m$ [figure].

The volume of the embankment will be equal to the volume of the earth dug out from the well. Now, the volume of the earth = volume of the cylindrical well

$\begin{array}{rl}& =\pi \times 1^2\times 14m^3\\ & =14\pi m^3\end{array}$

Also, the volume of the embankment

$=\pi (6^2-1^2)hc{m}^3=35\pi h{m}^3$

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Hence, we have

$35\pi h=14\pi$

$\Rightarrow h=\frac{14}{35}=\frac{2}{5}=0.4$

Hence, the required height of the embankment $=0.4m$

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Ex. 13 Water in a canal, $30dm$ wide and $12dm$ deep, is flowing with a speed of $10km/hr$. How much area will it irrigate in 30 minutes if $8cm$ of standing water is required from irrigation.

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Sol. Speed of water in the canal $=10km.h=10000m.60min=\frac{500}{3}m/min$.

$\therefore$ The volume of the water flowing out of the canal in 1 minute $=(\frac{500}{3}\times \frac{30}{10}\times \frac{12}{10})m^2=600m^3$

$\therefore$ In $30min$, the amount of water flowing out of the canal $=(600\times 30)m^3=600m^3$

If the required area of the irrigated land is $\times m^2$, then the volume of water to be needed to irrigate the land

$=(x\times \frac{8}{100})m^3$ $=\frac{2x}{25}{m}^3$

Hence, $\frac{2x}{25}=18000$

$\Rightarrow x=18000\times \frac{25}{2}=225000$

Hence, the required area is $225000m^2$.

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Ex. 14 A bucket is $40cm$ in diameter at the top and $28cm$ in diameter at the bottom. Find the capacity of the bucket in litters, if it is $21cm$ deep. Also, find the cost of tin sheet used in making the bucket, if the cost of tin is Rs. 1.50 per sq dm.

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Sol. Given : $r_1=20cm{r}_2=14cm$ and $h=21cm$

Now, the required capacity (i.e. volume) of bucket $=\frac{\pi h}{3}(r_1^2+r_1{r}_2+r_2^2)$

$\cong \frac{22\times 21}{7\times 3}({20}^2+20\times 14+{14}^2)c{m}^3=22\times 876c{m}^3=19272c{m}^3=\frac{19272}{1000}$ liters $=19.272$ liters.

Now, $I=\sqrt{(r_1-r_2{)}^2+h^2}=\sqrt{(20-14{)}^2+{21}^2}cm=\sqrt{6^6+{21}^2}cm=\sqrt{36+441}cm=\sqrt{477}cm\cong 21.84cm$.

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$\therefore$ Total surface area of the bucket (which is open at the top)

$=\pi \ell (r_1+r_2)+\pi r_2^2=\pi [(r_1+r_2)\ell +r_2^2]=\frac{22}{7}[(20+14)21.84+{14}^2]$

$=2949.76c{m}^3\therefore$ Required cost of the tin sheet at the rate of Rs. 1.50 per $d{m}^2$ i.e., per $100c{m}^2$

$=$ Rs $\frac{1.50\times 2949.76}{100}\cong Rs44.25$

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Ex. 15 A cone is divided into two parts by drawing a plane through a point which divides its height in the ratio 1 : 2 starting from the vertex and the place is parallel to the base. Compare the volume of the two parts.

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Sol. Let the plane $XY$ divide the cone $ABC$ in the ratio $AE:ED=1:2$, where $AED$ is the axis of the cone. Let $r_2$ and $r_2$ be the radii of the circular section $XY$ and the base $BC$ of the cone respectively and let $h_1-h$ and $h_1$ be their heights [figure].

Then, $\frac{h_1}{h}=\frac{3}{2}\Rightarrow h=\frac{3}{2}h$

And $\frac{r_1}{r_2}=\frac{h_1}{h_1-h}=\frac{\frac{3}{2}h}{\frac{1}{2}h}=3$

$\therefore r_1=3r_2$

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Volume of cone AXY

$=\frac{1}{32}\pi r_2^2(h_1-h)$

$=\frac{1}{3}\pi r_2^2(\frac{3}{2}h-h)=\frac{1}{6}\pi r_2^{2}h$

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Volume of frustum

$=\frac{1}{3}\pi h(r_1^2+r_2^2+r_1{r}_2)$

$=\frac{1}{3}\pi h(9r_2^2+r_2^2+3r_2^2)$

$=\frac{1}{3}\pi h(13r_2^2)$

So, $\frac{\text{ Volume of cone AXY }}{\text{ Volume of frustum XYBC }}=\frac{\frac{1}{6}\pi r_2^{2}h}{\frac{13}{3}\pi r_2^{2}h}$

$\frac{\text{ Volume of cone }AXY}{\text{ Volume of frustum }XYBC}=\frac{1}{26}.$

i.e. the ratio between the volume of the cone AXY and the remaining portion BCYX is $1:26$.

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DAILY PRACTIVE PROBLEMS 13

OBJECTIVE DPP - 13.1

1. If $BC$ passed through the centre of the circle, then the area of the shaded region in the given figure is

(A) $\frac{a^2}{2}(3-\pi )$

(B) $a^2(\frac{\pi }{2}-1)$

(C) $2a^2(\pi -1)$

(D) $\frac{a^2}{2}(\frac{\pi }{2}-1)$

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2. The perimeter of the following shaded portion of the figure is:

(A) $40m$

(B) $40.07m$

(C) $40.28m$

(D) $35m$

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3. If a rectangle of sides $5cm$ and $15cm$ is be divided into three squared of equal area, then the sides of the squares will be:

(A) $4cm$

(B) $6cm$

(C) $7cm$

(D) None

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4. The area of the shaded region in the given figure is :

(A) $\frac{\pi }{3}$ sq. units

(B) $\frac{\pi }{2}$ units

(B) $\frac{\pi }{4}$ sq. units

(D) ${\pi }^2$ sq. units

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5. The area of the shaded portion in the given figure is :

(A) $7.5\pi$ sq. units

(B) $6.5\pi$ sq. units

(C) $5.5\pi$ sq. units

(D) $4.5\pi$ sq. units

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6. In the adjoining figure, the radius of the inner circle, if other circles are of radii $1m$, is :

(A) $(\sqrt{2}-1)m$

(B) $\sqrt{2}m$

(C) $\frac{1}{\sqrt{2}}m$

(D) $\frac{2}{\sqrt{2}}m$

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7. The height of a conical tent of the centre is $5cm$. The distance of any point on its circular base from the top of the tent is $13m$. The area of the slant surface is :

(A) $144\pi$ sq m

(B) $130\pi$ sq m

(C) $156\pi sqm$

(D) $169\pi sqm$

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8. The radius of circle is increased by $1cm$, then the ratio of the new circumference to the new diameter is :

(A) $\pi +2$

(B) $\pi +1$

(C) $\pi$

(D) $\pi -\frac{1}{2}$

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9. A hemispherical bowl of internal diameter $36cm$ is full of some liquid. This liquid is to be filled in cylindrical bottles of radius $3cm$ and height $6cm$., Then no of bottles needed to empty the bowl.

(A) 36

(B) 75

(C) 18

(D) 144

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10. There is a cylinder circumscribing the hemisphere such that their bases are common. The ratio of their volume is

(A) $1:3$

(B) $1:2$

(C) $2:3$

(D) $3:4$

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11. A sphere of radius $3cms$ is dropped into a cylindrical vessel of radius $4cms$. If the sphere is submerged completely, then the height (in $cm$ ) to which the water rises, is

(A) 2.35

(B) 2.30

(C) 2.25

(D) 2.15

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12. If a rectangular sheet of paper $44cm\times 22cm$ is rolled along its length of form a cylinder, then the volume of cylinder in $c{m}^3$ is

(A) 1694

(B) 3080

(C) 3388

(D) none of these

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13. Two cones have their heights in the ratio $1:3$ and the radii of their bases are in the ratio $3:1$, then the ratio of their volumes is

(A) $1:3$

(B) $27:1$

(C) $3:1$

(D) $1:27$

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14. The total surface area of a cube is numerically equal to the surface area of a sphere then the ratio of their volume is

(A) $\frac{\pi }{6}$

(B) $\sqrt{\frac{\pi }{6}}$

(C) $\frac{\pi }{216}$

(D) $\sqrt{\frac{6}{\pi }}$

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15. A cone is dived into two parts by drawing a plane through the mid point of its axis parallel to its base then the ratio of the volume of two parts is

(A) $1:3$

(B) $1:7$

(C) $1:8$

(D) $1:9$

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SUBJECTIVE DPP - 13.2

1. The area of a circle inscribed in an equilateral triangle is $154c{m}^2$. Find the perimeter of the triangle.

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Sol. 1. $72.7cm$

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2. The radii of two circles are $8cm$ and $6cm$ respectively. Find the radius of the circle having its area egual to the sum of the areas of the two circles.

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Sol. 2. $10cm$

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3. Figure, shows a sector of a circle, centre $O$, containing an angle ${\theta }^0$. Prove that :

(i) Perimeter of the shaded region is $r(\tan \theta +\sec \theta +\frac{\pi \theta }{180}-1)$

(ii) Area of the shaded region is $\frac{r^2}{2}(\tan \theta -\frac{\pi \theta }{180})$

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4. The area of an equilateral triangle is $49\sqrt{3}c{m}^2$. Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in figure. Find the area of the triangle not included in the circle.

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Sol. 4. $7.77c{m}^2$

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5. Find the area of the shaded region in figure. where $ABCD$ is a square of side $10cm$. (use $\pi =3.14$ )

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Sol. 5. $57c{m}^2$

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6. A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is $\frac{8}{9}$ of the curved surface of whole cone, find the ratio of the line - segment into which the cone’s altitude is divided by the plane.

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Sol. 6. $1:2$

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7. A right - angled triangle whose sides are $15cm$ and $20cm$, is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. [Take $\pi \cong 3.14$ ]

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Sol. 7. $3768c{m}^3,1318.8c{m}^2$

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8. 50 persons took dip in a rectangular tank which is $80m$ long and $50m$ broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is $0.04m^3$ ?

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Sol. 8. $0.5cm$

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9. Water is flowing at the rate of $5km$ per hour through a pipe of diameter $14cm$ into a rectangular tank, which is $50m$ long and $44m$ wide. Find the time in which the level of water in the tank will rise by $7cm$.

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Sol. 9. $$ $2$ hrs.

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10. A circus tent is cylindrical to a height of $3m$ and conical above it. If its base radius is $52.5m$ and slant height of the conical portion is $53m$, find the area of the canvas needed to make the tent.

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Sol. 10. $9735c{m}^2$

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11. The diameters external and internal surfaces of a hollow spherical shell are $10cm$ and $6cm$ respectively. If it is melted and recast into a solid cylinder of length of $2\frac{2}{3}cm$, find the diameter of the cylinder.

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Sol. 11. $14$ $cm$

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12. A cylindrical container of radius $6cm$ and height $15cm$ is fulled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is four times the radius of its base, find the radius of the ice-cream cone.

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Sol. 12. $3cm$

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13. A hemi-spherical depression is cutout from one face of the cubical wooden block such that the diameter $\ell$ of the hemisphere is equal to the edge of the cube., Determine the surface are of the remaining solid.

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Sol. 13. $\frac{{\ell }^2}{4}(24+\pi )$

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14. In figure there are three semicircles, $A,B$ and $C$ having diameter $3cm$ each, and another semicircle $E$ having a circle D with diameter $4.5cm$ are shown. Calculate. (i) the area of the shaded region (ii) the cost of painting the shaded region of the 25 paisa per $c{m}^2$, to the nearest rupee.

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Sol. 14. $12.375c{m}^2$, Rs. 3

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15. The height of a cone is $30cm$. A small cone is cut off at the top by a plane parallel to the base. If its volume be $\frac{1}{27}$ of the volume of the given cone, at what above the vase is the section made?

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Sol. 15. $20cm$

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16. A solid cylinder of diameter $15cm$ and height $15cm$ is melted and recast into 12 toys in the shape of a right circular cone mounted on a hemisphere. Find the radius of the hemisphere and the total height of the to if height of the conical par is 3 times its radius.

[CBSE - 2005]

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Sol. 16. $$ radius $=3cm$ and height $=9cm$

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17. if the rail of the ends of bucket, $45cm$ high are $28cm$ and $7cm$, determine the capacity and total surface area of the bucket.

[CBSE - 2006]

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Sol. 17. $48510c{m}^3,5621c{m}^3$

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18. A tent is in the form of cylinder of diameter $4.2m$ and height $4m$, surmounted by a cone of equal base and height $2.8m$. Find the capacity of the tent and the cost of canvas for making the tent at Rs. 100 per sq. m. ?

[CBSE - 2006]

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Sol. 18. $68.376m^3$, Rs. 7590

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19. Water flows out through a circular pipe whose internal radius is $1cm$, at the rate of $80cm/second$ into an empty cylindrical tank, the radius of whose base is $40cm$. By how much will the level of water rise in the tank in half an hour?

[CBSE - 2007]

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Sol. 19. $90cm$

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20. A hemispherical bowl of internal radius $36cm$ is full of liquid. The liquid is to be filled into cylindrical shaped small bottles each of diameter $3cm$ and height $6cm$. How many bottles are need to empty the bowl?

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Sol. 20. $$ $2304$

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21. In figure $ABC$ is a right - angled triangle right-angled at $A$. Semicircles are drawn on $AB,AC$ and $BC$ as diameters. Find the area of the shaded region.

[CBSE - 2008]

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Sol. 21. $6sq$.

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22. Find the permetre of figure, where $AED$ is a semi-circle and $ABCD$ is a rectangle.

[CBSE - 2008]

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Sol. 22. $76cm$

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23. A tent consists of a frustum of a cone, surmounted by a cone. If the diameters of the upper and lower circular ends of the frustum $b$ $14m$ and $26m$ respectively, the height of the frustum be $8m$ and the slant height of the surmounted conical portion be $12m$, find the area of canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of surmounted conical portion are equal)

[CBSE - 2008]

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Sol. 23. $892.57m^2$