Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r₁ and r₂ be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.
In ΔABG and ΔADF, DF||BG
∴ ΔABG ∼ ΔADF
DF/BG = AF/AG =AD/AB
\(\begin{array}{l} \frac{r_{2}}{r_{1}}=\frac{h_{1}-h}{h_{1}}=\frac{l_{1}-l}{l_{1}} \\ \frac{r_{2}}{r_{1}}=1-\frac{h}{h_{1}}=1-\frac{1}{l_{1}} \\ l-\frac{l}{l_{1}}=\frac{r_{2}}{r_{1}} \\ \frac{l}{l_{1}}=1-\frac{r_{2}}{r_{1}}=\frac{r_{1}-r_{2}}{r_{1}} \\ \frac{l_{1}}{l}=\frac{r_{1}}{r_{1}-r_{2}} \\ l_{1}=\frac{r_{1} l}{r_{1}-r_{2}} \end{array}\)
CSA of frustum DECB = CSA of cone ABC − CSA cone ADE
\(\begin{array}{l} =\pi r_{1} l_{1}-\pi r_{2}\left(l_{1}-l\right) \\ =\pi r_{1}\left(\frac{l r_{1}}{r_{1}-r_{2}}\right)-\pi r_{2}\left[\frac{r_{1} l}{r_{1}-r_{2}}-l\right] \\ =\frac{\pi r_{1}^{2} l}{r_{1}-r_{2}}-\pi r_{2}\left(\frac{r_{1} l-r_{1} l+r_{2} l}{r_{1}-r_{2}}\right) \\ =\frac{\pi r_{1}^{2} l}{r_{1}-r_{2}}-\frac{\pi r_{2}^{2} l}{r_{1}-r_{2}} \\ =\pi l\left[\frac{r_{1}^{2}-r_{2}^{2}}{r_{1}-r_{2}}\right] \end{array}\)
CSA of frustum = Π(r₁ + r₂)l
Total surface area of frustum = CSA of frustum + Area of upper circular end + Area of lower circular end
\(\begin{array}{l} =\pi\left(r_{1}+r_{2}\right) l+\pi r_{2}^{2}+\pi r_{1}^{2} \\ =\pi\left[\left(r_{1}+r_{2}\right) l+r_{1}^{2}+r_{2}^{2}\right] \end{array}\)

Area of right-angled \(\Delta\)ABC = \(\frac { 1 }{ 2 } \times 10\times 10\)
= 50 cm²
Area of quadrant APR of the circle of radius 7 cm
\(\frac { 1 }{ 4 } \times \pi \times { (7) }^{ 2 }Area\quad of\quad quadrant=\frac { 1 }{ 4 } Area\quad of\quad circle\)
\(=\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 49=38.5{ cm }^{ 2 }\)
\(\therefore\) Area of base PBCR = Area of \(\Delta\)ABC - Area of quadrant APR
= 50 - 38.5 = 11.5 cm²

Given, edge of the cube=22cm
Volume of the cube=(Edge)³=(22)³=10648cm³
Also, given that diameter of a marble=0.5cm
Radius of a marble\(=\frac { 0.5 }{ 2 } =0.25cm\)  \(\left[ \because \quad radius=\frac { diameter }{ 2 } \right] \)
Volume of one spherical marble
\(=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \left( 0.5 \right) ^{ 3 } \ \ \ \ \left[ \because \ volume \ of \ sphere=\frac { 4 }{ 3 } \pi r^{ 3 } \right] \\ =\frac { 88 }{ 21 } \times 0.015625=\frac { 1375 }{ 21 } =0.0655cm^{ 3 }\)
Filled space of the cube with marble
=Volume of cube\(-\frac { 1 }{ 8 } \times \)Volume of the cube
\(=10648-\frac { 1 }{ 8 } \times 10648=10648-1331\\ =9317cm^{ 3 }\)
Required number of marbles
\(=\frac { Total \ space \ filled \ by \ marbles \ in \ a \ cube }{ Volume \ of \ one \ marble } \\ =\frac { 9317 }{ 0.0655 } =142244\)
Hence, the number of marbles that the cube accommodate is 142244.

Area of each sector = \(\frac{1}{3}\)\(\times\)Area of circle form by wings of fan
\(=\frac{1}{3} \times \frac{22}{7} \times(0.98)^2\)                             ...(i)
Area of each sector \(=\frac{1}{2} \times 0.98 \times l\)    ...(ii)
From Eqs.(i) and (ii), we get
\(\frac{1}{2} \times 0.98 \times l=\frac{1}{3} \times \frac{22}{7} \times(0.98)^2\)
Ans. 2.05 m

Area of the segment 
= area of the shaded portion 
= area of the sector - area of the triangle 
= \(\frac { 90° }{ 360° } \times \frac { 22 }{ 7 } \times { (10) }^{ 2 }-\frac { 1 }{ 2 } \times 10\times 10\)
= \(\left( \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 100-50 \right) \) cm²
= (78.57 - 50) cm²
= 28.57 cm²

Let radius of the base Of cylinder be r cm and height be h cm 
Total surface area = \(2\pi r(r+h)=231cm^{ 2 }\)
Curved surface area = \(2\pi rh\)
According to the question 
curved surface area = \(\frac { 2 }{ 3 } \) total surface area
\(\Rightarrow 2\pi rh=\frac { 2 }{ 3 } \times 2\pi r(r+h)\Rightarrow 2\pi rh=231\Rightarrow 2\pi rh=154cm^{ 2 }\)
Now \(2\pi r(r+h)=231\Rightarrow 2\pi r^{ 2 }+2\pi rh=231\)
\(\Rightarrow 2\pi r^{ 2 }+154=231\Rightarrow 2\pi r^{ 2 }=231\)
\(\Rightarrow 2\times \frac { 22 }{ 7 } \times r^{ 2 }=77\Rightarrow r^{ 2 }=\frac { 77\times 7 }{ 22\times 2 } \Rightarrow r=\frac { 7 }{ 2 } cm\)
\(2\pi rh=154\Rightarrow 2\times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times h=154\)
h= \(\frac { 154 }{ 22 } =7cm\)

A hemisphere and a cone.

18 cm

20 cm

176

Radius of the base of cone=21 cm

Volume of cone=\(\frac{2}{3}\)volume of hemisphere
\(\frac{1}{3}\pi{r}^{2}h=\frac{2}{3}\times{\frac{2}{3}}\pi{r}^{3}\)
⇒ h=\(\frac{4}{3}r=\frac{4}{3}\times{21}\)=28cm
l²=h²+r²=(28)²+(21)²
=784+441=1225
l=\(\sqrt{1225}\)=35cm
Surface area of the toy=πrl=2πr²
=\(\frac{22}{7}\times{21}\times{35}\)cm²+2x\(\frac{22}{7}\)x21x21 cm²
=2310 cm²+2772 cm²
=5082 cm²

We known that, if two circles touch externally, then the distance between their centres is equal to the sum of their radii.
Let the radii of the two circles be r₁ and r₂ , respectively and C₁ , C₂ be the centres of the given circles.
Then r₁ + r₂ = 14 [distance between the centres of two circles = 14cm, given]
Given, the sum of the areas of two circles is equal to 130\(\pi\) sq cm.
\(\therefore \quad { \pi r }_{ 1 }^{ 2 }+{ \pi r }_{ 2 }^{ 2 }=130\pi \Rightarrow \pi ({ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 })=130\pi \)
\(\Rightarrow { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }=130\)
Now, \(({ r }_{ 1 }+{ r }_{ 2 })^{ 2 }={ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+2{ r }_{ 1 }{ r }_{ 2 }\)
\(\Rightarrow { (14) }^{ 2 }=130+2{ r }_{ 1 }{ r }_{ 2 }\)  [from Eqs. (i) and (ii)]
\(\Rightarrow 196-130=2{ r }_{ 1 }{ r }_{ 2 }\)
\(\Rightarrow 66=2{ r }_{ 1 }{ r }_{ 2 }\)
\( \Rightarrow { r }_{ 1 }{ r }_{ 2 }=\frac { 66 }{ 2 } =33\)
We know that, 
\(({ r }_{ 1 }-{ r }_{ 2 })^{ 2 }={ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }-2{ r }_{ 1 }{ r }_{ 2 }\)
\(\Rightarrow ({ r }_{ 1 }-{ r }_{ 2 })^{ 2 }=130-2\times 33\)  [from Eq. (i)]
\(\Rightarrow ({ r }_{ 1 }-{ r }_{ 2 })^{ 2 }=130-66\)
\(\\ \Rightarrow ({ r }_{ 1 }-{ r }_{ 2 })^{ 2 }=64\)
\(\Rightarrow \quad \ { r }_{ 1 }-{ r }_{ 2 }=8\)
On adding Eqs. (i) and (iii), we get
\( { 2r }_{ 1 }=22\)
\(\Rightarrow \ \ { r }_{ 1 }=\frac { 22 }{ 2 } 11cm\)
On putting the value of r₁ in Eq. (i), we get
\(11+{ r }_{ 2 }=14\)
\( \Rightarrow \ { r }_{ 2 }=14-11=3cm\)
Hence, the radii of the circles are 11 cm and 3 cm.
The value depicted by the farmers, are of cooperative nature and mutual understanding.

(i) Rs. 3511.20  
(ii) Social value, cooperation, caring for old people

\(12\pi \ cm,\ 37.71{ cm }^{ 2 }\)

Radius of hemisphere = 4.2 cm
Volume of hemisphere  =\(\frac { 2 }{ 3 } \pi \) r³
=\(\frac { 2 }{ 3 } \pi \times (4.2)^{ 3 }{ cm }^{ 3 }\)   =49.329 \(\pi \) cm³
∴  volume of 2 hemispheres
=2 X  =98.392\(\pi \) cm³ = 98.784\(\pi \) cm³​​​​​​​
Height of cylinder =10 cm
Radius =4.2 cm
∴  volume of cylinder =\(\pi \) r²h
=\(\pi \) X (4.2)²X 10=176.4 \(\pi \)
∴  volume of metal left 
=176.4\(\pi \) -98.3784\(\pi \) =77.616\(\pi \) cm³
Radius of wire =0.7 cm
let length of wire be x
Volume of wire =\(\pi \) X 0.7 X 0.7 X x
  =0.49\(\pi \) x cm³
⇒   0.49\(\pi \)x =77.616 \(\pi \)
⇒ x= 158.4 cm
∴ length of wire =158.4 cm

(i) (b): capacity of first glass \(=\pi r^{2} h-\frac{2}{3} \pi r^{3}\)
\(=\pi r^{2}\left(h-\frac{2}{3} r\right)\)
\(=\pi(3)^{2}\left(10-\frac{2}{3} \times 3\right)\)
\(=9 \pi(10-2)\)
72\(\pi\) cm³
(ii) (a): capacity of second glass \(=\pi r^{2} \mathrm{H}-\frac{1}{3} \pi r^{2} h\)
\(=\pi r^{2}\left(\mathrm{H}-\frac{1}{3} h\right)\)
\(=\pi(3)^{2}\left(10-\frac{1}{3} \times 1.5\right)\)
\(=9 \pi(10-0.5)\)
= 85.5 \(\pi\)cm³
(iii) (a): \(\text { Ratio }=\frac{\text { Capacity of first glass }}{\text { Capacity of second glass }}=\frac{72 \pi}{85.5 \pi}=\frac{16}{19}=16: 19 .\)
(iv) (d): From part (i) and (ii)
Suresh got more quantity of juice.
\(=85.5 \pi \mathrm{cm}^{3}-72 \pi \mathrm{cm}^{3}=13.5 \pi \mathrm{cm}^{3}\)
(v) (a): Total quantity of juice is purchased by Suresh = (72 \(\pi\) + 85.5 \(\pi\) ) cm³
= 157.5\(\pi\) cm³

(i) (b):Curved surface area of two identical cylindrical parts \(=2 \times 2 \pi r h=2 \times 2 \times \frac{22}{7} \times \frac{2.5}{2} \times 5\)
= 78.57 cm²
(ii) (a): Volume of big cylindrical part \(=\pi r^{2} h\)
\(=\frac{22}{7} \times \frac{4.5}{2} \times \frac{4.5}{2} \times 12=190.93 \mathrm{~cm}^{3}\)
(iii) (b) :Volume of two hemispherical ends \(=2 \times \frac{2}{3} \pi r^{3}\)
\(=\frac{2 \times 2}{3} \times \frac{22}{7} \times\left(\frac{2.5}{2}\right)^{3}=8.18 \mathrm{~cm}^{3}\)
(iv) (c) : Curved surface area of two hemispherical ends \(=2 \times 2 \pi r^{2}=2 \times 2 \times \frac{22}{7} \times \frac{2.5}{2} \times \frac{2.5}{2}=19.64 \mathrm{~cm}^{2}\)
(v) (b): Difference of volume of bigger cylinder to two small hemispherical ends = 190.93 - 8.18 = 182.75 cm³

(i) (b): Volume of the earth taken out
\(=\pi\left(\frac{7}{2}\right)^{2} \times 12=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12=462 \mathrm{~m}^{3}\)
(ii) (d): Area of the rectangular field = 30 x 15 = 450 m²
(iii) (a): Area of top of the pit = \(=\pi\left(\frac{7}{2}\right)^{2} =\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
\(=\frac{77}{2}=38.5 \mathrm{~m}^{2}\)
(iv) (d): Area of the remaining field = Area of rectangular field - area of top of pit
= 450 - 38.5 = 411.5 m²
(v) (c): The rise in the level of field = \(=\frac{462}{411.5}=1.12 \mathrm{~m}\)