Let internal radius of the pipe = x cm
Speed of water = 2.52 km/h = 2520 m/h
\(\therefore \) Volume of water that flows in half an hour = \(\frac { 1 }{ 2 } \pi r^{ h }h\)
\(=\frac { 1 }{ 2 } \pi \times \frac { x }{ 100 } \times \frac { x }{ 100 } \times 2520=\frac { 126\pi x^{ 2 } }{ 1000 } m^{ 3 }\)
Volume of water cylindrical tank
= \(\pi \times \frac { 40 }{ 100 } \times \frac { 40 }{ 100 } \times 3.15\quad m^{ 3 }\)
\(\Rightarrow \frac { 126\pi x^{ 2 } }{ 1000 } =\pi \times \frac { 40 }{ 100 } \times \frac { 40 }{ 100 } \times 3.15\)
\(x^{ 2 }=\frac { 40 }{ 100 } \times \frac { 40 }{ 100 } \times 3.15\times \frac { 1000 }{ 126 } \)
\(x^{ 2 }=4\Rightarrow x=2cm\)
\(\therefore \) Internal diameter = 4cm

Diameter of cylinder = 4m \(\Rightarrow \) Radius of cylinder = 2m
Height of cylinder = 2.1 m
Slant height of cone = 2.8 m
Area of canvas used = C.S.A of cylinder + C.S.A of cone
\(=2\pi rh+\pi rl=\pi r\left[ 2h+l \right] \)
\(=\frac { 22 }{ 7 } \times 2\left[ 2\times 2.1+2.8 \right] =\frac { 22 }{ 7 } \times 2\left[ 4.2+2.8 \right] \)
\(=\frac { 22 }{ 7 } \times 2\times 7=44m^{ 2 }\)
Cost of 1 m² of canvas = Rs 500
\(\therefore \) Total cost = Rs 500 X 44 = Rs 22000
Volume of air = volume of cone + volume of cylinder 
= \(\frac { 1 }{ 3 } \pi r^{ 2 }H+\pi r^{ 2 }h\left[ \frac { 1 }{ 3 } H+h \right] \)
Height of cone = \(\sqrt { l^{ 2 }-r^{ 2 } } =\sqrt { (2.8)^{ 2 }-2^{ 2 } } =\sqrt { 7.84-4 } =\sqrt { 3.84 } =1.95\)
\(=\frac { 22 }{ 7 } \times 4\times 2.75=34.57m^{ 2 }\)​​​​​​​

Given, solid is a combination of a cone and a hemisphere and it is placed into a right circular cylinder.

Height of the cylinder, h = 180 cm = 1.8 m
\(\left[\because 1 \mathrm{~cm}=\frac{1}{100} \mathrm{~m}\right]\)
Radius of the cylinder, r = 60 cm = 0.6 m
\(\therefore\) Volume of water filled in a right circular cylinder = \(\pi\) r² h
\(\begin{aligned} & =\frac{22}{7} \times 0.6 \times 0.6 \times 1.8 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{14.256}{7} \mathrm{~m}^3 \end{aligned}\)
For conical portion,
Height, h₁ = 120 cm = 1.2 m
Radius, r₁ = 60 cm = 0.6 m
For hemispherical portion,
Radius, r₂ = 60 cm = 0.6 m
\(\therefore\) Volume of the solid = Volume of the cone + Volume of the hemisphere
\( =\frac{1}{3}\times \pi r_{1}^{2}k_{1}+\frac{2}{3}\pi r_{2}^{1}\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times(0.6)^2 \times(1.2)+\frac{2}{3} \times \frac{22}{7} \times(0.6)^3 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{21} \times(0.6)^2[1.2+2 \times 0.6)=\frac{22}{21} \times 0.36(1.2+1.2) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{21} \times 0.36 \times 2.4=\frac{19.008}{21}=\frac{6.336}{7} \mathrm{~m}^3 \end{aligned}\)
Clearly, volume of water left in the cylinder = volume of water filled in a right circular cylinder - volume of the solid
\(=\frac{14.256}{7}-\frac{6.336}{7}=\frac{7.92}{7}=\)1.131429 m³ = 1.131 m³

Let the side of the equilateral triangle be a.
Area of equilateral triangle = 17320.5 cm²
\(\sqrt3/4(s)^2= 17320.5\)
\(1.7320/4a^2 = 17320.5\)
a² = 4 x 10000
a = 200 cm
Each sector is of measure 60°
Area of sector ADEF \(=\frac{60^{\circ}}{360^{\circ}} \times \pi \times r^{2}\)
\(\begin{array}{l} =\frac{1}{6} \times \pi \times(100)^{2} \\ =\frac{3.14 \times 10000}{6} \\ =\frac{15700}{3} \mathrm{~cm}^{2} \end{array}\)
Area of shaded region = Area of equilateral triangle − 3 × Area of each sector
\(=17320.5-3 \times \frac{15700}{3}\)
= 17320.5-15700 = 1620.5 cm²

It can be observed that these designs are segments of the circle.
Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute 360º/6 = 60º at the centre of the circle.
In ΔOAB,
∠OAB = ∠OBA (As OA = OB)
∠AOB = 60°
∠OAB + ∠OBA + ∠AOB = 180°
2∠OAB = 180° − 60° = 120°
∠OAB = 60°
Therefore, ΔOAB is an equilateral triangle.
Area of ΔOAB  \(=\frac{\sqrt{3}}{4} \times(\text { side })^{2}\)
\(=\frac{\sqrt{3}}{4} \times(28)^{2}=196 \sqrt{3}=196 \times 1.7=333.2 \mathrm{~cm}^{2}\)
Area of sector OAPB  \(=\frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}\)
\(=\frac{1}{6} \times \frac{22}{7} \times 28 \times 28 \)
\(=\frac{1232}{3} \mathrm{~cm}^{2}\)
Area of segment APB = Area of sector OAPB − Area of ΔOAB
\(=\left(\frac{1232}{3}-333.2\right) \mathrm{cm}^{2}\)
Therefore are of designs \(=6 \times\left(\frac{1232}{3}-333.2\right) \mathrm{cm}^{2}\)
\(=(2464-1999.2) c m^{2}\)
= 464.8 cm²
Cost of making 1 cm² designs = Rs 0.35
Cost of making 464.76 cm² designs = 464.8 x  0.35 = Rs 162.68
Therefore, the cost of making such designs is Rs 162.68.

Given, side of a square = 15 m
\(\therefore\) Area of square = (15)² = 225 m² [\(\because\)area of square = (side)²]
also given, length of rope = 5 m
\(\therefore\) Radius of arc = 5 m
(i) Area of the field graze by the horse,
\(A_1=\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(5)^2\)
[\(\because\) each angle of a square is 90°]
\(=\frac{3.14 \times 25}{4}=\frac{78.5}{4}=19.625 \mathrm{~cm}^2\)
(ii) If length of rope = 10 m = r₁ (say)
then, area of the field graze by the horse,
\(\begin{aligned} A_2 & =\frac{\theta}{360^{\circ}} \times \pi r_1^2=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{3.14 \times 100}{4}=\frac{314}{4}=78.5 \mathrm{~cm}^2 \end{aligned}\)
\(\therefore\) Required increase in the grazing area
= A₂ - A₁
= 78.5 - 19.625 = 58.875 cm²

Let ABCD be a parallelogram circumscribing a circle.
To prove ABCD is a rhombus.
i.e. to prove AB = BC = CD= DA
Proof We know that the tangents to circle from an external point are equal in length.

\(\therefore\)   AM = AP, BM = BN, CO = CN and DO = DP
On adding all above equations, we get
(AM + BM) + (CO + DO) = AP + BN + CN + DP
\(\Rightarrow\) AB + CD = (AP + PD) + (BN + NC)
= AD + BC                          ....(i)
Given, ABCD is a parallelogram.
\(\therefore\) AB = CD and BC = AD   ....(ii)
[\(\because\) opposite sides of a parallelogram are equal]
Then, from Eq. (i), we get
2 AB = 2BC
\(\Rightarrow\)   AB = BC  ...(iii)
From Eqs. (ii) and (iii), we get
AB = BC = CD = DA
\(\Rightarrow\) ABCD is a rhombus.
Hence, the parallelogram circumscribing a circle is a rhombus.
Hence proved.

Let C₁ and C₂ be rwo circles of radii, r₁=3 cm and r₂ = 5 cm and having common centre O.
Now, let AB be the chord of circle C₂ such that it touches the circle C₁ at point D. Clearly, AB be the tangent to the circle C₁, at point D.

\(\therefore\) OD \(\perp\)AB   [\(\because\) radius is perpendicular to the tangent at the point of contact]
\(\Rightarrow\) AD = BD [ \(\because\) perpendicular from centre to the chord bisect the chord]
Now, in right angled \(\Delta\)ODB,
OB² = OD² + DB²   [by Pythagoras theorem]
\(\Rightarrow\) 5² = 3² + DB² \(\Rightarrow\) DB² = 25 - 9 = 16
\(\Rightarrow\) DB = 4 cm [ taking positive square root]
\(\therefore\) Length of chord = AB = 2 AD = 2 \(\times\)4 = 8 cm

OP = Radius of the circle OA = 5 cm; AP = 4 cm
o [RAdius and tangent are ⊥ ar]
OA² = AP² + OP²   [By pythagoras theorem]
5² = 4² + OP²
⇒ 25 = 16 + OP²  ⇒ 25 - 16 = OP² ⇒ 9 = OP² ⇒ OP = \(\sqrt9\) = 3
Radius = 3 cm

(i) (b): Let the radius of outer most circle be R.
Outer most circumference = 308 m [Given]
\(\Rightarrow 2 \pi R=308 \Rightarrow 2 \times \frac{22}{7} \times R=308\)
\(\Rightarrow \quad R=\frac{308 \times 7}{2 \times 22}=49 \mathrm{~m}\)
(ii) (c): Let the radius of inner most circle be r
Inner most circumference = 264 m [Given]
\(\Rightarrow 2 \pi r=264 \)
\(\Rightarrow \quad 2 \times \frac{22}{7} \times r=264 \Rightarrow r=\frac{264 \times 7}{2 \times 22}=42 \mathrm{~m}\)
(iii) (a): Width of the track = Radius of outer most track - Radius of inner most track = 49 - 42 = 7 m
(iv) (d): Area of the race track = Area of outer circle - Area of inner circle
\(=\pi\left(R^{2}-r^{2}\right)=\pi\left[(49)^{2}-(42)^{2}\right] \)
\(v=\frac{22}{7}[2401-1764]=2002 \mathrm{~m}^{2}\)
(v) (b): Cost of painting the whole race track
= Rs (6 x 2002) = Rs 12012.