3 cm, 11 cm

Let tangent touches the circles with centre O and O' at the points A and B respectively.

⇒ ㄥ1 = ㄥ2 = 90⁰
[Tangent makes 90⁰ angle with the radius at the point of contact] 
Through O' draw O'M I I AB meeting OA in M
⇒ O'M I I AB.Thus ABO'M is a rectangle with
AM = O'B = 4cm
In ΔOMO'
OM=OA-AM
= (12 — 4) cm = 8cm
and 00' = OP+ O'P= 12+4= 16cm
⇒ OO'² = 0M² + (O'M)²
⇒ 16² = 8² + (O'M)²
⇒ x² = 16²-8² = (16-8) (16 + 8)
= 8 x 24 = 8 x 8 x 3
\(⇒\ x=\sqrt{8\times8\times3}=8\sqrt3cm\)
Area ABO'O = area of rectangle ABO'M + area of AOMO'
=AM x O'M+\({1\over 2}\)x OM x O'M
\(=4\times8\sqrt3+{1\over 2}\times8\times8\sqrt3\)
\(=32\sqrt3+32\sqrt3=64\sqrt3cm\)
Now ΔOMO' is a right triangle with ㄥM=90⁰
\({O'M\over OO'}={8\sqrt3\over 16}=cos(ㄥ3)\)
\(⇒\ ㄥ3=30^0\ and\ {OM\over OO'}=cosㄥ4\)
\(⇒ {8\over 16}cos\angle4\)
⇒ ㄥ4=60⁰ ⇒ㄥAOP=60⁰
= angle of sector AOP
ㄥ3 = 30⁰ ⇒ ㄥMO'O = 30⁰ = 90⁰ + ㄥMO'O = 90⁰ + 30⁰
= 120⁰ = angle of sector BOP
Area of the portion enclosed between the circles and the tangents is
 = area of the trapezium ABO'O - 2 X area of the sectors
 = 64√3 cm² - (area of sector AOP + area of sector BO'P) 
=64√3 cm² - (area of sector AOP+area of sector BO'P)
\(=64\sqrt3cm^3-\left[{60^0\over 360^0}\times\pi\times12^2+{120^0\over 360^0}\times\pi\times 4^2\right]cm^2\)
\(=64\sqrt{3}cm^2-{\pi\over 6}\times8[18+2\times2]cm^2\)
\(=64\sqrt3cm^2-{4\pi\over 3}\times2[9+2]cm^2\)
\(=\left[64\sqrt3-{8\over 3}\times{22\over 7}\right]cm^2\)
\(=\left[64\times1.73-{88\over 3}\times{22\over 7}\right]cm^2\)
=[110.72-92.1904]=18.5296cm²

Side of the rhombus = radius of the circle
OP = OQ = PQ = x = OR = QR

⇒ ΔORQ and ΔOPQ are equilateral triangles
Area of the equilateral triangle OPQ=\({\sqrt3\over 4}(side)^2\)
Area of the equilateral ΔOPQ=\({\sqrt3\over 4}\times (x)^2\)
Area of the rhombus=2ar(ΔOPQ)
\(=2\times{\sqrt3\over 4}x^2={\sqrt3\over 2}x^2⇒{\sqrt3\over 2}x^2=32\sqrt3\)
⇒x²=64cm²⇒x=8cm

\(l=6\pi ,r=6cm,\theta =?\)
\(l=\frac { \theta \pi r }{ { 180 }^{ ° } } \Rightarrow 6\pi =\frac { \theta \times \pi \times 6 }{ { 180 }^{ ° } } \Rightarrow \theta ={ 180 }^{ ° }\)

Given, sector angle, \(\theta\)= 80°
and distance or radius, r = 16.5 km
\(\begin{aligned} \therefore \text { Area of sector } & =\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{80^{\circ}}{360^{\circ}} \times 3.14 \times(16.5)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{2 \times 3.14 \times 272.25}{9} \end{aligned}\)
\(\begin{aligned} =\frac{1709.73}{9} \end{aligned}\)
= 189.97 km²
which is the required area of the sea over which the ships are warned.

We know that area of sector of a circle =\(\frac{\theta}{360^{\circ}} \times \pi r^2\)
Given, radius of circle, r = 6 cm
and angle of sector, \(\theta\)= 60°
\(\therefore\) Area of sector of a circle \(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(6)^2=\frac{132}{7} \mathrm{~cm}^2\)

Given sector is OAPB 
Area of the sector \(=\frac{\theta}{360} \times \pi r^{2}\)
\(=\frac{30}{360} \times 3.14 \times 4 \times 4 \mathrm{~cm}^{2}\)
\(=\frac{12.56}{3} \mathrm{~cm}^{2}=4.19 \mathrm{~cm}^{2}(\text { approx. })\)
Area of the corresponding major sector
\(\begin{aligned} &=\pi r^{2}-\text { area of sector } \mathrm{OAPB}\\ &=(3.14 \times 16-4.19) \mathrm{cm}^{2}\\ &=46.05 \mathrm{~cm}^{2}=46.1 \mathrm{~cm}^{2}(\text { approx. }) \end{aligned}\)
Alternatively, area of the major sector  \(=\frac{(360-\theta)}{360} \times \pi r^{2}\)
\(\begin{array}{l} =\left(\frac{360-30}{360}\right) \times 3.14 \times 16 \mathrm{~cm}^{2} \\ =\frac{330}{360} \times 3.14 \times 16 \mathrm{~cm}^{2}=46.05 \mathrm{~cm}^{2} \\ =46.1 \mathrm{~cm}^{2}(\text { approx. }) \end{array}\)

(i) 2500 (ii) 19.8 m/s

Here PR² = PQ² + QR²
⇒ PR² = (42)² + (42)²
⇒PR=42√2m
⇒\(PO={42\sqrt2\over 2}=21\sqrt2m\)
Area of sector
\(={90^0\over 360^0}\times\pi(21\sqrt2)^2\)
\(={1\over4}\times{22\over 7}\times21\times21\times2\)
=693m²
Area of ΔPOS
\(={1\over 1}PO\times OS(∵ \ PO⊥OS)\)
\(={1\over 1}\times 21\sqrt2\times21\sqrt2=441m^2\)
∴ Area of one flower bed
=693-441=252m²
⇒ Area of two flower beds
=2 x 252 = 504m²

Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA
\(\begin{array}{l} =106+\frac{1}{2} \times 2 \pi r+106+\frac{1}{2} \times 2 \pi r \\ =212+\frac{1}{2} \times 2 \times \frac{22}{7} \times 30+\frac{1}{2} \times 2 \times \frac{22}{7} \times 30 \\ =212+2 \times \frac{22}{7} \times 30 \\ =212+\frac{1320}{7} \\ =\frac{1484+1320}{7}=\frac{2804}{7} m \end{array}\)
Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semi-circle HKI − Area of semi-circle BEC) + (Area of semi-circle GLJ − Area of semi-circle AFD)
\(\begin{array}{l} =106 \times 80-106 \times 60+\frac{1}{2} \times \frac{22}{7} \times(40)^{2}-\frac{1}{2} \times \frac{22}{7} \times(30)^{2}+\frac{1}{2} \times \frac{22}{7} \times(40)^{2}-\frac{1}{2} \times \frac{22}{7} \times(30)^{2} \\ =106(80-60)+\frac{22}{7} \times(40)^{2}-\frac{22}{7} \times(30)^{2} \\ =106(20)+\frac{22}{7}\left[(40)^{2}-(30)^{2}\right] \\ =2120+\frac{22}{7}(40-30)(40+30) \\ =2120+\left(\frac{22}{7}\right)(10)(70) \end{array}\)

=2120+2200
= 4320 m²
Therefore, the area of the track is 4320 m².

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

(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 = \(=\frac{462}{411.5}=1.12 \mathrm{~m}\)

(i) (c): Radius of circle representing red region
\(=\frac{22}{2}=11 \mathrm{~cm}\)  [\(\because\) Diameter = 22 cm (Given)]
(ii) (a): Area of red region = \(\pi r^{2}\)
\(=\frac{22}{7} \times 11 \times 11=380.28 \mathrm{~cm}^{2}\)
(iii) (b): Radius of circle formed by combining red and silver region = Radius of red region + width of silver sign
= (11 + 10.5) cm = 21.5 cm
(iv) (d): Area of silver region
= Area of combined region - Area of red region
\(=\frac{22}{7} \times 21.5 \times 21.5-380.28\)
= 1452.78 - 380.28 = 1072.50 crrr'
(v) (b): Area of circular path formed by two concentric circles = \(\pi\left(r_{1}^{2}-r_{2}^{2}\right)\) sq. units