Radius of the well = \(\frac { 7 }{ 2 } m=3.5\quad m\)
Volume of the earth taken out \(=\pi { \left( \frac { 7 }{ 2 } \right) }^{ 2 }\times 10\)
\(=\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times 10\)
= 385 m³
Area of the rectangular field = 20 \(\times\) 14
= 280 m²
Area of the top of the well = \(\\ \pi { \left( \frac { 7 }{ 2 } \right) }^{ 2 }\)
\(=\frac { 77 }{ 2 } { m }^{ 2 }\)
Area of the remaining field = \(280-\frac { 77 }{ 2 } \)
\(=\frac { 483 }{ 2 } { m }^{ 2 }\)
Let 'h' is the rise in the level of the field 
\(h=\frac { 385 }{ \frac { 483 }{ 2 } } =1.6m\) (approx)

Circumference of front wheel \(=2\times \frac { 22 }{ 7 } \times \frac { 80 }{ 2 } =\frac { 3520 }{ 14 } cm\)
Distance covered by front wheel in 800 revolutions \(=800\times \frac { 3520 }{ 14 } cm\)
Circumference of rear wheel = 2 \(\times\) \(\frac {22} {7}\) \(\times\) 100 cm
\(\therefore\) No. of revolutions made by rear wheel \(\\ =\frac { 800\times 3520\times 7 }{ 14\times 2\times 22\times 100 } =320\)

Let r cm be the radius of each circle.
Area of square - Area of 4 sectors = \(\frac { 24 }{ 7 } \)cm²
\(\Rightarrow \ { (2r) }^{ 2 }-4\left( \frac { 90° }{ 360° } \times \pi { r }^{ 2 } \right) =\frac { 24 }{ 7 } \)
\(\Rightarrow \ 4{ r }^{ 2 }-\frac { 22 }{ 7 } { r }^{ 2 }=\frac { 24 }{ 7 } \)
\(\Rightarrow \ \frac { 28{ r }^{ 2 }-22{ r }^{ 2 } }{ 7 } =\frac { 24 }{ 7 } \)
\(\Rightarrow\) 6r² = 24
\(\Rightarrow\) r² = 4
\(\Rightarrow\)  r = \(\pm \)2
\(\Rightarrow\) Radius of each circle is 2cm (r cannot be negative)

462 cm²

28.5 cm²

285.5 cm² , 28.5 cm²

150.9 cm²

\(45\frac { 5 }{ 6 } { cm }^{ 2 }\)

(i) 6250 (ii) 58.67 km/h

(i) 77 cm²
(ii) 44 cm

42 cm

\(\frac{6600}{7} cm\)

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

In right-angled triangle OAP,
\({OA\over OP}=cos\angle AOP\)

\(⇒\ \ {5\over 10}= cosㄥAOP\)
⇒ ㄥAOP=60⁰
Similarly, ㄥAOP=60⁰
ㄥAOB=60⁰+60⁰=120⁰
Length of arc (AB)=\({\theta\over 360^0}\times2\pi r\)
\(={120^0\over 360^0}\times2\times\pi\times5\)
\(={10\pi\over 3}cm\)
Length of the belt that is in the contact with the rim of pulley
= circumference of circle - length of arc AB
\(=2\pi\times5-{10\pi\over 3}={20\pi\over 3}cm\)
Area of sector
OAQB=\(={\theta\over 360^0}\times\pi r^2\)
\(={120^0\over 360^0}\times\pi5\times\times5cm^2={25\over 3}\pi cm^2\)
Area of quadrilateral OAPB
= 2 x area of triangle OAP
\(2\times{1\over 2}\times OA\times AP\)
\(=5\times5\sqrt3cm=25\sqrt3cm^2\)
Area of shaded region
\(=\left(25\sqrt3-{25\pi\over 3}\right)cm^2\)

Diameter of front wheel=80cm
∴ Radius=40cm
Distance covered in 1 revolution=\(2\pi r={2\times22\over 7}\times40={1760\over 7}cm\)
∴ Distance covered in 1400 revolution
\(={1400\times1760\over 7}=352000cm=3520m\)
Diameter of rear wheel=2m
∴ Radius of rear wheel=1m
Distance covered in 1 revolution=\(2\pi r={2\times\over 7}\times1m={44\over 7}m\)
∴ No. of revolutions to cover 3520m=\({3520\times7\over44}=560\)