250

\(\because\) Area of circle = \(\pi { r }^{ 2 }\)
\(\therefore\) Area of the shaded region = \(\pi \)(2)² -  \(\pi \)(1)²
4\(\pi \) - \(\pi \) = 3\(\pi \) sq units.

82 cm

154 cm²

20 cm

Let side of a square = acm²
Area= 784cm
A.T.Q. a²=784
⇒ \(a=\sqrt{784}=28cm\)
If two circles touch externally then centres of both circles and point of contact are in a same line.
Sum of diameters of two circles = 28cm
⇒ diameter of one circle = 14cm
⇒ radius of one circle = 7 cm 
⇒ area of one circle=π(7)²
\(={22\over 7}\times7\times7=154cm^2\)
Area of four circles = 4 X 154 = 616 cm².
∴ Area of square sheet not covered by the circular plates
= 784-616 = 168 cm².

Radius of circle=35cm
ㄥAOB=90⁰
Area of sector=\(\pi r^2\theta\over 360^0\)
\(={22\over 7}\times35\times35\times{90^0\over 360^0}={3925\over 2}cm^2\)
Area of minor segment = area of sector - area of ΔOAB
\(={1925\over 2}={1\over 2}\times35\times35\)
\(={1925\over2}-{1225\over2}={700\over 2}=350cm^2\)
Area of major segment APB = area of circle - area minor segment
\(={22\over 7}\times35\times35-350=3500cm^2\)

d of the pond=17.5m

r of the pond=\({17.5\over 2}m\)
Radius of pond including path,
\(R={17.5\over2}+3.5={17.5+7.0\over2}\)
\(={24.5\over2}m\)
Area of the pat=π(R²-r²)
\(={22\over 7}\left[\left(24.5\over 2\right)^2-\left(17.5\over2\right)^2\right]\)
\(={22\over 7}\left({24.5\over2}-{17.5\over2}\right)\left({24.5\over2}+{17.5\over2}\right)\)
\(={22\over7}\times{7.0\over 2}\times{42\over 2}=231m^2\)

l=length of the arc=\({\theta\pi r\over 180^0}\)
\(={45^0\over 18060}\times{22\over 7}\times7={11\over 2}cm\)

Permiter of the sector
\(=(r+r+l)=\left(7+7+{11\over2}\right)cm\)
\(\left(14+{11\over2}\right)cm=\left(28+11\over 2\right)cm\)
\(={39\over 2}cm=19.5cm\)

Perimeter = length of major arc + 2r
\(={270^o\over 306^o} \times 2 \times + 2r= {3\over 2}\times {22\over 7}\times 42+2\times 42+2\times 42 = 198 + 84=284\ cm\)

Long hand makes 24 rounds in 24 hours
Short hand makes 2 round in 24 hours
Distance travelled by long hand in 24 rounds
= 24 \(\times\) 12\(\pi\)
= 288\(\pi\)
Distance travelled by short hand in 2 round
= 2 \(\times\) 8\(\pi\)
= 16\(\pi\)
Sum of the distances = 288\(\pi\) + 16\(\pi\) = 304\(\pi\)
= 304 \(\times\) 3.14 = 954.56 cm.

Angle in 1 minute = 6\(°\)
\(\theta\) = angle in 35 minutes
=35 \(\times\) 6 = 210\(°\)
\(\therefore\) Area swept by the minute hand
= Area of a sector
\(=\frac { \pi { r }^{ 2 }\theta }{ 360° } =\frac { 22 }{ 7 } \times \frac { 14\times 14\times 210 }{ 360 } \)
\(\\ =\frac { 1078 }{ 3 } { cm }^{ 2 }\)

Area of shaded region = Area of  \(\Delta\)PQR - 3 (area of sector)
\(=\frac { \sqrt { 3 } }{ 4 } { (side) }^{ 2 }-3\left[ \frac { \theta }{ 360° } \times \pi { r }^{ 2 } \right] \)
\(=\frac { \sqrt { 3 } }{ 4 } \times 8\times 8-3\left[ \frac { 60 }{ 360 } \times 3.14\times 4\times 4 \right] \)
\(=16\sqrt { 3 } -3.14\times 8=16\times 1.732-25.12\)
= 27.712 - 25.12 = 2.59 cm².

Required area = area of two semi-circles + area of rectangle
=area of one circle + area of rectangle
\(=\pi { r }^{ 2 }+(l+b)\)  (where r = radius of semi-circle and l and b are length and breadth of rectangle)
\(=\frac { 22 }{ 7 } \times 7\times 7+(16\times 14)\)
= 154 + 224
= 378 m²

Area of square ABCD = 14 \(\times\) 14 = 196 cm²
Radius of the semi-circle formed inside = 2 cm
Area of 4 semi-circles = 4 \(\times \frac { 1 }{ 2 } \pi { r }^{ 2 }\)
\(=4\times \frac { 1 }{ 2 } \times 3.14\times 2\times 2\)
= 25.12 cm²
Length of the side of square formed inside the semi-circles = 4 cm.
Area of the square = 4 \(\times\) 4 = 16 cm²
Area of the shaded region = Area of square ABCD - (Area of 4 semi-circles + Area of square)
= 196 - (25.12 + 16)
= 196 - 41.12
= 154.88 cm²

Let radius of the circle is r cm
Diameter = 2r cm
Circumference = 2\(\pi\)r
Circumference = Diameter + 16.8
\(\Rightarrow\) 2\(\pi\)r = 2r + 16.8
\(\Rightarrow\)2\(\left( \frac { 22 }{ 7 } \right) \) r = 2r + 16.8
\(\Rightarrow\)44r = 14r + 16.8 \(\times\) 7
\(\Rightarrow\)30r = 117.6
\(\Rightarrow \quad r=\frac { 117.6 }{ 30 } \)
\(\therefore\)r = 3.92cm

10 m

Perimeter of square and circle formed by wire will be equal.154 cm²

Triangle AED is right-angled at E.
ஃ AD² = AE² + ED²
⇒ AD = \(\sqrt { { 9 }^{ 2 }+{ 12 }^{ 2 } } \) = \(\sqrt { 81+144 } \)
= \(\sqrt{225}\) = 15 cm
ஃ  BC = AD = 15 cm = Diameter of semicircle
Now area of shaded portion
= Area of rectangle + Area of semicircle - Area of triangle AED
= \(\left[ 20\times 15+\frac { 1 }{ 2 } \pi { \left( \frac { 15 }{ 2 } \right) }^{ 2 }-\frac { 1 }{ 2 } \times 9\times 12 \right] { cm }^{ 2 }\)
= \(\left[ 300+\frac { 1 }{ 2 } \times 3.14\times { \left( \frac { 15 }{ 2 } \right) }^{ 2 }-54 \right] { cm }^{ 2 }\)
= [246 + 88.31] cm² = 334.31 cm² 

In ΔCAP,
ㄥCAB - 90⁰ 
∴ BC²=AC²+AB²
⇒ BC²=(24)²+(7)²
⇒ Diameter of circle =25cm 
⇒ Radius=\({25\over 2}cm\)
Area of ΔACB=\({1\over 2}AB\times AC={1\over 2}\times7\times24=84cm^2\)
∵ ㄥBOD=90⁰
∴ ㄥCOD=90⁰
Area quadrant COD=\({1\over 4}\pi r^2\)
\(={1\over 4}\times3.14\times{25\over 2}\times{25\over 2}cm^2\)
\(={1962.5\over 16}cm^2\)
Area of shaded part=area of circle - Area of ΔCAB - area of quadrant COD
\(=3.14\times{25\over 2}\times{25\over 2}-84-{1962.5\over 16}\)
\(={1962.5\over 4}-{1962.5\over 16}-84={5887.5\over 16}-82=283.968cm^2\)