Perimeter of (semi-circular arc) + diameter
 = 72 cm
\(\pi\)r + 2r = 72 cm
\(\Rightarrow \quad r\left[ \frac { 22 }{ 7 } +2 \right] =72\quad cm\)
\(\Rightarrow \quad r\left[ \frac { 22+14 }{ 7 } \right] =72\)
\(\\ \Rightarrow \quad \frac { 36 }{ 7 } r=72\)
r = 14 cm
\(\therefore \quad Area\quad of\quad protractor=\frac { 1 }{ 2 } \pi { r }^{ 2 }=\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 14\times 14\)
= 308 cm².

Perimeter = \(\pi r+2r\)
\(=(\pi +2)r=36\)
\(\Rightarrow \quad \frac { 36 }{ 7 } r=36\Rightarrow r=7\)
\(\therefore\) Diameter = 14 cm.

\(\therefore \)  Circulference of a circle = \(2\pi r\)
\(=2\times \frac { 22 }{ 7 } \times 5=\frac { 220 }{ 7 } =31.43\quad cm\)

\(75\frac { 3 }{ 7 } { cm }^{ 2 }\)

7392 cm²

266 cm²

Here, time duration between 9 a.m. to 9.35 a.m is 35 minutes.

Angle made by minute hand in 60 minutes=360⁰
∴ Angle made by minute hand in 1 minute \(={360^0\over 60}=6^0\)
∴ Angle made by minute hand in 35 minutes = 6⁰ x 35=210⁰
Length of minute hand=12cm
∴ Area described by the minute hand between 9 a.m. and 9.35 a.m = Area of sector AOB
\(={210^0\over 360^0}\times{22\over 7}\times12\times12\)
=264cm²

17.6 sq.cm (a) Pollution is very harmful
(1) Affects human life and surroundings.
(2) Causes diseases.
(3) Causes ecological unbalance.
(i) Save fuel use resources judiciously.
(ii) Use biodegradable and ecofriendly material
(iii) More plantation of trees.

535.71 cm²

Here, radius of the circle (r)=21cm
Angle of the sector (θ)=120⁰

∴ Area of the sector=\({\theta\over 360^0}\times\pi r^2\)
\(={12^0\over 360^0}\times{22\times7}\times21\times21\)
=462cm²
Area of major sector
\(={360^0-120^0\over 360 ^0}\times{22\over 7}\times21\times21\)
\(={240^0\over 360^0}\times22\times3\times21\)
=924cm²
Thus, the required difference=924-462
=462cm²

Here, radius of the circular pond
\(r={17.5\over 2}=8.75m\)
Width of the path=2m

∴ Radius of the outer circle (R)=8.75+2=10.75
Area of the path=πR²-πr²
=π​​​​​​​(R+r)(R-r)
\(={22\over 7}(10.75+8.75)(1.75-8.75)\)
\(={22\over 7}\times19.5\times2=122.57m^2\)
Cost of constructing the path at the rate of Rs.25 per m²
=Rs.25 x 122.57
=Rs.3064.25

462 cm²

Here, \({PQ\over PR}=sin30^0\)
\(⇒\ {PQ\over 10}={1\over 2}⇒\ PQ=5cm\)
And \({QR\over PR}=cos30^0\)
\(⇒\ {QR\over 10}={\sqrt3\over 2}⇒ QR=5\sqrt3cm\)

∴ Area of ΔPQR\(={1\over 2}\times PQ\times QR\)
\(={2\over 2}\times5\times5\sqrt3\)
\(={25\over 2}\times1.73={43.25\over 2}\)
=21.625cm²

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will rotate = 360^@/60xx5 = 30^@
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area of sector of angle θ = \(\frac{\theta}{360^{\circ}} \times \pi r^{2}\)
Area of sector of 30° \(=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14\)
\(\begin{array}{l} =\frac{22}{12} \times 2 \times 14 \\ =\frac{11 \times 14}{3} \end{array}\)
=154/3 cm²
Therefore, the area swept by the minute hand in 5 minutes is 154/3 cm²

Let, Radius of the circle = r
\(\therefore\)  Circumference of the circle = \(2\pi r\)
A.T.Q \(2\pi r\) = 22 cm
\(\Rightarrow\) \(2\times {22\over 7}\times r=22\ \ \ \Rightarrow\ \ \ \ r={{22\times 7\over 2\times 22}}={7\over 2}cm\)
Area of quadrant of the circle \(={\pi r^2\theta\over 360^o}={22\over 7}\times {7\times 7\over 2\times 2}\times {90^o\over 360^o}={22\times 7\over 2\times 2\times 4}={77\over 8}cm^2\)