428.75 cm²

10 cm, 4 cm

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
\(=\left(25\sqrt3-{25\pi\over 3}\right)cm^2\)

Given, chord AB subtends an angle 60° at the centre.
\(\therefore\) \(\angle\)AOB = 60° and OA = OB     [radii of circle]
\(\therefore\)  \(\angle\)OBA = \(\angle\)OAB = x                            [say]
[\(\because\) angles opposite to equal sides are also equal]
In \(\Delta\)AOB,
\(\angle\)AOB + \(\angle\)OBA + \(\angle\)OAB = 180°
[by angle sum property of triangle]
\(\Rightarrow\) 60° + x + x = 180° \(\Rightarrow\) 2x = 180° - 60°
\(\begin{array}{ll} \Rightarrow & x=\frac{120^{\circ}}{2}=60^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \angle O B A=\angle O A B=60^{\circ}=\angle A O B \end{array}\)
\(\therefore\)\(\Delta\)AOB is an equilateral triangle.
Now, area of \(\Delta A O B=\frac{\sqrt{3}}{4} \times(15)^2\)
\(\left[\because \text { area of equilateral triangle }=\frac{\sqrt{3}}{4}(\text { side })^2\right]\)
= 56.25 \(\times\)1.73 = 97.3125 cm²
Area of sector OACBO =\(\begin{aligned} & =\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{60^{\circ}}{360^{\circ}} \times 3.14 \times(15)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{3.14 \times 225}{6}=\frac{706.5}{6}=117.75 \mathrm{~cm}^2 \end{aligned}\)
\(\therefore\) Area of minor segment ACBDA
= Area of sector OACBO - Area of \(\Delta\)AOB
= 117.75 - 97.3125 = 20.4375 cm²
Now, area of major segment
= Area of circle - Area of minor segment
= \(\pi\)(15)² - 20.4375 = 3.14 \(\times\) 225 - 20.4375
= 706.5 - 20.4375 = 686.0625 cm²

Circumference of a circle = 44 cm
Sum of two sides of a square is 22 cm.
Radius of the circle = \(\frac { 44 }{ 2\times \frac { 22 }{ 7 } } =7\quad cm\)
Area of the circle = \(\pi { r }^{ 2 }=\frac { 22 }{ 7 } \times 7\times 7\)
= 154cm²

Perimeter of the circle = Perimeter of square 
Let side of square be x cm.
\(\begin{aligned} & 2 \pi \mathrm{r}=4 \mathrm{x} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad 2 \times \frac{22}{7} \times 77=4 x \\ \end{aligned}\)
\(\begin{aligned} & \therefore \quad x=\frac{2 \times 22 \times 11}{4}=121 \end{aligned}\)
Side of the square = 121 cm.

Perimeter of the sector = r + r + \(\frac { 2\pi r\theta }{ { 360 }^{ ° } } \)
= 10.5 + 10.5 + 2 \(\times \frac { 22 }{ 7 } \times \frac { 10.5\times 60 }{ 360 } \)
= 21 + 11 = 32 cm.

4 : 9

Perimeter of a quadrant = \(r+r+\frac { 1 }{ 4 } \times 2\pi r\)
\(=2r+\frac { 1 }{ 2 } \pi r=\frac { r }{ 2 } (\pi +4)\)

9.8 cm

\((i) \ \frac { 32 }{ 7 } { cm }^{ 2 } \\ (ii) \ \frac { 16 }{ 7 } { cm }^{ 2 }\)

Perimeter ofa sector OAB = 16.4cm
⇒ OA+AB + 0B 16.4cm
⇒ 5.2 cm + 5.2 cm + AB = 16.4cm
⇒ 10.4 cm + AB 16.4cm
⇒ AB = (16.4- 10.4)cm
Length of the arc = 6.0 cm
Area of the sector=\({lr\over 2}={6.0\times5.2\over2}\)
=6.0 x 2.3cm²
=15.6cm²

Let radius of the circle = R 
∴ Circumference = 2πR and area = πR²
New circumference
\(=2\pi R+\frac { 50 }{ 100 } 2\pi R=3\pi R\)
∴ New Radius \(=\frac { 3 }{ 2 } R\)
New area \(=\pi \times \left( \frac { 3 }{ 2 } R \right) ^{ 2 }=\frac { 9 }{ 4 } \pi { R }^{ 2 }\)
Increase in area\(=\frac { 9 }{ 4 } \pi { R }^{ 2 }-\pi { R }^{ 2 }=\frac { 5 }{ 4 } \pi { R }^{ 2 }\)
%increase in area\(=\frac { \frac { 5 }{ 4 } \pi { R }^{ 2 } }{ \pi { R }^{ 2 } } \times 100\)=125%

Diagonal = \(BD=\sqrt {DC^2+BC^2}\)
\(=\sqrt{12^2+5^2}=13cm\)
⇒ Diameter of circle = 13cm
⇒ Radius=\({13\over 2}cm=6.5cm\)
Area of circle=πr²=3.14 x (6.5)²
=3.14 x 42.25=132.665cm²
Area of rectangle ABCD=5 x 12=60cm²
∴ Area of shaded portion=132.665-60=72.665cm²

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

Short hand completes I round in 12 hours 
ஃ Distance covered by tip of short hand in 12 hours 
= circumference of circle = 2 x  \(22\over7\) x 4 = \(176\over7\) cm 
ஃ Distance covered in 2 days (48 hours) = 4 x cm x
Long hand completes 1 round in I hour 
ஃ Distance covered by the tip of long hand in 1 hour = 2 x \(22\over7\) x 6 =\(264\over7\) cm
Distance covered in 2 days = 48 x \(264\over7\) cm
Total distance covered = \({4\times176\over7}+{48\times264\over7}\) = 1910.86 cm

Time taken by minute hand to make one circle = 60 minutes.
\(\therefore\) Angle described in 60 minutes = \(360^o\)
Angle described in 5 minutes [i.e., from 7.00 a.m. to 7.05 a.m.] = \({360^o\over 60^o}\times 5=30^o\)
Radius of circle = length of minute hand = \(\sqrt {21}cm\)
Area swept = \({\theta \over 360^o}\times r^2={30^oover 360^o}\times {22\over 7}\times \sqrt{21}\times \sqrt {21}={1\over 12}\times {22\over 7}\times 21 = {11\over 2}\ cm^2=5.5 cm^2\)

Area of the sector = \({\theta \pi r^2\over 360^O}1={90^o\over 360^o}\times \pi \times (14)^2\) 
= \({1\over 4}\times {22\over 7}\times 14\times 14\)
= \(154\ cm^2\)

4 cm and 6 cm

Here ABCD is a square of side = 28 cm
There are four small squares each of side = 14 cm
Now, in each small square, we have a quadrant of radius 14 cm and a semicircle of radius 7 cm.
= 4 x {area of quadrant - area of semicircle}
= 4 x  \(\left\{ \frac { 1 }{ 4 } \times \pi \times 14\times 14-\frac { 1 }{ 2 } \times \pi \times 7\times 7 \right\} { cm }^{ 2 }\)

= \(4\pi\left\{ 49-\frac { 49 }{ 2 } \right\} \)cm²
= 4 x \({22\over7}\times{49\over2}\)
= 308 m²

Here, AC = 24 cm, BC = 10 cm and AOB is a diameter. 
Since angle in semicircle is a right angle i.e., \(\angle \)ACB = 90°
ஃ  AB = \(\sqrt{AC^2+BC^2}\)
= \(\sqrt{24^2+10^2}\) = \(\sqrt{576+100}\)
= \(\sqrt{676}\) = 26 cm
ஃ RAdius, area of circle = \(26\over2\) = 13 cm
Now, Area of semicircle
= Area of semicircle - Area of \(\triangle\)ABC
 = \(1\over2\)\(\pi\)r² - \(1\over2\) x BC x BC
= \(1\over2\) x 3.14 x 13 x 13 - \(1\over2\) x 24 x 10
= \(530.66\over2\) - 120
= 265.33 - 120 = 145.33 cm²

Diameter of Semicircle PSR, PAQ and QTR are 10 cm, 7 cm and 3 cm respectively.

Perimeter of shaded region
= length of arc PSR + length of arc PAQ + length of arc QTR
= [\(\pi\)(5) + \(\pi\)(3.5) + \(\pi\)(1.5)] cm
= x 10 cm = 3.14 x 10 cm = 31.4 cm

Length of semicircle ADC = \(\pi\times{4.2\over 2}cm=2.1\pi cm\)
Length of semicircle AEB\(=\pi\times{2.8\over2}cm=1.4\pi cm\)
Length of semicircle BFC=\(\pi\times{1.4\over2}cm=0.7\pi cm\)
∴ Perimeter of shaded region = 2.1π+1.4π+0.7π=4.2π cm=13.2cm