Areas Related To Cricles - Class 10 Maths - Extra Questions

We have,

Let the radius of circle $$r=4\,cm.$$

Angle $$\theta ={{30}^{o}}$$

Area of sector of circle

$$=\dfrac{\theta }{{{360}^{o}}}\times \pi {{r}^{2}}$$

$$=\dfrac{{{30}^{o}}}{{{360}^{o}}}\times \pi \times 4\times 4$$

$$=\dfrac{\pi }{12}\times 16$$

$$=\dfrac{4\pi }{3}$$

$$=\dfrac{4}{3}\times 3.14$$

$$=4.186\,c{{m}^{2}}$$

Area of major sector = Area of circle-area of minor sector

$$=\pi {{r}^{2}}-4.186$$

$$=3.14\times 4\times 4-4.186$$

$$=3.14\times 16-4.186$$

$$=50.24\,cm{{.}^{2}}$$

Area of a circle of radius $$r$$ $$=\pi {r}^{2} = 1386  { cm }^{ 2 }$$

$$\dfrac { 22 }{ 7 } \times { r }^{ 2 }=1386 { cm }^{ 2 }$$
$${ r }^{ 2 }=441 { cm }^{ 2 }$$
$$r = 21 cm$$

Circumference of a circle of radius $$r$$ $$= 2\pi r$$
Circumference of circle of radius $$21 cm = 2 \times \dfrac {22}{7} \times 21 = 132  cm$$

Length of an arc subtending an angle $$\theta  = \dfrac { \theta  }{ 360 } \times 2\pi R$$ where R is the radius of the circle.

So, $$6.05  = \dfrac { \theta }{ 360 } \times 2 \times \dfrac {22}{7} \times 5.5$$

Hence, Angle subtended $$= {63}^{0}$$

Length of an arc subtending an angle $$\theta  = \dfrac { \theta  }{ 360 } \times 2\pi R$$ where R is the radius of the circle.

So, $$78.50 = \dfrac { \theta }{ 360 } \times 2 \times \dfrac {22}{7} \times 20$$

Hence, Angle subtended $$= {225}^{0}$$

$${\textbf{Step - 1: Calculate circumference of figure (a)}}{\text{.}}$$

                  $${\text{For part (a) divide the figure in two parts one is circumference of semicircular path }}$$

                  $${\text{and another is the straight line , diameter part}}{\text{.}}$$

                  $${\text{So here radius}} = \dfrac{{{\text{Diameter}}}}{2}.$$

                  $$= \dfrac{{2.8}}{2}1.4{\text{ cm}}{\text{.}}$$

                  $${\text{So , circumference of a half circle}}$$

                  $$= \dfrac{1}{2} \times {\text{circumference of a circle}}$$

                  $$= \dfrac{1}{2} \times 2\pi r{\text{ cm}}{\text{.}}$$

                  $$= \pi  \times 1.4{\text{ cm}}{\text{.}}$$

                  $$= \dfrac{{22}}{7} \times 1.4{\text{ cm}}{\text{.}}$$

                  $$= 4.4{\text{ cm}}{\text{.}}$$

                  $${\text{So circumference of figure (a)}} = 4.4 + 2.8{\text{ cm}}{\text{.}}$$

$${\textbf{Step - 2: Calculate circumference of figure (b)}} = 7.2{\text{ cm}}{\text{.}}$$

                    $${\text{Draw the figure clearly and add name to it}}{\text{.}}$$

                    $${\text{Like here as given AB}} = {\text{CD}} = 2.8{\text{ cm}}{\text{.}}$$

                    $${\text{AD}} = {\text{BC}} = 1.5{\text{ cm}}{\text{.}}$$

                    $${\text{So here total distance that is moved by ant }}$$

                    $$= {\text{AB}} + {\text{BC}} + {\text{AD}} + {\text{ circumference of half circle}}{\text{.}}$$

                    $$= 1.8 + 1.5 + 1.5 + \dfrac{{2\pi r}}{2}{\text{ cm}}{\text{.}}$$

                    $$\left[ {{\text{For circumference of half circle }}} \right]$$

                    $${\text{r}} = \dfrac{{2.8}}{2} = 1.4{\text{ cm}}{\text{.}}$$

                    $$= 2.8 + 3 + \pi  \times 14{\text{ cm}}{\text{.}}$$

                    $$= 5.8 + \dfrac{{22}}{7} \times 14{\text{ cm}}{\text{.}}$$

                    $$= 5.8 + 4.4{\text{ cm}}{\text{.}}$$

                   $$= 10.2{\text{cm}}{\text{.}}$$

                   $${\text{For figure (b) perimeter is 10}}{\text{.2 cm}}{\text{.}}$$

$${\textbf{Step - 3: Calculate circumference of (c)}}$$

                  $${\text{Here the circumference will of consist of both the lines and circumference of half circle}}{\text{.}}$$

                    $${\text{So , circumference}} = 2 + 2 + \dfrac{{2\pi r}}{2}{\text{ cm}}{\text{.}}$$

                    $$= 4 + \dfrac{{22}}{7} \times 1.4{\text{ cm}}{\text{.}}$$

                    $$= 4 + 4.4{\text{ cm}}{\text{.}}$$

                    $$= 8.4{\text{ cm}}{\text{.}}$$

$${\textbf{Step - 4: Find which is longest}}{\text{.}}$$

                    $${\text{So , for figure (a) , (b) and (c) only (b) has longest circumference of 10}}{\text{.2 cm}}$$  

                  $${\text{. So ant will have to take longer round}}{\text{.}}$$

$${\textbf{Hence, for figure (b) the ant have to take longer round}}{\text{.}}$$

Given:

Radii of inner circle $$= 21\ cm = r$$

Radii of outer circle $$= 42\ cm = R$$

$$\angle AOB = \theta = 60^o$$

Area of ring $$= \pi (R^2-r^2)$$

Area of sector $$= \dfrac{\theta}{360}\pi r^2$$

The area of shaded region $$=$$ Area of ring $$–$$ Area of $$ABDC$$

$$=$$ Area of ring $$–$$ [Area of sector of outer circle $$–$$ Area of sector of inner circle]

$$=\pi(R^2-r^2)-\dfrac{\pi(R^2-r^2)}{1}\times \dfrac{\theta}{360}$$

$$=\pi[R^2-r^2]\left(1-\dfrac{\theta}{360}\right)$$

$$= \dfrac{22}{7}[42^2-21^2]\left(1-\dfrac{60}{360}\right)$$

$$= 3465$$ $$cm^2$$

In $$\Delta ABC$$ ,

Angle in a semicircle, $$\angle C = 90^o$$

Therefore, by Pythagoras theorem

$$BC^2+AC^2=AB^2$$

$$BC^2+12^2=13^2$$

$$BC^2=169-144=25$$

$$BC = 5$$

Area of shaded region = Area of semicircle – Area of triangle

= $$\dfrac{1}{2}\pi r^2-\dfrac{1}{2}bh$$

= $$\dfrac{1}{2}\times 3.14 \times \left(\dfrac{13}{2}\right)^2-\dfrac{1}{2}\times 5 \times 12$$

= $$\dfrac{3.14 \times 13\times 13}{2\times 4}-30$$

Area of shaded region $$= 36.39$$ cm $$^2.$$

Given: Radius of inner circle $$= 7$$ cm

Radius of outer circle $$= 14$$ cm

Angle made by sector = $$40^o$$

Area of sector OAC = $$\dfrac{40^o}{360^o}\times \pi r^2$$

$$=\dfrac{1}{9}\times \times \dfrac{22}{7} \times 14^2=68.44$$

Area of sector OBD = $$\dfrac{40^o}{360^o}\times \pi r^2$$

$$=\dfrac{1}{9}\times \times \dfrac{22}{7} \times 7^2=17.11$$

Area of the small region $$ABDC$$
$$=$$ Area of the small sector $$OAC$$ $$–$$ Area of the small sector $$OBD$$ $$=(68.44-17.11)\,cm^2$$

Area of shaded region $$ABDC =$$ Area of outer circle   $$-$$  Area of inner circle $$-$$ area of small region $$ABCD$$

Area of shaded region $$ABDC = \pi\times{14}^2 - \pi\times7^2 - (68.44-17.11)\,cm^2$$  

Area of shaded region $$ABDC = 410.484$$ cm $$^2$$

Length of any arc is given by

$$l = r\theta$$

Given, $$l = 11\ cm$$

and $$\theta = 30^{\circ} = \dfrac{\pi}{6}$$

$$\implies r = \dfrac{11}{22} \times 7 \times 6 \quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$$

$$= 21 \, cm$$

Hence, the length of the pendulum is $$21\ cm$$ .

the formula is given as:

$$\dfrac{\theta}{360^{\circ}}\times 2\pi r=l$$

where, $$r=$$ radius of the arc

            $$l=$$ length of the arc

            $$\theta=$$ central angle

Given that $$r=10.5\ cm$$ and $$l=22\ cm$$ .

by substituting the values in the formula we get,

$$\Rightarrow \dfrac{\theta}{360^{\circ}}\times 2\times \dfrac{22}{7}\times 10.5=22$$

$$\Rightarrow \theta=22\times 360^{\circ} \times \dfrac{1}{2}\times \dfrac{7}{22}\times \dfrac{1}{10.5}$$

$$\therefore \theta =120^{\circ}$$

Given that $$r=14\ cm , area =77\ cm^2$$

Area of the sector $$= \dfrac{\theta}{360^{\circ}}\times \pi r^2$$

                          $$77 = \dfrac{\theta}{360^{\circ}}\times \dfrac{22}{7}\times 14\times 14$$

                        $$\therefore \theta =\dfrac{77\times 360^{\circ} \times 7}{22\times 14\times 14}=45^{\circ}$$

$$BC = 38$$ cm, $$CD = 25$$ cm and $$BP=27$$ cm

$$\therefore$$ From the figure, 

$$BP = BQ =27$$ cm [Tangents from an external point are equal]

Also, $$BC = 38$$ cm

$$\Rightarrow BQ + QC=38$$

$$\Rightarrow 27 + QC=38$$

$$\Rightarrow QC = 38-27$$

$$\Rightarrow QC = 11$$ cm

$$\therefore QC = 11$$ cm $$= CR$$          [ Tangents from an external point are equal ]

$$CD = 25$$ cm

$$CR + RD=25$$

$$\Rightarrow 11+ RD = 25$$

$$\Rightarrow RD = 14$$ cm

Also, 

$$RD = DS = 14$$ cm   [ Tangents from an external point are equal ]

$$\therefore OR$$ and $$OS$$ are radii of the circle.

From tangents $$R$$ and $$S$$ ,  $$\angle ORD = \angle OSD = 90^{\circ}$$

Now, $$ORDS$$ is a square.

$$\therefore OR = DS = 14$$ cm

$$diameter\>of\>circle\>=14cm\\\therefore\>radius\>=7cm\\Area\>of\>square\>design\>=14\cdot\>14\\=196cm^2\\Area\>of\>circular\>parts\>\\=4\cdot (\frac{\pi r^2}{2})\\=2\cdot(\frac{22}{7})\cdot\>7^2\\=308cm^2\\\therefore\>Total\>area\>=504cm^2$$

In triangle $$ABC$$ , applying Pythagoras theorem,

$$A{B^2} = A{C^2} + B{C^2}$$

$$A{B^2} = {24^2} + {10^2}$$

$$AB^{2}= 576 + 100$$

$$AB^{2}= 676$$

$$AB = 26$$

Area of the triangle $$ABC=\dfrac{1}{2}\times BC \times AC$$

$$= \dfrac{1}{2} \times 24 \times 10$$

$$= 120$$

Area of the semi circle $$=\dfrac{{\pi {r^2}}}{2}$$

here $$r=\dfrac{AB}{2}$$

$$=\dfrac{26}{2}$$

$$=13$$

So,

Area of semi circle $$=\begin{array}{l}\dfrac{{\pi {{\left( {13} \right)}^2}}}{2}\ = \dfrac{{22}}{7} \times 169 \times \dfrac{1}{2}\ = 265.57\end{array}$$

Area of shaded part = Area of semicircle - Area of triangle ABC

$$=265.57-120$$

$$=145.57$$

Thus area of shaded part is $$145.57\;\text{units}^{2}$$

The Radius of smaller circle is $$21cm$$

The Radius of larger circle is $$42cm$$

The area of Ring is $$\pi(R^2-r^2)\\\pi(1764-441)=1323\times \dfrac {22}7=4158$$

The major angle $$\angle AOB =360-60=300$$

The area of sector in Ring is given as $$\dfrac{300}{360}\times 4158 \\3465cm^2$$

$${\textbf{Step 1: }}{\mathbf{ Find}}\;{\mathbf{angle}}\;{\mathbf{protruded}}\;{\mathbf{by}}\;{\mathbf{each}}\;{\mathbf{rib}}$$

                 $${\text{No}}.{\text{of ribs}} =8.$$

                 $${\text{The center of an umbrella protrudes an angle of }}360^\circ$$

                 $${\text{Hence}},{\text{one rib protrudes an angle of}} = {\text{}}\dfrac{{{\text{}}360}}{8}$$

                 $${\text{}} = {\text{}}45^\circ$$

$${\textbf{Step 2: }}{\mathbf {Find}}\;{\mathbf{the}}\;{\mathbf{area}}\;{\mathbf{consecutive}}\;{\mathbf{between}}\;{\mathbf{two}}\;{\mathbf{ribs}}$$

                $${\text{As we know that area of the sector = }  \dfrac{\theta}{360}\times\pi \times{{\text{r}}^2}}$$

                 $$\therefore {\text{Area between two consecutive ribs}}$$

                 $$= \dfrac{45}{360}\times\pi \times{{\text{r}}^2}$$

                 $$= \dfrac{45}{360}\times \pi \times{{\text{45}}^2}$$

                 $$= \dfrac{{22275}}{{28}}$$

$${\textbf{Hence,}}{\textbf{ the area}}\;{\mathbf{between}}\;{\mathbf{two}}{\textbf{ consecutive}}\;{\textbf{ribs is}} \; \mathbf{\dfrac{{22275}}{{28}}}\;{\mathbf{c}}{{\mathbf{m}}^2}$$

The diameter of circle is $$35mm$$

The circumference is $$2\pi r\\ 2\times \dfrac{22}7\times 35=220mm$$

The 5 diamteres is of length $$5\times 35=175mm$$

Total wire is $$220+175=395mm$$

The circel is divides into 10 equal parts so the angle of sector is $$\dfrac{360}{10}=36^{\circ}$$

The afrea of secto is given as $$\dfrac{36}{360}\times \pi r^2\\\dfrac 1{10}\times {22}{7}\times 35\times 35=385cm^2$$

$$\angle{POQ} = \theta = 90^\circ$$
Let the radius of the circle be r cm.
Area of the shaded region = Area of the segment $$PRQ = 114\, cm^2$$

$$[\therefore r^2 \left( \dfrac{\pi\theta}{360° } - \dfrac{\sin\theta}{2} \right) = 114]$$

$$[ \Rightarrow r^2 \left( \dfrac{3 . 14 \times 90° }{360°  } - \dfrac{\sin90° }{2} \right) = 114]$$

$$[ \Rightarrow r^2 \left( \dfrac{3 . 14}{4} - \dfrac{1}{2} \right) = 114]$$

$$[ \Rightarrow r^2 \times \left( 0 . 785 - 0 . 5 \right) = 114]$$

$$[ \Rightarrow r = \sqrt{\dfrac{114}{0 . 285}}]$$

$$[ \Rightarrow r = \sqrt{400} = 20\,cm]$$

Thus, the radius of the circle is $$20\,cm$$ .