Three identical balls fit snugly into a cylindrical can. The radius of the spheres is equal to the radius of the can and the balls just touch the -Maths 9th

Three identical balls fit snugly into a cylindrical can. The radius of the spheres is equal to the radius of the can and the balls just touch the -Maths 9th
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A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th

Description : A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th

Last Answer : Let radius of iron rod = r ∴∴ Height = 8r ∴∴ Volume of iron rod =π×(r)2×8r⇒8πr3=π×(r)2×8r⇒8πr3 ⇒⇒ Radius of spherical ball =r2=r2 Volume of spherical ball =43π(r2)3=43π(r2)3 Let n balls are casted ∴n×43π(r38)=8πr3∴n×43π(r38)=8πr3 ⇒n6=8⇒n=48

A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

Description : A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

Last Answer : Let radius of cylindrical rod =r ⇒ height =4r Volume of cylindrical rod =πr2h =πr2(4r) =4πr3 Volume of spherical balls of radius r=34​πr3 No. of balls =34​πr34πr3​=

A spherical metal of radius 10 cm is melted and made into 1000 smaller spheres of equal sizes. In this process the surface area of the -Maths 9th

Description : A spherical metal of radius 10 cm is melted and made into 1000 smaller spheres of equal sizes. In this process the surface area of the -Maths 9th

Last Answer : Option (C) is correct. Solution: Let the radius of the small spheres be r' cm. Volume of metal remains the same in both cases. So, vol of the spherical metal of radius 10 cm = total ... Total Surface area of 1000 smaller spheres: 1000*4π12 = 4000π Hence, the surface area increased by 10 times.

Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Last Answer : Volume of rectangular box=lbh=16(64)=1024cm3 Volume of sphere=34​πr3=33.5238cm3 16 sphere=16(33.5238)=536.3808 Volume of liquid=1024−536.3808=488cm3

Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Last Answer : According to question find the volume of this liquid.

A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Last Answer : NEED ANSWER

A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Last Answer : Solution of the question

The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th

Description : The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th

Last Answer : answer:

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

Three circles of radius a, b, c touch each other externally. The area of the triangle formed by joining their centres is: -Maths 9th

Description : Three circles of radius a, b, c touch each other externally. The area of the triangle formed by joining their centres is: -Maths 9th

Last Answer : (a) \(\sqrt{(a+b+c).a.b.c}\)As shown in the figure, AB = a + b, BC = b + c, CA = a + c∴ Area of ΔABC = \(\sqrt{s(s-AB)(s-BC)(s-CA)}\)where, s = \(rac{1}{2}\) (AB + BC + CA)= \(rac{a+b+b+c+c+a}{2}\) = a + b + ... (\sqrt{(a+b+c)[(a+b+c)-(a+b)][(a+b+c)-(b+c)][(a+b+c)-(c+a)]}\)= \(\sqrt{(a+b+c).a.b.c}\)

Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. -Maths 9th

Description : Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. -Maths 9th

Last Answer : NEED ANSWER

Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. -Maths 9th

Description : Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. -Maths 9th

Last Answer : Solution of this question

The volume of two spheres are in the -Maths 9th

Description : The volume of two spheres are in the -Maths 9th

Last Answer : Let r1 and r2 be the radii of two spheres . Then, the ratio of their volumes is given by 4/3πr13/4/3πr23 = 64/27 (r1/r2)3 = (4/3)3 ⇒ r1/r2 = 4/3 Now, ratios of surface areas of two spheres = 4/3πr12/4/3πr22 = (r1/r2)2 = (4/3)2 = 16/9 ∴ Required ratio = 16 : 9

Twenty-seven solid iron spheres, -Maths 9th

Description : Twenty-seven solid iron spheres, -Maths 9th

Last Answer : (i) Volume of 27 solid sphere, each of radius, r = 27 x 4/3 πr3 = 36 πr3 According to statement, Volume of sphere of radius r' = Volume of 27 solid spheres ⇒ 4/3 π(r'3 ) = 36 πr3 ⇒ (r')3 = 27r3 = (3r)3 ⇒ r' = 3r ( ... πr'2 = 4 π(3r)2 = 36 πr2 ∴ S/S' = 4 πr2 /36 πr2 = 1/9 ⇒ S : S' = 1 : 9

Two solid spheres made of the same metal -Maths 9th

Description : Two solid spheres made of the same metal -Maths 9th

Last Answer : Let r and R be the radii of the smaller and larger spheres respectively. We have, r = 5/2 cm Volume of the smaller sphere = 4/3πr3 = 4/3π(5/2)3 cm3 = 4/3 x π x 125/8 cm3 Density of metal = Mass/Volume = 740/4/3 x 125/8π ... ⇒ R3 = 5920 x 125/740 x 8 = 125 ⇒ R3 = 53 ⇒ R = 5 cm

The volumes of two spheres are in the ratio 64 : 27. Find the difference of their surface areas, if the sum of their radii is 7 cm. -Maths 9th

Description : The volumes of two spheres are in the ratio 64 : 27. Find the difference of their surface areas, if the sum of their radii is 7 cm. -Maths 9th

Last Answer : answer:

The sum of the radii of two spheres is 10 cm and the sum of their volumes is 880 cm^3. What will be the product of their radii ? -Maths 9th

Description : The sum of the radii of two spheres is 10 cm and the sum of their volumes is 880 cm^3. What will be the product of their radii ? -Maths 9th

Last Answer : answer:

What is the answer. There are three identical metal spheres labeled A, B and A has a charge of 1C while B and Care neutral. If A touches B and then is removed, and then B touches C and is removed what is the charge on C?

Description : What is the answer. There are three identical metal spheres labeled A, B and A has a charge of 1C while B and Care neutral. If A touches B and then is removed, and then B touches C and is removed what is the charge on C?

Last Answer : C are***

Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th

Description : Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th

Last Answer : (c) 450 cm3. Required difference in capacities = 227227 x (5)2 x 21~ (10)2 x 21 = (1650 ~ 2100) cm3 = 450 cm3

From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

Description : From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

Last Answer : answer:

An urn contains nine balls, of which three are red, four are blue and two are green. -Maths 9th

Description : An urn contains nine balls, of which three are red, four are blue and two are green. -Maths 9th

Last Answer : (b) \(rac{2}{7}\)Let S be the sample space having 9 balls (3R + 4B + 2G) Then n(S) = Total number of ways in which 3 balls can be drawn out of the 9 balls= \(rac{9 imes8 imes7}{3 imes2}\) = 84Let A : Event of drawing three ... 2 = 24.∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{24}{84}\) = \(rac{2}{7}\).

If from each of the three boxes containing 3 blue and 1 red balls, 2 blue and 2 red balls, -Maths 9th

Description : If from each of the three boxes containing 3 blue and 1 red balls, 2 blue and 2 red balls, -Maths 9th

Last Answer : (a) \(rac{13}{32}\)The three boxes B1, B2 and B2 contain the different coloured balls as follows :BlueRedBox 131Box 222Box 313There can be three mutually exclusive cases of drawing 2 blue balls and 1 red balls in the ways as given :Box ... {64}\) + \(rac{2}{64}\) = \(rac{26}{64}\) = \(rac{13}{32}\).

A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Description : A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Last Answer : (d) \(rac{1}{4}\)The box contains 100 balls numbered from 1 to 100. Therefore, there are 50 even and 50 odd numbered balls. The sum of the three numbers drawn will be odd, if all three are odd or one is even and 2 are odd. ∴ Required probability = P(odd) × P(odd) × P(odd) + P(even) × P(odd) × P(odd)

Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways -Maths 9th

Description : Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways -Maths 9th

Last Answer : According to the question, we have 5 balls to be placed in 3 boxes where no box remains emptyHence, we can have the following kinds of distribution firstly, where the distribution will be (3,1,1) that is, one box gets three ... go in second box is = 4 C 2 . Total no. of ways =90 Total:60+90=150.

Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Description : Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Last Answer : (d) \(rac{1}{36}\)Total number of outcomes when three identical dice are rolled, n(S) = 6 6 6 = 216 Let A : Event of rolling same number or each dice ⇒ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), ... 6)} ⇒ n(A) = 6 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{6}{216}\) = \(rac{1}{36}\).

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

Description : A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

Last Answer : answer:

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Description : A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Last Answer : Diameter of the pillar = 50 cm ∴ Radius (r) = 502m = 25 m = 14m and height (h) = 3.5m Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2 ∴ Cost of painting of 1 m2 pillar = Rs. 12.50 ∴ Cost of painting of 112 m2 pillar = Rs. ( 112 x 12.50 ) = Rs. 68.75.

Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th

Description : Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th

Last Answer : Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of cylindrical tank is 2πrh = 2 (22/7) 2.1 4.5 m2 = (44 0.3 ... ) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was used in actual while making such a tank.

In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Description : In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Last Answer : Height of cylindrical pipe = Length of cylindrical pipe = 28m Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of ... = 2 (22/7) 0.025 28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2.

It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th

Description : It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th

Last Answer : Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface are of tank = 2πr( ... [2 (22/7) 0.7(0.7+1)] = 7.48 Therefore, 7.48 square meters of the sheet are required.

Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Description : Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Last Answer : Required sheet = T.S.A. of cyclinder = 2πr (h+r) = 2 × 22 / 7 × 70 / 100(1 + 70 / 100) = 2 × 22 × 0.1 × 1.7 = 7.48 m2

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Description : Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Last Answer : Required sheet = T.S.A. of cyclinder = 2πr (h+r) = 2 × 22 / 7 × 70 / 100(1 + 70 / 100) = 2 × 22 × 0.1 × 1.7 = 7.48 m2

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Volume of milk in 1 glass =πr2h =π×(3.5)2×12=461.58cm2​for 1600 students milk needed is =1600×461.58=738258litre​

A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : According to question find the litres of milk is needed to serve 1600 students.

Mukta had to make a model of a cylindrical -Maths 9th

Description : Mukta had to make a model of a cylindrical -Maths 9th

Last Answer : Radius of the base of the cylindrical kaleidoscope = r = 3.5 cm Height of kaleidoscope = h = 25 cm Chart paper required = curved surface area of kaleidoscope = 2 πrh = 2 x 22/7 x 3.5 x 25 = 550 cm2

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th

Description : It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th

Last Answer : Radius of the closed cylindrical tank = 140/2 cm = 70 cm = 0.7 m Height of the closed cylindrical tank = 1 m Area of metal sheet required = 2πr(r + h) = 2 x 22/7 x 0.7 (1 + 0.7) = 7.48 m2

The capacity of a closed cylindrical -Maths 9th

Description : The capacity of a closed cylindrical -Maths 9th

Last Answer : Height of the cylindrical vessel (h) = 1 m Capacity of the cylindrical vessel = 15.4 L = 15.4/1000 m 3 = 0.0154 m3 Let 'r' m be the radius of the base of the cylindrical vessel Volume of the cylindrical vessel = πr2h ⇒ πr2h = 0 ... 22/7 x 0.07(0.07 + 1) m2 = 1/7 x 44 x 0.07 x 1.07 m2 = 0.4708 m2

The circumference of the base of a cylindrical -Maths 9th

Description : The circumference of the base of a cylindrical -Maths 9th

Last Answer : Height of the cylindrical vessel (h) = 25 cm Let r cm be the radius of the base of the cylindrical vessel. Circumference of the base = 2 πr = 132 cm ⇒ 2 x 22/7 x r = 132 ⇒ r = 132 x 7/2 x 22 = 21 ... 21 x 21 x 25 = 34650 cm3 ∴ Volume of water which vessel can hold = 34650 L/ 1000 L = 34.65 L

A cylindrical tube opened at both the ends -Maths 9th

Description : A cylindrical tube opened at both the ends -Maths 9th

Last Answer : Outer radius of the cylindrical tube (R) = 16/2 cm = 8 cm Inner radius of the cylindrical tube (r) = (8 - 2) cm = 6 cm Length of the tube (h) = 100 cm Volume of iron used in making the cylindrical tube = πR2h - πr2h = πh(R2 - r2) = 22/7 x100(82 – 62) = 22/7 x 100 x 14 x 2 = 8800 cm3

The inner diameter of a cylindrical wooden -Maths 9th

Description : The inner diameter of a cylindrical wooden -Maths 9th

Last Answer : Inner radius of the cylindrical pipe (r) = 24/2 cm = 12 cm Outer radius of the cylindrical pipe (R) = 28/2 cm = 14 cm Length of the cylindrical pipe (h) = 35 cm Volume of the wood used in making the cylindrical pipe ... x 0.6) g = 3432 g = 3432/1000 kg = 3.432 kg Thus, the mass of pipe = 3.43 kg

A person donates cylindrical bowls of diameter 7 cm -Maths 9th

Description : A person donates cylindrical bowls of diameter 7 cm -Maths 9th

Last Answer : Radius of cylindrical bowl = 7/2 cm = 3.5 cm Height of the bowl filled with soup (h) = 4 cm Volume of soup for 1 patient = πr2h = 22/7 x 3.5 x 3.5 x 4 = 154 cm3 ∴ Volume of soup for ... ( ∴ 1L = 1000 cm3) = 38.5 L The person is kind hearted, caring and contributing for the welfare of society.

A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

Description : A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

Last Answer : answer:

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. -Maths 9th

Description : A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. -Maths 9th

Last Answer : Given, AB is a chord of a circle, which is equal to the radius of the circle, i.e., AB = BO …(i) Join OA, AC and BC. Since, OA = OB= Radius of circle OA = AS = BO