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

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
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According to question find the volume of this liquid.
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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

A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Description : A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Last Answer : When squares of 8 cm is cutt-off from rectangulare sheet then, Length of box (l) = (98 - 8 - 8) = 32 cm Breadth of box (b) = (36 - 8 - 8) = 20 cm Height of box (h) = 8cm ∴ Volume of box = lbh = 32 x 20 x 8 = 5120 cm3

A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Description : A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Last Answer : When squares of 8 cm is cutt-off from rectangulare sheet then, Length of box (l) = (98 - 8 - 8) = 32 cm Breadth of box (b) = (36 - 8 - 8) = 20 cm Height of box (h) = 8cm ∴ Volume of box = lbh = 32 x 20 x 8 = 5120 cm3

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.

A metallic sheet is of rectangular shape with dimensions 28m × 36m. From each of its corners, a square is cut off so as to make an open box. -Maths 9th

Description : A metallic sheet is of rectangular shape with dimensions 28m × 36m. From each of its corners, a square is cut off so as to make an open box. -Maths 9th

Last Answer : R.E.F image Volume of box =l×b×h From the diagram l=48−2(8) ∵ Two square formed side =32m b=36−2(8) =20m Also h=8m from question ∴ Volume =32×20×8 =5120m3

A rectangular box has dimensions x, y and z units, where x < y < z. If one dimension is only increased by one unit, -Maths 9th

Description : A rectangular box has dimensions x, y and z units, where x < y < z. If one dimension is only increased by one unit, -Maths 9th

Last Answer : answer:

The volume of a certain rectangular solid is 8 cm^3. Its total surface area is 32 cm^2 and its three dimensions are in geometric progression. -Maths 9th

Description : The volume of a certain rectangular solid is 8 cm^3. Its total surface area is 32 cm^2 and its three dimensions are in geometric progression. -Maths 9th

Last Answer : (b) 32 Let the edges of the solid be a, ar, ar2. Then, Volume = a x ar x ar2 = a3r3 = (ar)3. Given (ar)3 = 8 ⇒ ar = 2 Also, surface area = 2(a x ar + ar x ar2 + a × ar2) = 2(a2r + ... Given, 2ar (a + ar + ar2) = 32 ⇒ 4(a + ar + ar2) = 32 ; Sum of lengths of all edges = 32.

A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

Description : A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

Last Answer : According to question find the the total area of the design and the remaining area of the tiles.

A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

Description : A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

Last Answer : According to question find the the total area of the design and the remaining area of the tiles.

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

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

Description : 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

Last Answer : hope its clear

(1) they contain free electrons Explanation: Metals typically consist of closepacked atoms, meaning that the atoms are arranged like closely packed spheres. Two packing motifs are common, one being bodycentered cubic wherein each metal atom is surrounded by eight equivalent atoms. The other main motif is face-centered cubic where the metal atoms are surrounded by six neighboring atoms. Several metals adopt both structures, depending on the temperature.

Description : (1) they contain free electrons Explanation: Metals typically consist of closepacked atoms, meaning that the atoms are arranged like closely packed spheres. Two packing motifs are common, one ... surrounded by six neighboring atoms. Several metals adopt both structures, depending on the temperature.

Last Answer : Helium gas is filled in ballons because (1) its atomic number is 2 (2) it is lighter than air (3) it is one of the constitutents of water (4) it is a noble gas

If two rectangular sheets each of dimensions 2x and 2y form the curved surfaces of two different cylinders, then the ratio -Maths 9th

Description : If two rectangular sheets each of dimensions 2x and 2y form the curved surfaces of two different cylinders, then the ratio -Maths 9th

Last Answer : answer:

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:

(i) The base of an open rectangular box is square and its volume is ` 256 cm^(3)`. Find the dimensions of this box if we want to use least material fo

Description : (i) The base of an open rectangular box is square and its volume is ` 256 cm^(3)`. Find the dimensions of this box if we want to use least material fo

Last Answer : (i) The base of an open rectangular box is square and its volume is ` 256 cm^(3)`. ... dimensions of this window from which maximum light can admit.

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

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

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 whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high -Maths 9th

Description : A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high -Maths 9th

Last Answer : From the question statement, we have Edge of a cube = 10cm Length, l = 12.5 cm Breadth, b = 10cm Height, h = 8 cm (i) Find the lateral surface area for both the figures Lateral surface ... . Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2

A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

Description : A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 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

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:

The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Description : The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Last Answer : According to question find the lengths QC and PD.

The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Description : The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Last Answer : According to question find the lengths QC and PD.

The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Last Answer : Length of the plank=4m=400cm Breadth=50cm Height=20cm Volume of the plank=L*B*H =400*50*20 =400000cm^3 Length of the pit=16m=1600cm Breadth=12m=1200cm Height=4m=400cm Volume of the pit= L ... *1200*400 =768000000cm^3 Number of planks that can be fitted= 768000000/400000 =1920 planks is the answer.

The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Last Answer : Solution of this question

Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Last Answer : Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter 2 Radius = 0.14 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume ... hours needed: Number of hours = 154 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

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 cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Last Answer : NEED ANSWER

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Last Answer : According to question find the radius of the sphere

A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Description : A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Last Answer : Height of cone = 3 times height of cylinder = 3 3 = 9 cm Volume of cylinder = volume of cone r2 = 8 8 r = 8 cm l2 = h2 + r2 = (9)2 + (8)2 l = = 12 cm C.S.A (cone) = = 301.71 cm2

A cubical box has each edge 10 cm -Maths 9th

Description : A cubical box has each edge 10 cm -Maths 9th

Last Answer : (i) Lateral surface area of cubical box = 4a2 = 4 x 102 = 400 cm2 Lateral surface area of cuboidal box = 2h (l + b) = 2 x 8(12.5 +10) = 16 X 22.5 = 360 cm2 Thus, lateral surface area of ... x 305cm2 = 610 cm2 Thus, total surface area of cuboidal box is greater by (610 - 600) cm2 = 10 cm2

A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm

30 circular plates, each of radius 14 cm -Maths 9th

Description : 30 circular plates, each of radius 14 cm -Maths 9th

Last Answer : Height of the cylinder formed (h) = 30 x 3 = 90 cm Radius of the base of the cylinder formed (r) = 14 cm (i) Total surface area of the cylinder = 2 πr(r + h) = 2 x 22/7 x 14(14 + 90) = 2 x 22/7 x 14 x 104 = 9152 cm2 (ii) Volume of the cylinder formed = πr2h = 22//7 x 14 x 14 x 90 = 55440 cm3

27 drops of water form a big drop of water. If the radius of each smaller drop is 0.2 cm, then what is the radius of the bigger drop ? -Maths 9th

Description : 27 drops of water form a big drop of water. If the radius of each smaller drop is 0.2 cm, then what is the radius of the bigger drop ? -Maths 9th

Last Answer : no of small drops of water =27 radius of per smaller drop of water =0.2cm therefore , radius of the big drop =(0.2 x 27) cm =5.4cm

The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

The dimensions of a rectangle ABCD are 51 cm × 25 cm. -Maths 9th

Description : The dimensions of a rectangle ABCD are 51 cm × 25 cm. -Maths 9th

Last Answer : Area of rectangle ABCD = AB x BC = 51 x 25 = 1275 cm2 Area of trapezium PBCQ = 5/6 x 1275 = 6375/6 cm2 Let QC = 9x cm and PB = 8x cm ∴ Area of trapezium PBCQ = 1/2(QC + PB) x BC ⇒ 6375/6 = 1/2(9x + 8x) x 25 ⇒ 17x ... 6375/6 x 2/17 x 25 ⇒ x = 5 ∴ QC = 9 x 5 cm = 45 cm and PB = 8 x 5 cm = 40 cm

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm