A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made ? -Maths 9th

A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made ? -Maths 9th
by

1 Answer

Answer :

Radius of a big spherical laddoo = 5 cm ∴ Volume of the big spherical laddoo = 4 / 3 π × 5 × 5 × 5 cm3 Radius of a small spherical laddoo = 2.5 = 5 / 2 cm ∴ Volume of the small spherical laddoo =  4 / 3 π × 5 / 2 × 5 / 2  × 5 / 2  cm3  Let n be the number of small laddoos that can be made using the same amount (volume) of big laddoo.  Therefore , n  4 / 3 π × 5 / 2 × 5 / 2  × 5 / 2 =  4 / 3 π × 5 × 5 × 5   ⇒ n = 2 ×  2 ×  2 = 8 Hence , with the same amount of big laddoo, 8 small laddoos can be made.   
by

Related questions

A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made ? -Maths 9th

Description : A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made ? -Maths 9th

Last Answer : Radius of a big spherical laddoo = 5 cm ∴ Volume of the big spherical laddoo = 4 / 3 π 5 5 5 cm3 Radius of a small spherical laddoo = 2.5 = 5 / 2 cm ∴ Volume of the small spherical laddoo = 4 / ... = 2 2 2 = 8 Hence , with the same amount of big laddoo, 8 small laddoos can be made.

A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made? -Maths 9th

Description : A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made? -Maths 9th

Last Answer : NEED ANSWER

A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made? -Maths 9th

Description : A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made? -Maths 9th

Last Answer : Solution of the question

A shopkeeper has one spherical laddoo -Maths 9th

Description : A shopkeeper has one spherical laddoo -Maths 9th

Last Answer : Radius of larger spherical laddoo (R) = 5 cm Radius of smaller spherical laddoo (r) = 2.5 cm Number of laddoos = Volume of larger laddoo/ Volume of smaller one = 4/3 πR3/4/3πr3 = 4/3π(5)3/ 4/3π(2.5)3 = 5 x 5 x 5/2.5 x 2.5 x 2.5 = 8 تنش بن

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.

The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : Surface area of a spherical balloon whose radius is 6 cm. = 4π × 6 × 6 cm2 Surface area of a spherical balloon whose radius is 12 cm. = 4π × 12 × 12 cm2 ∴ Ration of surface areas = 4π × 6 × 6 / 4π × 12 × 12 = 1 / 4 = 1 : 4

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

The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : Surface area of a spherical balloon whose radius is 6 cm. = 4π × 6 × 6 cm2 Surface area of a spherical balloon whose radius is 12 cm. = 4π × 12 × 12 cm2 ∴ Ration of surface areas = 4π × 6 × 6 / 4π × 12 × 12 = 1 / 4 = 1 : 4

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

Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water. -Maths 9th

Description : Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water. -Maths 9th

Last Answer : The amount of water displaced by a solid spherical ball when it is completely immersed in water is equal to its volume. Volume of a sphere of radius r is 34​πr3 As the diameter of the ball is 4.2 cm, its radius r=2.1 cm Hence, volume of water displaced =34​πr3=34​×722​×2.1×2.1×2.1=38.808 cm3

Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water. -Maths 9th

Description : Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water. -Maths 9th

Last Answer : Solution of this question

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​=

The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Description : The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Last Answer : Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So r1 = 7cm r2 = 14 cm Now, Required ratio = (initial surface area)/(Surface area after pumping air into ... = (7/14)2 = (1/2)2 = ¼ Therefore, the ratio between the surface areas is 1:4.

The radius of a spherical balloon increases -Maths 9th

Description : The radius of a spherical balloon increases -Maths 9th

Last Answer : Radius of the spherical balloon = r1 = 7 cm Surface area s1 of the balloon = 4 πr12 = 4 x 22/7 x 72 = 616 cm2 Radius of the spherical balloon when air is pumped into it = r2 = 14 cm Surface area S2 of the balloon = 4 πr22 = 4 x ... S 1/S 2 = 4 x 22/7 x 7 2/4 x 22/7 x14 2 = 1/4 ∴ S1 : S2 = 1 : 4

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 spherical iron shell with external diameter 21 cm weighs 22775 x 5/21 grams. Find the thickness of the shell if the metal weighs 10 gms per cu cm. -Maths 9th

Description : A spherical iron shell with external diameter 21 cm weighs 22775 x 5/21 grams. Find the thickness of the shell if the metal weighs 10 gms per cu cm. -Maths 9th

Last Answer : answer:

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. What is the volume of steel used in making the bowl ? -Maths 9th

Description : A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. What is the volume of steel used in making the bowl ? -Maths 9th

Last Answer : answer:

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Description : A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Last Answer : Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of ... 22/7) (5.25)2 = 173.25 Therefore, the outer curved surface area of the bowl is 173.25 cm2.

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

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

Description : How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

Last Answer : Here, radius (r) = 3.5 cm and height (h) = 12 cm ∴ Amount of ice cream = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 3.5 × 3.5 × 12 = 154 cm3

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 chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

Description : How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

Last Answer : Here, radius (r) = 3.5 cm and height (h) = 12 cm ∴ Amount of ice cream = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 3.5 × 3.5 × 12 = 154 cm3

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.

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

The surface area of a sphere of radius 5 cm -Maths 9th

Description : The surface area of a sphere of radius 5 cm -Maths 9th

Last Answer : Radius of the sphere (r1) = 5 cm Radius of the base of cone (r2) = 4 cm Let r сm be the height of the cone. Surface area of sphere = 4 πr2 ⇒ 4 π(5)2 = 100 π cm2 Curved surface area of cone = πrl = 4 πl ... ∴ Volume of cone = 1/3 πr2h = 1/3 x 22/7 x 42 x 3 = 352/7 cm3 = 50.29 cm3 (Approximately)

If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Description : If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Last Answer : Semi-perimeter of triangle (s) = \(rac{3+4+5}{2}\)cm = 6 cm∴ Area of triangle A = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{6 imes3 imes2 imes1}\) cm2 = 6 cm2∴ Radius of circum-circle = \(rac{abc}{4( ext{Area of}\,\Delta)}\) = \(rac{3+4+5}{4 imes60}\) cm = 2.5 cm

If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Description : If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Last Answer : (b) 7.5 cm.Area of a triangle = \(rac{1}{2}\)x base x height= In-radius x semi-perimeter of the Δ \(\big[ ext{Using r =}rac{\Delta}{s}\big]\)Let the sides of triangle be 4x, 5x and 6x respectively. Given: In-radius = 3 ... x \(rac{15x}{2}\) = \(rac{6x}{2}\) x h ⇒ h = \(rac{45}{6}\) = 7.5 cm.

A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th

Description : A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th

Last Answer : 92239223% Volume of cone = 1313πr2h = 13×227×1×7=22313×227×1×7=223 cu. cm Volume of cubical block = (10 × 5 × 2) cm3 = 100 cm3 ∴ Wastage of wood = (100−227)100×100(100−227)100×100 = 27832783% = 92239223%

Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Description : Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Last Answer : STEP1: Draw a circle with centre at point O and radius 5 cm. STEP2: Draw its cord AB. STEP3: With centre A as centre and radius more than half of AB, draw two arcs, one on each side ... is perpendicular bisector of AB which is chord of circle, Hence, it passes through the centre of the circle.

Find the total surface area of a hemisphere of radius 10 cm -Maths 9th

Description : Find the total surface area of a hemisphere of radius 10 cm -Maths 9th

Last Answer : Radius of hemisphere, r = 10cm Formula: Total surface area of hemisphere = 3πr2 = 3×3.14×102 = 942 The total surface area of given hemisphere is 942 cm2.

A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.

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

A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

Last Answer : Volume of cone = Volume of sphere 1 / 3π(2.1)2 × 8.4 = 4 / 3 πr3 ⇒ r3 = (2.1)2 × 8.4 / 4 = (2.1)3 ⇒ r = 2.1 cm ∴ Radius of the sphere = 2.1 cm

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

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 .

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

A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

Last Answer : Volume of cone = Volume of sphere 1 / 3π(2.1)2 × 8.4 = 4 / 3 πr3 ⇒ r3 = (2.1)2 × 8.4 / 4 = (2.1)3 ⇒ r = 2.1 cm ∴ Radius of the sphere = 2.1 cm

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

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 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

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : NEED ANSWER

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 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

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : Find the ratios of the surface areas of the balloon in the two cases.

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.

Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th

Description : Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th

Last Answer : Let AB be a chord of circle with centre O and radius 13cm. Draw OM perpendicular AB and join OA. In the right triangle OMA, we have OA2 = OM2 + AM2 ⇒ 132 = 122 + AM2 ⇒ AM2 = 169 - 144 ... . As the perpendicular from the centre of a chord bisects the chord.Therefore, AB = 2AM = 2 x 5 = 10cm.