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.
Description : A hemispherical bowl is made of steel 0.25 cm thick. -Maths 9th
Last Answer : Outer radius of hemispherical bowl = R = r + 0.25 = 5 + 0.25 = 5.25 cm Outer curved surface area of bowl = 2 πR2 = 2 x 22/7 x (5.25)2 = 2 x 22 x 5.25 x 0.75 = 173.25 cm2
Description : A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th
Last Answer : Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm Formula for Surface area of hemispherical bowl = 2πr2 = 2 (22/7) (5.25)2 = 173.25 Surface area of hemispherical ... of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.
Description : A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th
Last Answer : External radius of hemispherical bowl = 5 cm Internal radius of the bowl = (5 – 1) cm = 4 cm Surface area of external portion = 2π(5)2 = 50 p sq. cm Surface area of internal portion = 2π(4)2 = ... = 91π sq. cm = (91×227)(91×227) sq. cm = 13 × 22 sq. cm = 286 cm2
Description : A hemispherical bowl is 176 cm round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical -Maths 9th
Last Answer : Let the radius of the hemispherical bowl be r cm Then, 2πr=176⇒r=176×72×22⇒r=282πr=176⇒r=176×72×22⇒r=28 Volume of liquid in the bowl : =12×( ... : =Volume of liquid in the bowlVolume of 1 glass=(14×28×282×2×2)= 1372
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
Last Answer : Find the ratios of the surface areas of the balloon in the two cases.
Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th
Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2
Description : A hemispherical tank is made up of an iron -Maths 9th
Last Answer : Inner radius of the hemispherical tank (r) = 1 m Outer radius of the hemispherical tank (R) = 1 + 0.01 = 1.01 m Volume of iron used to make the hemispherical tank = 2/3 πR3 - 2/3πr3 = 2/3π(R3 - r3) = 2/3 x 22/7 [(1.01)3 - 13] = 44/21(1.0303 - 1) = 44/21 x 0.0303 = 0.06349 m3
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
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:
Description : Two hemispherical domes are to be painted -Maths 9th
Last Answer : Let radii of the bases of two domes be r and R. ∴ 2πr = 17.6 ⇒ 2 x 22/7 x r = 17.6 ⇒ r = 17.6 x 7/2 x 22 = 2.8 cm and 2 πR = 70.4 ⇒ 2 x 22/7 x R = 70.4 ⇒ R = 70.4 x 7/2 x 22 = ... .2 x 11.2 = 49.28 + 788.48 cm2 = 837.76 cm2 Cost of painting at the rate of ₹10 per cm2 = 837.76 x 10 = ₹8377.6
Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th
Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400
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.
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
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
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.
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
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
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
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.
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
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)
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
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.
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%
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.
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
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.
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.
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.
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
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
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
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
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 .