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Solution: Here, Slant height(h) =13cm and length of square base = a cm Now, The area of triangular surfaces of pyramid= 260 cm2 Or, 2×ah = 260 Or, 2×a×13 .....
We have given, OB = 25cm = e OP = 24cm = l AB = a First of all, PB2 = OB2 – OP2 Or, PB = √[252 - 242] .....
Given, Surface area of the sphere (TSA) = 154 cm2 We know, Surface area of the sphere (TSA) = 4πr2 Or, 154 cm2 = 4⨉ 3.14 ⨉ r2 Or, 154 cm2 = 12.56 ⨉ r2 Or, .....
Given, Volume of the sphere (V) = 43 πcm3 Or, 43r3 = 43 π cm3 Or, r3 = 1 cm3 Or, r = ∛(1cm3) Or, r = 1 cm So, the radius of the sphere is 1 cm. Now, Surface area of the sphere (TSA) = .....
Given, Radius of sphere (r) = 21cm. Volume =? We have, Volume of sphere = 43πr3 .....
Given, Radius of the metallic sphere (r) = 3 cm Then, Volume of the metallic sphere (V) = 43πr3 .....
Let r be the radius of sphere. Given, surface area : volume=3:7 Using formula, we get, or, (4πr2):(43 πr3 ) = 3:7 or, 3r = 37 or, r = 7cm ∴ r = 7cm. Therefore, radius of sphere is .....
Here, Surface area of the sphere (A) = 400cm2 Or, 4πr2 = 400 cm2. Or, πr2 =100 cm2. Now, TSA of hemisphere = 3πr2 .....
Given, Area of the great circle of the sphere (A) = 49π cm2 Or, πr2 = 49π cm2 Or, r2 = 49 cm2 Or, r2 = 49 cm2 Or, r = √(49 cm2 ) Or, r = 7 cm So, the radius .....
Given, Height of the cylinder (h) = 10 cm Radius of the base (r) = 3.5 cm. Now, Total surface area of the article .....
Given, Volume of the cylinder (V) = 770 cm3 Area of curved surface (CSA) = 440 cm2 We know, Volume of cylinder (V) = πr2h Or, 770 cm3 = πr2h Or, 770 cm3 = πrh .....
Here, Total number of circular plates (N) = 35 Diameter of each plate (d) = 14 cm Radius of each plate (r) = 7 cm .....
Given, Curved surface area of the cylinder (CSA) = 462 cm2 Height of the cylinder (h) = 14 cm. We know, Curved surface area of the cylinder (CSA) = .....
Given, Diameter of glass (d) = 5 cm Radius of glass (r) = 2.5 cm Height of the glass (h) = 10 cm. Now, Apparent capacity of glass = Volume of cylindrical .....
Given, Volume of the object (V) = 7,70,000 cm3 Diameter of the object (d) = 14 cm Radius of the object (r) =7 cm. We know, Volume of the .....
Given, Diameter of the object (d) = 24 cm Radius of the object (r) = 12 cm Height of the object (h) = 12 cm. We know, Volume of the .....
Given, Volume of the solid object (V) = 1234 cm3 Radius of the solid object (r) = 14 cm We know, Volume of the half of cylinder (V) = 12πr2h Or, 1234 cm3 = 2214(14cm)2 ⨉ h Or, 1234 cm3 = 308 cm2 ⨉ h Or, h = 4 cm. So, .....
Given, Length of the cylinder (h) = 7 cm Total surface area of each half (TSA) = 406 cm2 Radius of the cylinder (r) = ? We know, Total surface area of half of .....
Given, Diameter of the object (d) = 49 cm Radius of the object (r) = d2 = 24.5 cm Volume of the object (V) = 3,773 cm3. We know, .....
Given, Diameter of the cylinder (d) = 28 cm Radius of the cylinder (r) = 14 cm Total surface area of each half (TSA) = 1624 cm2 .....
Given, Radius of the base of the cylinder (r) = 35 cm Height of the cylinder (h) = 60 cm. We know, (i) Area of the curved surface of cylinder (CSA) = 2πrh .....
Given, Area of the base of cylinder (A) = 38.5 cm2 Or, πr2 = 38.5 cm2 Or, 3.14 ⨉ r2 = 38.5 cm2 Or, r2 = 12.25 .....
Given, Length of the pipe (h) = 14 cm Rate of pipe (C) = Rs.7 per cm3 Total cost of pipe (T) = Rs. 154. We know, Volume of pipe (V) = TC .....
Given, Volume of the cylinder (V) = 7700 cm3 Area of the base (A) = 3850 cm2. We know, Volume of the cylinder (V) .....
Given, Length of the water pipe (h) = 126 m Outer diameter of the pipe (d1) = 8 cm Outer radius of the pipe (r1) = 4 cm Inner diameter of the pipe (d2) = 5 .....
Given, Volume of the object (V) = 4,400 cm3 Length of the object (h) = 14 cm. We know, Volume of the cylindrical half (V) = ½ ⨉ πr2h Or, 4400 cm3 .....
Given, Diameter of the cylinder (d) = 14 cm Radius of the cylinder (r) = 7 cm Height of the cylinder (h) = 30 cm. We know, Curved surface area of cylinder (CSA) = .....
Given, Depth of the cylindrical well (h) = 20 m Diameter of the cylindrical well (d) = 3.5 m Radius of the cylindrical well (r) = d2 .....
Given, Curved surface area of the cylinder (CSA) = 308 cm Radius of the cylinder (r) = Height of the cylinder (h). We know, Curved surface area of cylinder .....
Given, Volume of cylindrical can (V) = 1.54 l = .....
Given, Length of the iron pipe (h) = 140 cm Internal diameter of the iron pipe (d2) = 5 cm Or, internal radius of the iron pipe (r2) = d22 = 2.5 cm .....
Given, Thickness of the cylindrical plates (t) = 0.4 cm Total number of cylindrical plates (N) = 35 So, total height of the cylinder so formed (h) = N ⨉ t .....
Here, Height (h) =14cm, diameter (d) = 14 cm. So, radius (r) = 142cm = .....
Given, Diameter of the cylindrical log of wood (d) = 42 cm Radius of the cylindrical log of wood (r) = 21 cm. Area of the curved .....
Given, Radius of the sphere (R) = 18 cm Radius of cylinder (r) = 7 cm. Now, Volume of sphere (V) = 43 πR3 .....
Given, Length of the cylinder (h) = 10 cm Diameter of the cylinder (d) = 14 cm Radius of the cylinder (r) = 7 cm. We know, Surved .....
Given, Total number of circular plates (N) = 35 Thickness of each plate (t) = 0.8 cm So, the height of the cylinder so formed (h) = N ⨉ .....
Given, Diameter of water vessel (d) = 3.5 m Radius of water vessel (r) = d2 = 1.75 m Capacity of the vessel (V) = 770 l .....
Given, Volume of the sphere (V) = 4851 cm3 Let r be the radius of sphere. We know, Volume of the sphere (V) = 43πr3 Or, 4851 cm3 = 4 3⨉ .....
Given, Length of wooden log (h) = 1.5m Radius of base (r) = 42cm = 0. 42m Volume of log = ? We know that, Volume .....
Given, Curved surface area = CSA = 7392cm2. Diameter = d = 28cm. We know, CSA = 2πrh = πdh Or, h = CSAπd .....
Let r be the radius of sphere. Given, Total surface area of the sphere (TSA) = 616 cm2 Or, 4πr2 = 616 cm2 Or, πr2 = 154cm2. Now, Total surface area of the hemisphere (TSA) = 3πr2 .....
Given, Internal radius of the hollow sphere (r) = 5 cm External radius of the hollow sphere (R) = 8 cm. We know, Volume of the hollow sphere (V) = 43 π (R3 .....
Given, Total surface area of iron ball (TSA) = 1810.28 cm2 Rate of iron (C) = Rs. 15 per cm3 We know, Total surface area of the iron ball (TSA) = .....
Given, Circumference of the circle (C) = 44 cm Or, 2πr = 44 cm [Since,C = 2πr ] Or, 2 ⨉ 3.14 ⨉ r = 44 cm Or, 6.28 ⨉ r = 44 cm Or, r = 7 cm So, the .....
Given, External radius of the spherical shell (R) = 14 cm Internal radius of the spherical shell (r) = 12 cm. Then, Thickness of the spherical shell (t) = R – .....
Here, Given, Total number of circular plates (N) = 50 Diameter of each plate (d) = 15 cm Radius of each plate (r) = 7.5 cm .....
Given, Length of the pipe (h) = 15 cm Rate of the pipe (C) = Rs. 5 per m3 Total cost of the pipe (T) = Rs. 9428.57 Thickness of the .....
Given, Length of the pipe (h) = 15 cm Outer radius of the pipe (r1) = 4 cm Inner radius of the pipe (r2) = 2 cm. We know, Total surface area of the pipe .....
Given, Volume of the cylinder (V) = 3080 cm3 Diameter of the base (d) = 28 m Radius of the base (r) =14 m Depth of the cylinder (h) = .....