Students and Teachers Forum

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) = .....