# Surface Areas and Volumes

Topics: Volume, Area, Sphere Pages: 119 (20981 words) Published: November 20, 2012
Question Bank In Mathematics Class X (Term–II)

13

SURFACE AREAS AND VOLUMES
A. SUMMATIVE ASSESSMENT

(c) Length of diagonal =

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(a) Lateral surface area = 4l2 (b) Total surface area = 6l2 (c) Length of diagonal = 3 l 3. Cylinder : For a cylinder of radius r and height h, we have : (a) Area of curved surface = 2πrh

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2. Cube : For a cube of edge l, we have :

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TEXTBOOK’S EXERCISE 13.1 Unless stated otherwise, take  = 22 . 7

Q.1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid. [2011 (T-II)] 1

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l 2  b2  h2

5. Sphere : For a sphere of radius r, we have : Surface area = 4πr2 6. Hemisphere (solid) : For a hemisphere of radius r we have : (a) Curved surface area = 2πr2 (b) Total surface area = 3πr2

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(a) Lateral surface area = 2h(l + b) (b) Total surface area = 2(lb + bh + lh)

(d) Total surface area of hollow cylinder = 2πh(R + r) + 2π(R2 – r2) 4. Cone : For a cone of height h, radius r and slant height l, we have : (a) Curved surface area = πrl = r h 2  r 2 (b) Total surface area = πr2 + πrl = πr (r + l)

Sol. Let the side of cube = y cm
Volume of cube = 64 cm3 Then, volume of cube = side3 = y3 As per condition  y3 = 64  y3 = 4 3

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13.1 SURFACE AREA OF A COMBINATION OF SOLIDS 1. Cuboid : For a cuboid of dimensions l, b and h, we have :

(b) Total surface area = 2πr2 + 2πrh = 2πr(r + h) (c) Curved surface area of hollow cylinder = 2πh(R – r), where R and r are outer and inner radii

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Q.3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. [2011 (T-II)] Sol. Radius of the cone = Radius of hemisphere = 3.5 cm Total height of the toy = 15.5 cm  Height of the cone = (15.5 – 3.5) cm = 12 cm Slant height of the cone (l ) =

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= Diameter of the hollow cylinder = 14 cm 14 Radius of the hollow hemisphere = cm = 7 cm 2  Radius of the base of the hollow cylinder = 7 cm Total height of the vessel = 13 cm  Height of the hollow cylinder = (13 – 7) cm = 6 cm Inner surface area of the vessel = Inner surface area of the hemisphere + Inner surface area of the hollow cylinder = 2 (7)2 cm2 + 2 (7) (6) cm2 = 98cm2 + 84cm2 = (98 + 84) cm2 22 = 182 cm2 = 182 × cm2 = 26 × 22 cm2 7 = 572 cm2.

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Q.2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. [2011 (T-II)] Sol.  Diameter of the hollow hemisphere

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(3.5)2  (12) 2 cm

= 156.25 cm = 12.5 cm Total surface area of the toy = Curved surface area of the hemisphere + Curved surface area of the cone = 2 (3.5)2 cm2 + (3.5) (12.5) cm2 = 24.5 cm2 + 43.65cm2 = 68.25 cm2 =

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Q.4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. [2011 (T-II)] Sol. Side of cubical block = 7 cm Side of cubical block = Diameter of hemisphere = 7 cm  2R = 7 7  R = cm 2 Surface area of solid = Surface area of the cube – Area of base of hemisphere + C.S.A. of hemisphere 2 – R2 + 2R2 = 6 × side = 6 (7)2 cm2 + R2 22 7 7 2 = 6 × 7 × 7 cm2 + × × cm 7 2 2 7  = 6  49  11  cm2 2  77    588  77  2 =  294   cm2 =   cm . 2 2   

r 2  h2

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665 cm2 = 332.50 cm2 2

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68.25  22 cm2 = 214.5 cm2. 7

 y = 4 cm Hence, side of cube is 4 cm. For the resulting cuboid length (l ) = 4 + 4 = 8 cm breadth (b) = 4 cm height (h) = 4 cm  Surface area of the resulting cuboid = 2(lb + bh + hl ) = 2(8 × 4 + 4 × 4 + 4 × 8) cm2 = 2(32 + 16 + 32) cm2 = 2(80) cm2 = 160 cm2.

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Q.5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the...