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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 curved surface area of the vessel.

Let r cm be the radius of the cylinder and h cm be the height of the cylinder, then
r = 7 cm,
and    h = (13–7) cm
= 6 cm.
Let r1 cm be the radius of the hemisphere, then
r1 = 7 cm
Now,
the inner curved surface area of the vessel
= C,S.A of hemisphere
+ C,S.A of cylinder
= (2πr12 + 2 π rh) cm2
= (2 π r2 + 2 π rh) cm2 [∵ r1 = r]
= [2 π r (r + h)] cm2

equals open square brackets open parentheses 2 space straight x space 22 over 7 space straight x space 7 close parentheses open parentheses 7 space plus space 6 close parentheses close square brackets space cm squared

= (44 x 13) cm2
= 572 cm2.

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A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

It is given that,
Edge of the cube = l
Then, Surface area = 6 (edge)2 = 6l2
Now, the greatest diameter of hemisphere
= length of an edge of the cube
= l
So, curved surface area of the hemisphere
                  
                   equals space 2 space straight pi space open parentheses 1 half close parentheses squared
equals space 2 space straight pi space straight l squared over 4 equals πl squared over 2
And, Area of base of the hemisphere
               
                    equals straight pi space open parentheses l over 2 close parentheses squared
equals space straight pi l to the power of italic 2 over 4

Required surface area
= Surface area of the cubical wooden block – area of the base of the hemisphere + curved surface area of the hemisphere

equals space 6 l to the power of italic 2 italic minus pi over italic 4 l to the power of italic 2 italic equals italic 6 l to the power of italic 2 italic plus pi over italic 4 l to the power of italic 2

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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.

Let r cm be the radius, h cm be the height and l cm be the slant height of the cone, then


Let r cm be the radius, h cm be the height and l cm be the slant heig

r = 3.5 cm,
h = (15.5 – 3.5) cm = 12 cm.

Now,   l = square root of straight r squared plus straight h squared end root
rightwards double arrow space space straight l space equals space square root of left parenthesis 3.5 right parenthesis squared plus left parenthesis 12 right parenthesis squared end root
space space space space space space space equals space square root of 12.25 plus 144 end root
space space space space space space space equals space square root of 156.25 end root
space space space space space space space equals space 12.5 space cm.
space space space space space space space space

 Let r1 cm be the radius of the hemisphere.
Then, r1 = 3.5 cm    [∵ r = r1]
Now,
The total surface area of the toy
= CSA of hemisphere
+ CSA of cone
= 2 π r12 + πrl
= 2π r2 + πrl    [ ∵ r1= r]
= π r [2r + l]

equals space 22 over 7 space straight x space 3.5 space left square bracket 2 space straight x space 3.5 space plus space 12.5 right square bracket
equals space space left square bracket 11 space left parenthesis 7 space plus 12.5 right parenthesis right square bracket space cm squared
equals space space left square bracket space 11 space straight x space 19.5 right square bracket space cm squared space equals space 214.5 space cm squared
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2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.


Let ‘a’ be the length of a side of a cube. Then
Volume of one cube = 64 cm3
⇒    n3 = 65 cm3
⇒    n = 4 cm
On joining the cubes, a cuboid is formed.
Then The length of the resulting cuboid (l) = 2
The breadth of the resulting cuboid (b) = a cm
The thickness of the resulting cuboid (h) = a cm
Now,
Surface area of the resulting cuboid
= 2(lb + bh + hl)
= 2 (2a. a + a. a + a. 2a)
= 2 (2a2 + a2 + 2a2)
= 2 (5a2) = 10a2 = 10 (4)2
= 160 cm2.

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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.

Let a be the length of an edge of the cube. Then
a = 7 cm
Greatest diameter of the hemisphere
= Length of an edge of the cube
= 7 cm
Now,
Surface area of the cube
= 6 (edge)2
= 6 x 72
= 6 x 49 = 294 cm2
Let r be the radius of the hemisphere.

Then,       r = 7 over 2 space cm
Now,  
Curbed surface area of hemisphere

equals space 2 πr squared
equals space open parentheses 2 space straight x space 22 over 7 straight x 7 over 2 straight x 7 over 2 close parentheses
equals space 77 space cm squared

And,  Base area = πr squared equals open parentheses 22 over 7 straight x 7 over 2 straight x 7 over 2 close parentheses space cm squared

                              space space space equals space space 77 over 2 space cm squared
Total surface area
= Surface area of the cube + curved surface area of the hemisphere – base area of the hemisphere

equals space open parentheses 294 space plus space 77 minus 77 over 2 close parentheses space cm squared
equals space open parentheses 294 plus 77 over 2 close parentheses space cm squared equals space left parenthesis 294 plus 38.5 right parenthesis space cm squared
equals space 332.5 space cm squared

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