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Volume and Surface Area of Combined Solids - Solved Example Problems | Mensuration | Mathematics | Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail |

Chapter: 10th Mathematics : Mensuration

Volume and Surface Area of Combined Solids

To calculate the surface area of the combined solid, we should only calculate the areas that are visible to our eyes. For example, if a cone is surmounted by a hemisphere, we need to just find out the C.S.A. of the hemisphere and C.S.A. of the cone separately and add them together.

Volume and Surface Area of Combined Solids

Observe the shapes given (Fig.7.37).


The  shapes  provided  lead  to  the following definitions of ‘Combined Solid’.

A combined solid is said to be a solid formed by combining two or more solids.

The  concept  of  combined  solids  is useful in the fields like doll making, building construction, carpentry, etc.

To calculate the surface area of the combined solid, we should only calculate the areas that are visible to our eyes. For example, if a cone is surmounted by a hemisphere, we need to just find out the C.S.A. of the hemisphere and C.S.A. of the cone separately and add them together. Note that we are leaving the base area of both the cone and the hemisphere since both the bases are attached together and are not visible.

But, the volume of the solid formed by joining two basic solids will be the sum of the volumes of the individual solids.

 

Example 7.24

A toy is in the shape of a cylinder surmounted by a hemisphere. The height of the toy is 25 cm. Find the total surface area of the toy if its common diameter is 12 cm.

Solution

Let r and h be the radius and height of the cylinder respectively.

Given that, diameter d = 12 cm, radius r = 6 cm


Total height of the toy is 25 cm

Therefore, height of the cylindrical portion = 25 − 6 = 19 cm

T.S.A. of the toy = C.S.A. of the cylinder + C.S.A. of the hemisphere +Base Area of the cylinder

= 2πrh + 2πr 2 + πr 2

= πr (2h + 3r)  sq. units

= (22/7) × 6 × (38 + 18)

= (22/7) × 6 ×56 = 1056

Therefore, T.S.A. of the toy is 1056 cm2

 

Example 7.25 

A jewel box (Fig. 7.39) is in the shape of a cuboid of dimensions 30 cm × 15 cm ×10 cm surmounted by a half part of a cylinder as shown in the figure. Find the volume and T.S.A. of the box.


Solution 

Let l, b and h1  be the length, breadth and height of the cuboid. Also let us take r and h2 be the radius and height of the cylinder.

Now,  Volume of the box = Volume of the cuboid + 1/2 (Volume of cylinder)


= 4500 + 2651. 79 = 7151. 79

Therefore, Volume of the box = 7151.79 cm3

Now, T.S.A. of the box = C.S.A. of the cuboid+ 1/2 (C.S.A. of the cylinder)


= 900 + 707. 14 = 1607. 14

Therefore, T.S.A. of the box = 1607. 14 cm2

 

Example 7.26 Arul has to make arrangements for the accommodation of 150 persons for his family function. For this purpose, he plans to build a tent which is in the shape of cylinder surmounted by a cone. Each person occupies 4 sq. m of the space on ground and 40 cu. meter of air to breathe. What should be the height of the conical part of the tent if the height of cylindrical part is 8 m?

Solution 

 Let h1  and h2 be the height of cylinder and cone respectively.

Area for one person = 4 sq. m

Total number of persons = 150


Therefore total base area = 150×4

πr2 = 600


Volume of air required for 1 person = 40 m3

Total Volume of air required for 150 persons = 150 ×40 = 6000 m3


Therefore, the height of the conical tent h2  is 6 m

 

Example 7.27 A funnel consists of a frustum of a cone attached to a cylindrical portion 12 cm long attached at the bottom. If the total height be 20 cm, diameter of the cylindrical portion be 12 cm and the diameter of the top of the funnel be 24 cm. Find the outer surface area of the funnel.

Solution 

 Let R, r be the top and bottom radii of the frustum.

Let h1h2  be the heights of the frustum and cylinder respectively.

Given that, R = 12 cm, r = 6 cm, h2 = 12 cm


Now, h1 = 20–12 = 8 cm

Here, Slant height of the frustum


 l = 10 cm

Outer surface area = 2πrh2 + π (R + r )l sq. units

= π [2rh2 + (R + r )l]

= π [ (2 × 6 ×12) + (18 ×10)]

= π [144 + 180]

= (22/7) ×324 = 1018.28

Therefore, outer surface area of the funnel is 1018.28 cm2

 

Example 7.28 

A hemispherical section is cut out from one face of cubical block (Fig.7.42) such that the diameter l of the hemisphere is equal to side length of the cube. Determine the surface area of the remaining solid.


Solution  

Let r be the radius of the hemisphere.

Given that, diameter of the hemisphere = side of the cube = l

Radius of the hemisphere = l/2

TSA of the remaining solid = Surface area of the cubical part + C.S.A. of the hemispherical part − Area of the base of the hemispherical part

= 6 ×(Edge)2  + 2πr2πr2

= 6 ×(Edge)2  + πr2


Total surface area of the remaining solid = 1/4 (24 + π)l2  sq. units


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