10th Class Mathematics Surface Areas and Volumes Volume and Surface Area of Solids

Volume and Surface Area of Solids

Category : 10th Class

 

Volume and Surface Area of Solids

           

S. no.

SOLID

Lateral/ Curved Surface area

Total Surface Area

Volume

1.

Cube

\[Each\text{ }side\text{ }=\text{ }a\]

\[4{{a}^{2}}\]

\[6{{a}^{2}}\]

\[{{A}^{3}}\]

2.

Cuboid

\[Length\text{ }=\text{ }l\] \[Breadth\text{ }=\text{ }b\] \[Height\text{ }=\text{ }h\]

 

\[2\left( l+b \right)\times h\]

 

\[2(lb+bh+lh)\]

 

\[(l\times b\times h)\]

3.

Right Circular

Cylinder

\[Radius\text{ }of\text{ }base\text{ }=\text{ }r\]

\[Height\text{ }=\text{ }h\]

 

\[2\pi rh\]

 

\[2\pi r\,(h\,+\,r)\]

 

\[\pi {{r}^{2}}h\]

4.

Cone

\[Radius\text{ }of\text{ }base\text{ }=\text{ }r\] \[Height\text{ }=\text{ }h\]

\[Slant\text{ }height\text{ }=\]

\[l\,\,=\,\,\sqrt{{{r}^{2}}\,\,+\,\,{{h}^{2}}}\]

 

 

\[\pi rl\]

 

 

\[\pi r(l+r)\]

\[\frac{1}{3}\pi {{r}^{2}}h\]

5.

Sphere

\[Radius\text{ }=\text{ }r\]

 

__________

 

\[4\pi {{r}^{2}}\]

\[\frac{4}{3}\pi {{r}^{3}}\]

 

6.

Hemisphere

\[Radius\text{ }=\text{ }r\]

 

\[2\pi {{r}^{2}}\]

 

\[3\pi {{r}^{2}}\]

\[\frac{2}{3}\pi {{r}^{3}}\]

 

7.

Hollow Cylinder

\[Inner\text{ }radius\text{ }=~r\] \[Outer\text{ }radius\text{ }=\text{ }R\]

\[Height\text{ }=\text{ }h\]

 

\[2\pi h\,(R\,+\,r)\]

\[2\pi h\,(R\,+\,r)\,+\,\,2\pi ({{R}^{2}}-{{r}^{2}})\]

\[Or\,\,\,2\pi \,(R\,+\,r)\,(R+h-r)\]

 

\[\pi h\,({{R}^{2}}\,\,+\,\,{{r}^{2}})\]

8.

Spherical Shell

\[Inner\text{ }radius\text{ }=~r\]\[Outer\text{ }radius\text{ }=\text{ }R\]

 

\[2\pi ({{R}^{2}}+{{r}^{2}})\]

 

\[\pi (3{{R}^{2}}\,\,+\,\,{{r}^{2}})\]

\[\frac{4}{3}\pi ({{R}^{3}}-{{r}^{3}})\]

9.

Frustum of a cone

\[Radius\text{ }of\text{ }base\text{ }=\text{ }R\] \[Radius\text{ }of\text{ }top\text{ }=\text{ }r\]

\[Height\text{ }=\text{ }h\]

\[Slant\text{ }height\text{ }=\text{ }l\]

\[{{L}^{2}}=\text{ }{{h}^{2}}+\text{ }{{\left( R\text{ }-\text{ }r \right)}^{2}}\]

 

 

\[\pi l\,(R\,\,+\,\,r)\]

 

 

\[\pi [{{R}^{2}}+{{r}^{2}}+l\,(R+r)]\]

 

\[\frac{\pi h}{3}[{{R}^{2}}+{{r}^{2}}+Rr]\]

 

Snap Test

 

  1. Three cubes of each side 5 cm are joined end to end. Find the surface area of the resulting cuboid.

            (a) \[350\text{ }c{{m}^{2}}\]               

(b) \[355\text{ }c{{m}^{2}}\]   

            (c) \[340\text{ }c{{m}^{2}}\]               

(d) \[345\text{ }c{{m}^{2}}\]   

            (e) None of these

Ans.     (a)

Explanation: \[Length\text{ }of\text{ }the\text{ }cuboid\text{ }=\text{ }15\text{ }cm,\text{ }breadth\text{ }=\text{ }5\text{ }cm,\text{ }height\text{ }=\text{ }5\text{ }cm\]  

 

           

          

\[\therefore \] Surface area of the cuboid   \[=\text{ }2\text{ }\left( Ib\text{ }+\text{ }bh\text{ }+\text{ }Ih \right)\]           

= \[2\text{ }\left[ 15\text{ }\times \text{ }5\text{ }+\text{ }5\text{ }\times \text{ }5\text{ }+\text{ }1\text{ }5\text{ }\times \text{ }5 \right]\text{ }c{{m}^{2}}\]

                                                            \[=\,\,\,350\text{ }c{{m}^{2}}.\]

 

  1. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

            (a) \[2\text{ }:\text{ }2\] (b) \[3\text{ }:\text{ }1\]

            (c) \[2\text{ }:\text{ }1\]  (d) \[4\text{ }:\text{ }5\]

            (e) None of these

Ans.     (c)

            Explanation: Area of base of the cone = radius of base of the hemisphere

            \[\Rightarrow \,\,p{{r}^{2}}\,\,=\,p{{R}^{2}}\,\,\,\,\,\,\,\,\,\Rightarrow \,\,r=R\]

            Thus, the radius of base of the cone = radius of base of the hemisphere = r.

            Now, let h be the height of the cone.

            Then, volume of the cone = volume of the hemisphere

 

\[\Rightarrow \,\,\,\frac{1}{3}\pi {{r}^{2}}h\,\,\,=\,\,\,\frac{2}{3}\pi {{r}^{3}}\,\,\,\Rightarrow \,\,\,\frac{h}{2}\,\,=\,\,\frac{2}{1}\,\,\,\Rightarrow \,\,\,\frac{height\,of\,the\,cone}{height\,of\,the\,hemisphere}\,\,\,=\,\,\,\frac{2}{1}\]

 

            \[\therefore \] \[Ratio\text{ }of\text{ }their\text{ }height\text{ }=\text{ }2\text{ }:\text{ }1.\]

 

  1. The volume of a right circular cylinder is \[\mathbf{1100}\text{ }\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}}\] and the radius of its base is 5 cm. Find its curved surface area.

(a) \[450\text{ }c{{m}^{2~~~~~~~~~~~~~~~~~~~~~~~~}}\]                     

(b) \[440\text{ }c{{m}^{2}}\]      

                        (c) \[420\text{ }c{{m}^{2}}\]               

(d) \[430\text{ }c{{m}^{2}}\]   

            (e) None of these

Ans.     (b)

            Explanation: Radius of the base of the cylinder \[r\text{ }=\text{ }5\text{ }cm\]. Let its height be h cm.

            \[Then,\text{ }volume\text{ }of\text{ }the\text{ }cylinder\text{ }=\text{ }1100\text{ }c{{m}^{3}}\]

\[\Rightarrow {{r}^{2}}h\,\,=\,1100\,\,\Rightarrow \,\,\frac{22}{7}\times 5\times 5\times h\,\,=\,\,1100\,\,\Rightarrow \,\,h\,\,=\,\,\frac{1100\times 7}{22\times 5\times 5}\,\,=\,\,14\,cm.\]

\[\therefore \]  Curved surface area of the cylinder \[=\,2\pi rh\,\,=\,\,\left( 2\times \frac{22}{7}\times 5\times 14 \right)\,c{{m}^{2}}\,=\,\,440\,c{{m}^{2}}\]

 

  1. If a right circular cone of height. 24 cm, has a volume of \[\mathbf{1232}\text{ }\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}}\], then find its curved surface area

            (a) \[154\text{ }c{{m}^{2}}\]               

(b) \[270\text{ }c{{m}^{2}}\]   

            (c) \[550\text{ }c{{m}^{2}}\]               

(d) \[740\text{ }c{{m}^{2}}\]   

            (e) None of these

Ans.     (c)

            Explanation: Height of the cone \[h\text{ }=\text{ }24\text{ }cm.\]          

            Let radius of the base of the cone be r cm

            Then, volume of the cone = \[\frac{1}{3}\pi {{r}^{2}}h\,\,=\,\,1232\,c{{m}^{2}}\]

            \[\Rightarrow \,\,\,\,\,\frac{1}{3}\times \frac{22}{7}\times {{r}^{2}}\times 24\,\,=\,\,1232\,c{{m}^{2}}\]

\[\Rightarrow \,\,\,\,\,{{r}^{2}}\,\,=\,\,\frac{1232\,\,\times \,3\,\,\times \,\,7}{22\,\,\times \,\,24}\,\,\,\Rightarrow \,\,r=7\,cm\]    

So, Slant height \[l\,\,=\,\,\sqrt{{{h}^{2}}+{{r}^{2}}}\,\,=\,\,\sqrt{{{24}^{2}}+{{7}^{2}}}\,\,=\,\,25\,cm\]

            Therefore curved surface area \[=\,\,\pi rl\,\,~=\text{ }550\text{ }c{{m}^{2}}.\]

 

  1. The volume of a sphere of radius r is obtained by multiplying its surface area by:

            (a) \[\frac{4}{3}\]           (b) \[\frac{r}{3}\]

            (c) \[2r\]                         (d) \[\frac{2}{3}\,r\]

            (e) None of these

Ans.     (b)

Explanation: Volume of sphere \[=\,\,\frac{4}{3}\pi r3\,\,=\,\,\frac{r}{3}(4\pi r2)\,\,\frac{r}{3}\times (surface\,\,area)\]

 

Notes - Volume and Surface Area of Solids
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