# 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 $=\,\,\frac{4}{3}\pi r3\,\,=\,\,\frac{r}{3}(4\pi r2)\,\,\frac{r}{3}\times (surface\,\,area)$

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