Tag: surface areas and volumes of solids

Questions Related to surface areas and volumes of solids

A right pyramid is on a regular hexagonal base. Each side of the base is 10 m. Its height is 60 m.The volume of the pyramid is

  1. 5196 $m^3$

  2. 5200 $m^3$

  3. 5210 $m^3$

  4. 51220$m^3$


Correct Option: A
Explanation:

Volume of pyramid $= \displaystyle \frac{1}{3}$ [Area of hexagonal base of side 10 m $\times$ Height]
$= \displaystyle \frac{1}{3} \left [ \frac{3 \sqrt 3}{4} (10)^2 \times 60 \right ] = 5196 m^3$.

A right pyramid on a regular hexagonal base is of height $60$ m. Each side of the base is $10$ m. The volume of the pyramid is

  1. $\displaystyle 4500\ \text{m}^{3}$

  2. $\displaystyle 5000\ \text{m}^{3}$

  3. $\displaystyle 5196\ \text{m}^{3}$

  4. $\displaystyle 6196\ \text{m}^{3}$


Correct Option: C
Explanation:

Volume of the pyramid
$\displaystyle =\frac{1}{3}\times $ base area $\displaystyle \times $height
$\displaystyle =\frac{1}{3}\times \left ( \frac{3}{2}\sqrt{2}\times 10^{2} \right )\times 60m^{2}$,
since $\displaystyle \sqrt{3}=1.732$
=$\displaystyle =5196m^{3}$

A regular square pyramid is $3$ m height and the perimeter of its base is $16$ m. Find the volume of the pyramid.

  1. 1212 $cu. m$

  2. 1414 $cu. m$

  3. 1616 $cu. m$

  4. 1818 $cu. m$


Correct Option: C
Explanation:

Given, height of regular square pyramid is $3$ m and the perimeter of its base is $16$ m

Let the base side of pyramid is $l$ m
Then perimeter of base $=4a=16$
So, $ a=4$
Then volume of pyramid $=$ $\dfrac{1}{3}l^{2}h=\dfrac{1}{3}\times (4)^{2}\times 3=16 $ $cu. m$

The altitude of the frustum of a regular rectangular pyramid is $5\ m$ the volume is $140\ cu.\ m.$ and the upper base is $3\ m$ by $4\ m$. What are the dimensions of the lower base in $m$?

  1. $9\times10$

  2. $6\times8$

  3. $4.5\times6$

  4. $7.5\times10$


Correct Option: B
Explanation:

Given : Height of pyramid$(h)=5\ m$, Volume$(V)=140\ cu. m$

Edge of upper base are $length(L)=3\ m, breadth(B)=4\ m$
And let $l,h$ be the length and height of lower base

$\therefore$ Area of upper base$(A _{1})=L\times B=3\times 4 = 12\ m^2$
And Area of lower base$A _{2}=l\times b$
Volume of frustum$(V)=\dfrac{h}{3}(A _{1} + A _2 + \sqrt{A _1 \times A _2})$
$\implies$$140=\dfrac{5}{3}(12 + A _2 + \sqrt{12 \times A _2})$
$\implies 84=A _2 + 12 + \sqrt{12}\ \sqrt{A _2}$
$\implies A _2+\sqrt{12}\ \sqrt{A _2}-72=0$
$\implies \left(\sqrt{A _2} + \dfrac{\sqrt{12}}{2}\right)^2 - \dfrac{12}{4} - 72=0$
$\implies \left(\sqrt{A _2} + \dfrac{\sqrt{12}}{2}\right)^2 =75$
$\implies \left(\sqrt{A _2} + \dfrac{\sqrt{12}}{2}\right) =\sqrt{75}$
$\implies \sqrt{A _2}=6.928$
$\therefore A _2 = 47.9971=48\ m^2=6\times 8$ 
Hence, dimensions of lower base is $6 \times 8$.

The length of the base of a square pyramid is $2\ cm$ and the height is $6\ cm$. Calculate the volume.

  1. $8\ cm^3$

  2. $6\ cm^3$

  3. $4\ cm^3$

  4. $2\ cm^3$


Correct Option: A
Explanation:

Volume of square pyramid  $=\dfrac { 1 }{ 3 } \times { a }^{ 2 }\times h=\dfrac { 1 }{ 3 } \times 2\times 2\times 6=8{ cm }^{ 3 }$

The base of a right pyramid is an equilateral triangle of perimeter 8 cm and the height of the pyramid is $30\sqrt 3$ cm. The volume of the pyramid is

  1. $160 cm^3$

  2. $1600 cm^3$

  3. $\dfrac {160}{3} cm^3$

  4. $\dfrac {5}{4} cm^3$


Correct Option: A
Explanation:

Volume of right pyramid $=$ $\dfrac { 1 }{ 3 } \times area\quad of\quad base\times height\quad of\quad pyramid$

Now, base is equilateral $\triangle $, therefore,
area $=\dfrac { \sqrt { 3 }  }{ 4 } \times { \left( side \right)  }^{ 2 }$
Perimeter of triangle $=8cm$
$\therefore \quad \quad 3a=8\Rightarrow a=\dfrac { 8 }{ 3 } cm$
$\therefore \quad \quad area=\dfrac { \sqrt { 3 }  }{ 4 } \times \dfrac { 8 }{ 3 } \times \dfrac { 8 }{ 3 } =\dfrac { 16\sqrt { 3 }  }{ 9 } { cm }^{ 2 }$
Now,  Volume $=\dfrac { 1 }{ 3 } \times \dfrac { 16\sqrt { 3 }  }{ 9 } \times 30\sqrt { 3 } $
                        $=\dfrac { 160\times 3 }{ 9 } =\dfrac { 160 }{ 3 } { cm }^{ 3 }$

A right pyramid is on a regular hexagonal base. Each side of the base is $10$ m. Its height is $60$ m. The volume of the pyramid is

  1. $5196 m^3$

  2. $5200 m^3$

  3. $5210 m^3$

  4. $5220 m^3$


Correct Option: A
Explanation:
Volume of a hexagonal Pyramid $=$ $\displaystyle \frac{\sqrt{3}}{2}a^2h$
where $a$ $=$ side of the base. and $h$ $=$ height of the pyramid

$\therefore $Volume of a pyramid $=$ $\displaystyle \frac{\sqrt{3}}{2}{(10^2)}\times {60}$ $=$ $5196$ $m^3$

If a regular square pyramid has  a base of side 8 cm and height  of 30 cm, then its volume is

  1. 120 c.c.

  2. 240 c.c.

  3. 640 c.c.

  4. 900 c.c.


Correct Option: C
Explanation:

Side of square base $=$ 8 cm.
Height of pyramid $=$ 30 cm
$\therefore$ Volume of square pyramid
$= \frac{1}{3}\times $area  of  base $\times$ height
$= \frac{1}{3}\times 82\times 30 = 640$ c.c.

If the volume of a prism is $1920$ $\sqrt{3} cm^3$ and the side of the equilateral base is $16$ $cm$, then the height (in cm) of the prism is?

  1. $19$

  2. $20$

  3. $30$

  4. $40$


Correct Option: C
Explanation:
Volume of prism $=$ Area of equilateral triangle $\times $ height
Now, Area of triangle $= \dfrac{\sqrt{3}}{4}a^2$
= $\dfrac{\sqrt{3}}{4} 16^2$ = $64 \sqrt{3}$
$1920 \sqrt{3} = 64\sqrt{3}\times $ height

$\therefore $ height $=$ $\displaystyle \dfrac{1920}{64}$
$= 30 \ cm$

The corner of a cube_has  been cut by the plane passing through mid-point of  the three edges meeting at that corner. If the edge of  the cube is of 2 cm length,  then the volume of the  pyramid thus cut off is

  1. $\dfrac{1}{24}cm^3$

  2. $\dfrac{1}{6}cm^3$

  3. $\dfrac{1}{48}cm^3$

  4. $6cm^3$


Correct Option: B
Explanation:

The base of the pyramid thus out off will be a right angled triangle whose sides containing the right angle will be each equal to 1 cm. The height of the pyramid will also be equal to 1 cm. Hence, the volume will be equal to $\frac{1}{6}  cm^3$.