Tag: volume of prism and pyramid

Questions Related to volume of prism and pyramid

A frustum of a pyramid has an upper base $100\ m$ by $10\ m$ and a lower base of $80\ m$ by $8\ m$. if the altitude of the frustum is $5\ m$, find its volume (in cu. m).

  1. $4567.67$

  2. $3873.33$

  3. $4066.67$

  4. $2345.98$


Correct Option: C
Explanation:

Give : Height of pyramid$(h)=5\ m$ 

It is a pyramid with rectangular base.
Edge of upper base $length(L)=100\ m, breadth(B)=10\ m$

Edge of lower base $length(l)=80\ m, breadth(b)=8\ m$
Area of upper base$(A _{1})=L\times B=100\times 10 = 1000\ m^2$
Area of lower base$A _{2}=l\times b=640\ m^2$
Volume of frustum$(V)=\dfrac{h}{3}(A _{1} + A _2 + \sqrt{A _1 \times A _2})$
$\implies$$(V)=\dfrac{5}{3}(1000 + 640 + \sqrt{1000 \times 640})$
                 $=\dfrac{5}{3}(1640+800)=\dfrac{5}{3}(2440)=4066.666667$
Hence, volume$(V)=4066.67\ cu. m$

A regular triangular pyramid has an altitude of $9\ m$ and a volume of $187.06\ cu.\ m$. What is the base edge in meters?

  1. $12$

  2. $13$

  3. $14$

  4. $15$


Correct Option: A
Explanation:

Given : Altitude$(height\quad (h))=9\ m$, volume $=187.06\ cu.\ m$

We know that, Volume $=\dfrac{1}{3}Bh$, where $B=x^2 \sin \theta$
$\implies 187.06=\dfrac{1}{3} \left(\dfrac{1}{2}x^2 (\sin \theta)\right) (9)$,  where $x$ is base edge
$\implies 187.06=\dfrac{1}{3} \left(\dfrac{1}{2}x^2 \sin60\right)9$
$\implies x^2=143.9988=144$
$\therefore\ x=12\ m$

The frustum of a regular triangular pyramid has equilateral triangles for its bases. The lower and upper base edges are $9\ m$ and $3\ m$, respectively. If the volume is $118.2\ cu.\ m$, how far apart (m) are the base?

  1. $9$

  2. $8$

  3. $7$

  4. $10$


Correct Option: C
Explanation:

Given : Volume $=118.2\ cu. m$

Upper base edge $=9\ m$, lower base edge $=3\ m$
We know that, 
Volume $=\dfrac{h}{3}(A _{1}+A _{2}+\sqrt{A _{1} A _{2}})$ ....... $(1)$, where $A _{1}, A _{2}$ are area of upper and lower bases.
$A _{1}=\dfrac{\sqrt{3}}{4}\times 9^2=35.074$
$A _{2}=\dfrac{\sqrt{3}}{4}\times 3^2=3.897$
From $(1)$ we get,
$118.2 = \dfrac{h}{3}(35.074+3.897+\sqrt{35.074\times 3.897})$
$\implies 118.2=\dfrac{h}{3}(38.971+11.6911)$
$\implies 118.2\times 3=h(50.6621)$
$\implies h=7\ m$
Hence, the bases are $7\ m$ far from each other.

A regular hexagonal pyramid whose base perimeter is $60\ cm$ has an altitude of $30\ cm$, the volume of the pyramid (in cu. cm)is:

  1. $2958$

  2. $2598$

  3. $2859$

  4. $2589$


Correct Option: B
Explanation:

Given : Perimeter of regular hexagon $=60\ cm$ and height $=30\ cm$

We know that, regular hexagon has all its sides of equal length
$\therefore$ Perimeter $=6a=60\ cm$,  where $a$ is side of hexagon
$\implies a=10\ cm$
There are exactly $6$ equilateral triangle
Area of one equilateral triangle $=\dfrac{\sqrt{3}}{4}a^2$
                                                     $=\dfrac{\sqrt{3}}{4}\times 10^2=25\sqrt{3}$
$\therefore$ Area of $6$ equilateral triangles $=6\times 25\sqrt{3}=150\sqrt{3}$
$\therefore\ Area\ of\ base = 150\sqrt{3}$
Volume $=\dfrac{1}{3}\times base \times height$
              $=\dfrac{1}{3}\times 150\sqrt{3}\times 30$
              $=1500\sqrt{3}=2598\ cu. m$
Hence, volume of pyramid is $2598\ cu. m$.

A pyramid whose base is a regular pentagon of area $42\ {cm}^2$ and whose height is $7$ cm. What is the volume (in ${cm}^3$) of the pyramid?

  1. $98$

  2. $105$

  3. $126$

  4. $147$


Correct Option: A
Explanation:

Given : Area of base of pyramid $=42\ cm^2$, height $=7\ cm$

We know that, Volume of pyramid $=\dfrac{1}{3}\times Area\ of\ base \times height$
                                                          $=\dfrac{1}{3}\times 42\times 7$
                                                          $=98\ cm^3$
Hence, volume of pyramid is $98\ cm^3$.

General formula of volume of a prism is:

  1. Area of base $\times$ height

  2. Area of triangle $\times$ height

  3. Area of square $\times$ height

  4. Area of rectangle $\times$ height


Correct Option: A
Explanation:

The volume of a general 3d figure is simply = Area of  Base $\times$ Its height.

This applies for prism also.

If base and height of a prism and pyramid are same, then the volume of a pyramid is:

  1. $\dfrac{1}{3}\times$ Volume of prism

  2. ${3}\ \times$ Volume of prism

  3. $\dfrac{1}{2}\times$ Volume of prism

  4. $2\ \times$ Volume of prism


Correct Option: A
Explanation:

If a pyramid and a prism have the same base and height, their volumes are always in the ratio of $1:3\times $Volume of prism.  

General formula to find volume of a pyramid is:

  1. $\dfrac{\text{Base Area} \times \text{Height}}{2}$

  2. $2(\text{Base Area} \times \text{Height})$

  3. $\dfrac{\text{Base Area} \times \text{Height}}{3}$

  4. $3(\text{Base Area} \times \text{Height})$


Correct Option: C
Explanation:

Pyramid is a structure whose outer surfaces are triangular and converge to a 

single point at the top.

Its base is a polygon i.e triangle, rectangle, pentagon etc.

So, volume is found by finding area of its base times height 

$\therefore$ Volume of a pyramid $=\dfrac{\text{Base Area} \times \text{Height}}{3}$

The base of the right pyramid is a square of side 16 cm and height 15 cm. Its volume $(cm^{3})$ will be

  1. $3840$

  2. $1920$

  3. $1280$

  4. $960$


Correct Option: C
Explanation:

Area of the base $=$ $(16 \times 16) cm^2$
Volume of the pyramid $=$ $\frac{1}{3} \times B h$
Volume of the pyramid $=$ $\frac{1}{3}\left ( 16\times 16 \right )\times 15$
$= 1280 cm^2$

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

  1. $1600$ cm$^{3}$

  2. $16000$ cm$^3$

  3. $\displaystyle \frac{16000}{3} cm^3$

  4. $\displaystyle \frac{5}{4} cm^3$


Correct Option: B
Explanation:
Base of pyramid is an equilateral triangle of parameter
$8dm=80cm$
Let the side of the equilateral triangle be $'a'cm$
$\therefore $Parameter of equilateral triangle$=3a$
$\Rightarrow 3a=80\Rightarrow a=\cfrac { 80 }{ 3 } cm$
Height of pyramid$=30\sqrt { 3 } cm$
Volume of pyramid=Area of base $\times $ height
$=\cfrac { \sqrt { 3 }  }{ 4 } { a }^{ 2 }\times 30\sqrt { 3 } $

$=\cfrac { \sqrt { 3 }  }{ 4 } \times \cfrac { 80 }{ 3 } \times \cfrac { 80 }{ 3 } \times 30\sqrt { 3 } $

$=\cfrac { 3 }{ 4 } \times \cfrac { 80 }{ 3 } \times 80 \times 10 $

$=\cfrac { 1 }{ 4 } \times 80 \times 80 \times 10 $

$=20 \times 80 \times 10 $

$=16000cm^3$