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Volume of prism and pyramid - class-X

Description: volume of prism and pyramid
Number of Questions: 22
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Tags: maths perimeter, area and volume mensuration
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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).

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $4567.67$

  2. $3873.33$

  3. $4066.67$

  4. $2345.98$

Show answer
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?

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $12$

  2. $13$

  3. $14$

  4. $15$

Show answer
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?

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $9$

  2. $8$

  3. $7$

  4. $10$

Show answer
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:

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $2958$

  2. $2598$

  3. $2859$

  4. $2589$

Show answer
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?

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $98$

  2. $105$

  3. $126$

  4. $147$

Show answer
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:

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. Area of base $\times$ height

  2. Area of triangle $\times$ height

  3. Area of square $\times$ height

  4. Area of rectangle $\times$ height

Show answer
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:

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  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

Show answer
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:

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

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

Show answer
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

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $3840$

  2. $1920$

  3. $1280$

  4. $960$

Show answer
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

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

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

  2. $16000$ cm$^3$

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

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

Show answer
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 $

$=16000cm^3$

If the areas of the adjacent faces of a rectangular block are in the ratio $2:3:4$ and its volume is $9000{cm}^{3}$, then the length of the shortest edge is

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $30cm$

  2. $20cm$

  3. $15cm$

  4. $10cm$

Show answer
Correct Option: C
Explanation:

Let the edge of the cuboid be $a\,cm,\,b\,cm$ and $c\,cm$.

And, $a<b<c$
The areas of the three adjacent faces are in ratio $2:3:4$
So,
$ab:ca:bc=2:3:4$ and its volume is $9000\,cm^3$
We have to find the shortest edge of the cuboid
Since,
$\dfrac{ab}{bc}=\dfrac{2}{4}$

$\dfrac{a}{c}=\dfrac{1}{2}$

$\therefore$  $c=2a$
Similarly,
$\dfrac{ca}{bc}=\dfrac{3}{4}$

$\dfrac{a}{b}=\dfrac{3}{4}$

$\therefore$  $b=\dfrac{4a}{3}$

Volume of cuboid,
$V=abc$
$\Rightarrow$  $9000=a\left(\dfrac{4a}{3}\right)(2a)$

$\Rightarrow$  $27000=8a^3$

$\Rightarrow$  $a^3=\dfrac{27\times 1000}{8}$

$\Rightarrow$  $a=\dfrac{3\times 10}{2}$

$\therefore$  $a=15\,cm$
Now, $b=\dfrac{4a}{3}=\dfrac{4\times 15}{3}=20$
$c=2a=2\times 15=30\,cm$
$\therefore$  The length of the shortest edge is $15\,cm.$

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

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. 5196 $m^3$

  2. 5200 $m^3$

  3. 5210 $m^3$

  4. 51220$m^3$

Show answer
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

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

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

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

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

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

Show answer
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.

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. 1212 $cu. m$

  2. 1414 $cu. m$

  3. 1616 $cu. m$

  4. 1818 $cu. m$

Show answer
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 $m$?

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $9\times10$

  2. $6\times8$

  3. $4.5\times6$

  4. $7.5\times10$

Show answer
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.

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $8\ cm^3$

  2. $6\ cm^3$

  3. $4\ cm^3$

  4. $2\ cm^3$

Show answer
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

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $160 cm^3$

  2. $1600 cm^3$

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

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

Show answer
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

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $5196 m^3$

  2. $5200 m^3$

  3. $5210 m^3$

  4. $5220 m^3$

Show answer
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

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. 120 c.c.

  2. 240 c.c.

  3. 640 c.c.

  4. 900 c.c.

Show answer
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?

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. $19$

  2. $20$

  3. $30$

  4. $40$

Show answer
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

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

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

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

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

  4. $6cm^3$

Show answer
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 $\frac{1}{6}  cm^3$.

Each side of the base of a square pyramid is reduced by $20%$. By what percent must the height be increased so that the volume of the new pyramid is the same as the volume of the original pyramid?

maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

  1. 20

  2. 40

  3. 46.875

  4. 56.25

  5. 71.875

Show answer
Correct Option: D
Explanation:
Let $a$ be the side of the square.
Length of side of square when reduced by $20\% = a-\dfrac{20a}{100}=0.8a$
Let $a _1=0.8a$
Volume of pyramid $V=\dfrac { 1 }{ 3 } \times $ Area of base $\times height=\dfrac{1}{3}A\times h$
Area of base with side $a = { a }^{ 2 }$ 
${ a }^{ 2 }=0.8a$

${V} _{ 1 }=\dfrac { 1 }{ 3 } \times { \left( { a } _{1} \right)  }^{ 2 }\times { h } _{ 1 }$ 
${V} _{ 1 }=V$ ....... [Given]

$\Rightarrow \dfrac { 1 }{ 3 } { a }^{ 2 }\times h=\dfrac { 1 }{ 3 } { \left( 0.8 \right)  }^{ 2 }{ a }^{ 2 }\times { h } _{ 1 }$

$\therefore { h } _{ 1 }=1.5625h$ 

$\implies \dfrac { { h } _{ 1 }-h }{ h } =1.5625-1=56.25%$

$\therefore$  'h' need to be increase by $56.25$ 
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