Tag: perimeter, area and volume

Questions Related to perimeter, area and volume

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

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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

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

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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. $960$

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

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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. $10cm$

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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 $6 \times 8$.