Tag: perimeter, area and volume

Questions Related to perimeter, area and volume

The volume of triangular prism whose adjacent sides are $\bar a,\ \bar b,\ \bar c$ each of magnitude $4$ units and each is inclined at an angel $\dfrac {\pi}{3}$ with other two is  

maths perimeter, area and volume surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

  1. $16\sqrt {2}$

  2. $16\sqrt {3}$

  3. $8\sqrt {3}$

  4. $8\sqrt {2}$

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Correct Option: B

If angle of minimum deviation through an equilateral prism is $40^o$, angle of incidence (being equal to angle of emergence) would be

maths perimeter, area and volume surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

  1. $50^o$

  2. $60^o$

  3. $40^o$

  4. none of these

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Correct Option: A
Explanation:
For minimum deviation $ i=ei=e $
$a+δm=i+ea+δm=i+e $
$a+δm=2ia+δm=2i $
$60^o+40^o=2i$
$i=50^o$

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 surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

  1. $120 cm^3.$

  2. $240 cm^3.$

  3. $640 cm^3.$

  4. $900 cm^3.$

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Correct Option: C
Explanation:
Given that:
Side $a=8\ cm$, Height $h=30\ cm$
As we know that
Volume of regular square pyramid 
$\Rightarrow a^2\dfrac{h}{3}$
$\Rightarrow 8^2\dfrac{30}{3}$
$\Rightarrow 64\times 10$
$\Rightarrow 640\ cm^3$
This is the required solution.

A square pyramid can contain $16\ m^3$ of water. The height of the pyramid is $3\ m$. Calculate the length of base of the square pyramid.

maths perimeter, area and volume surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

  1. $16\ m$

  2. $12\ m$

  3. $4\ m$

  4. $2\ m$

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Correct Option: C
Explanation:

Volume of square pyramid  $=16{ m }^{ 3 }$

$\Rightarrow \quad \dfrac { 1 }{ 3 } \times { a }^{ 2 }\times h=16{ m }^{ 3 }\Rightarrow \dfrac { 1 }{ 3 } \times { a }^{ 2 }\times 3=16\Rightarrow a=\sqrt { 16 } =4cm$
$\therefore $  Length of base $= 4m$

$VPQRS$ is rectangle based pyramid where $PQ = 30\ cm, QR = 20\ cm$ and volume is $2000\ {cm}^3$, then height (in cm) is

maths perimeter, area and volume surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

  1. $20$

  2. $40$

  3. $10$

  4. $30$

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Correct Option: C
Explanation:

Given : Length of base$(l)=30\ cm$, width of base$(h)=20\ cm$, Volume of pyramid$=2000\ cm^3$

Let $h$ be the height of the pyramid
We know that, volume of pyramid $=\dfrac{l\times w\times h}{3}$
$\implies 2000\ cm^3 = \dfrac{30 cm\times 20 cm\times h}{3}$
$\implies h=\dfrac{2000\times 3}{30\times 20} cm=10 cm$
Hence, height is $10 cm$.

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. $2345.98$

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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. $15$

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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. $10$

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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. $2589$

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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 $98\ cm^3$.