Surface Areas And Volumes - Class 9 Maths - Extra Questions

Total surface area of a cuboid $$ = 2(l \times b + b

\times h + l \times h) $$

Let $$A$$ be the total surface area of the cuboid with length $$ = 16  $$cm, breadth $$ = 14 $$ cm and height $$ = 20$$ cm 

Thus,

$$A= 2(16 \times 14 +

14 \times 20 + 20 \times 16) $$

    $$= 2(224 + 280 + 320) $$

    $$= 1648  {cm}^{2} $$

Given, diameter $$d=28\ cm$$.

$$\therefore$$ radius $$ = \dfrac {d}{2} = \dfrac {28}{2} = 14\ cm $$.

Then, surface area of a sphere of radius $$(r)$$ $$ = 4\pi { r }^{ 2 }$$

$$= 4 \times \dfrac {22}{7} \times 14\times 14 $$

$$= 2464 \  {cm}^{2} $$.

Therefore, the required surface area of the sphere is $$2464  \ {cm}^{2} $$.

$$Volume =length\times breadth\times height$$

$$Height  = \dfrac{Volume}{length\times breadth} $$

$$\Rightarrow \dfrac{14175cm^3}{35cm\times 15cm} $$

$$\Rightarrow 27cm$$

$$\therefore  Height = 27 \ cm$$

Consider the following question.

Given  data.

Length of cuboidal vessel=$$ 10m $$

Width of cuboidal vessel $$= 8m $$

Volume of cuboidal vessel  $$=80{{m}^{3}} $$

Volume of cuboidal vessel $$=L*B*H $$

$$H=\dfrac{80}{L*B} $$

$$=\dfrac{80}{10*8} $$

$$=1m $$

Hence,  This is the required answer.

Inner radius, $$r = 6\ cm$$

Thickness, $$t = 1\ cm$$

Outer radius, $$R = 6+1 = 7\ cm$$

Therefore,

Volume of steel required = $$\dfrac{2}{3} \pi R^{3}-\dfrac{2}{3}\pi r^{3}$$

$$=\dfrac{2}{3} \times \dfrac{22}{7} \times 7^{3} -\dfrac{2}{3} \times \dfrac{22}{7} \times 6^{3}$$

$$=\dfrac{5588}{21}$$ $$cm^{3}$$

Inner radius $$R$$ of the bowl  = $$ 9$$cm

Height, $$h$$ of bottles =$$4$$cm

Let no. of bottles required be $$ x$$

 Vol of $$x$$  bottle = vol. of hemisphere

$$\pi r^2h x=\cfrac{1}{2}\times \cfrac{4}{3} \pi R^3$$

$$x=\cfrac{324}{13.5}=54$$

The tank is in the shape of a cuboid.
To find the area of the paper required, we need to calculate the surface area of the cuboidal tank and subtract the area of the top side and the front side.

Total surface area(TSA) of a cuboid of length $$l$$, breadth $$b$$ and height $$h$$ $$ = 2(l \times b + b
\times h + l \times h) $$
 Given length $$ l= 80\ cm $$ ,breadth $$ b= 40\ cm $$  and height $$ h= 30\ cm$$  

Total Surface Area of Cuboid

$$= 2(80 \times 40 + 40
\times 30 + 30 \times 80)\\ = 2(3200 + 1200 + 2400)\\ = 13600\  {cm}^{2} $$

Area of the top side $$ = 80 \times 40 = 3200\  {cm}^{2} $$

Area of the front side $$ = 80 \times 30 = 2400\  {cm}^{2} $$

Hence. area of the paper required $$ = 13600 - 3200 - 2400 = 8000\ {cm}^{2} $$

Total surface area of a cube of edge a $$ = 6{a}^{2} $$
So, the total surface area of the cube of edge a $$ = 25  cm $$ is $$ =

6{(25)}^{2} = 3,750\   {cm}^{2} $$

$${\textbf{Step  - 1: Calculating volume of earth and moon in terms of diameter}}$$

                    $${\text{We know , volume of sphere in terms of diameter  =  }}\dfrac{{{\text{ }\pi}{{\text{d}}^{\text{3}}}}}{{\text{6}}}$$

                    $$ \Rightarrow {\text{ Volume of earth  =  }}\dfrac{{{\text{  }\pi}{{\text{d}}_{\text{e}}}^{\text{3}}}}{{\text{6}}}$$

                    $$ \Rightarrow {\text{ Volume of moon  =  }}\dfrac{{{\text{  }\pi}{{\text{d}}_{\text{m}}}^{\text{3}}}}{{\text{6}}}$$

$${\textbf{Step  - 2: Calculating the ratio of volume of moon to volume of earth}}$$

                    $$ \Rightarrow {\text{ }}\dfrac{{{\text{Volume of moon}}}}{{{\text{Volume of earth}}}}{\text{  =  }}\dfrac{{\dfrac{{{\text{ }\pi}{{\text{d}}_{\text{m}}}^{\text{3}}}}{{\text{6}}}}}{{\dfrac{{{\text{ }\pi}{{\text{d}}_{\text{e}}}^{\text{3}}}}{{\text{6}}}}}$$

                    $$ \Rightarrow {\text{ }}\dfrac{{{\text{Volume of moon}}}}{{{\text{Volume of earth}}}}{\text{  =  }}\dfrac{{{{\text{d}}_{\text{m}}}^{\text{3}}}}{{{{\text{d}}_{\text{e}}}^{\text{3}}}}$$

                    $$ \Rightarrow {\text{ }}\dfrac{{{\text{Volume of moon}}}}{{{\text{Volume of earth}}}}{\text{  =  }}{\left( {\dfrac{{{{\text{d}}_{\text{m}}}}}{{{{\text{d}}_{\text{e}}}}}} \right)^{\text{3}}}$$

                    $${\text{According to question, }}$$

                    $$\text{The diameter of the moon is approximately one-fourth of the diameter of the earth.}$$

                    $$\Rightarrow{{\text{d}}_{\text{m}}}{\text{  =  }}\dfrac{{{{\text{d}}_{\text{e}}}}}{{\text{4}}}$$

                    $$\therefore {\text{ }}\dfrac{{{{\text{d}}_{\text{m}}}}}{{{{\text{d}}_{\text{e}}}}}{\text{  =  }}\dfrac{{\text{1}}}{{\text{4}}}$$

                    $$ \Rightarrow {\text{ }}\dfrac{{{\text{Volume of moon}}}}{{{\text{Volume of earth}}}}{\text{  =  }}{\left( {\dfrac{{\text{1}}}{{\text{4}}}} \right)^{\text{3}}}$$

                    $$ \Rightarrow {\text{ }}\dfrac{{{\text{Volume of moon}}}}{{{\text{Volume of earth}}}}{\text{  =  }}\dfrac{{\text{1}}}{{{\text{64}}}}$$

$$\mathbf{{\text{Hence, the volume of moon is }}\dfrac{{\text{1}}}{{{\text{64}}}}{\text{th times volume of earth}}{\text{.}}}$$

$$\textbf{Step - 1: Compare both volume and find the radius}$$

                   $$\text{Volume of twenty seven solid spheres =  } 27\times \dfrac{{4}}{{3}}\pi r^3$$

                                                                                         $$ = 36 \pi r^3 .............(1)$$

                   $$\text{Volume of new spheres  }= \dfrac{{4}}{{3}}\pi r'^3 ..........(2)$$

                   $$\text{Compare } (1) \text{ & } (2)$$

                   $$36 \pi r^3  = \dfrac{{4}}{{3}}\pi r'^3$$

                   $$27 r^3 = r'^3$$

                   $$\text{Taking, cube root on both sides}$$

                   $$3r=r'$$

                   $$r'=3r$$

$$\textbf{Step - 2: Ratio of Surface area}$$

                   $$\text{Surface area of the sphere,S is } 4\pi r^2$$

                   $$4\pi r^2 = 4\pi r'^2$$

                   $$ r^2 = (3r)^2$$

                  $$\text{The ratio of } S \text{ and } S' = 1:9$$

$$\textbf{Hence , the radius of new sphere is } \mathbf{3r}, \textbf{ and the ratio of } \mathbf{S} \textbf{ and } \mathbf{S'} \textbf{ is 1:9}$$

Diameter of the spherical marble $$= 1.4$$ cm

Radius of the marble $$= 0.7$$ cm

Volume of each marble = $$\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\times \dfrac{22}{7}\times (0.7)^3=1.44$$

Volume of $$150$$ marbles $$= 1.44 x 150 = 216$$ $$cm^3$$

Let the rise in level of water in the cylindrical vessel $$= h$$ cm

Diameter of the vessel $$= 7$$ cm

Radius of the vessel $$= 3.5$$ cm

Volume of increased level of the water = $$\pi r^2h =\dfrac{22}{7}\times 3.5^2\times h$$

Volume of increased level of the water = Volume of $$150$$ marbles

$$\dfrac{22}{7}\times 3.5^2\times h=216$$

$$h = 5.6$$ cm

Therefore, the rise in the level of water in the vessel $$= 5.6$$ cm.

$$  {\textbf{Step 1: Finding Curved surface Area}} $$

               $$  {\text{Given,}} $$

               $$  {\text{The diameter of a garden roller is 1}}{\text{.4 m and it is 2 m long}}{\text{.}} $$

               $$  {\text{So, radius}(r) = }\dfrac{{{\text{diameter}(d)}}}{{\text{2}}} $$

               $$   = \dfrac{{1.4}}{2} $$

               $$   = 0.7\text{ m} $$

               $$  {\text{and }h = 2\text{ m}} $$

               $$  {\text{Curved surface  =  2}}\pi {\text{rh}} $$

               $$  {\text{ = 2}} \times \frac{{22}}{7} \times 0.7 \times 2 $$

               $$   = 8.8{\text{ m}^2} $$

$$  {\textbf{Step 2: Finding required area}} $$

               $$  {\text{Hence,}} $$

               $$  {\text{Area covered  =  5}} \times 8.8 $$

               $$   = 44{\text{ m}^2} $$

$$  {\textbf{Hence, the area covered is 44}}{{\text{m}}^2}. $$

T.S.A of cube $$= 6a^2$$

                        $$= 6 \times  3.5 \times  3.5$$

                        $$ = 6 \times  12.25= 73.5 m^2$$

Since we have to paint inside and outside therefore total area $$= 73.5 \times  2 = 147 m^2$$

Now we know that cost of painting  $$= Rs. 75$$  per $$m^2$$

Cost of painting for $$147m^2$$ $$= 147 \times  75 = Rs. 11025$$

Volume of new cube $$=$$ Sum of volume of three cubes

                                   $$= 3^2 + 4^2 + 5^2$$

                                   $$= 216\ cm^3$$

i.e.,              $$ (edge) ^3= 216\ cm^3$$

So, Edge of the single cube $$= 6\ cm$$

And, Surface area of the cube  $$= 6(edge)^2= 6\times 6^2 = 216\ cm^2$$ 

Let $$l$$ be the length, $$b$$ be the breadth and $$h$$ be the depth of the cuboid.

Then according to the question,

$$l + b + h = 19$$

And the length of the diagonal is,

$$\sqrt {{l^2} + {b^2} + {h^2}}  = 11$$

$${l^2} + {b^2} + {h^2} = {\left( {11} \right)^2}$$

$${l^2} + {b^2} + {h^2} = 121$$

The surface area of the cuboid is,

$$A = 2\left( {lb + bh + lh} \right)$$

$$ = {\left( {l + b + h} \right)^2} - \left( {{l^2} + {b^2} + {h^2}} \right)$$

$$ = {\left( {19} \right)^2} - 121$$

$$ = 361 - 121$$

$$ = 240$$

Therefore, the surface area of the cuboid is $$240\;{\rm{c}}{{\rm{m}}^2}$$.

$${\textbf{Step  - 1 : Find the area of wall}}.$$

                     $${\text{Given ,the length of the wall is}}(l) = 4.5{\text{m and the breadth}}(b) = 3.6{\text{ m}}.$$

                     $${\text{So the area of wall = }}l \times b.$$

                                                        $$ = 4.5 \times 3.6$$

                                                        $$ = 16.2{\text{ }}{{\text{m}}^2}$$

$${\textbf{Step  - 2 : Find the area of door frame}}{\text{.}}$$

                     $${\text{Given that , The door is of length(}}{l_d}{\text{) = 2 m}}{\text{ and breadth}}({b_d}) = 1\text{m}.$$

                     $${\text{Area of the door in total  = }}{l_d} \times {b_d} = 2 \times 1 = 2{\text{m}^2}.$$

$${\textbf{Step  - 3 : Find the area to be white washed and the cost for it}}{\text{.}}$$

                     $${\text{The area to be white washed  =  The total area of wall  -  the area covered by door}}{\text{.}}$$

                     $$ \Rightarrow {\text{Area to be white washed  = 16}}{\text{.2 - 2}} = 14.2{\text{m}^2}.$$

                     $${\text{Given tha the rate white washing the wall is Rs}}{\text{. 20 per }}{\text{m}^2}.$$

                     $${\text{So total cost will be = }}14.2 \times 20 = \text{Rs.}284.$$

$${\textbf{ So , total cost for white washing the wall is Rs}}{\textbf{. 284}}{\text{.}}$$

The surface area of sphere $$=616\,c{{m}^{2}}$$

Since, the surface area of sphere $$=4\pi {{r}^{2}}$$

Therefore,

$$ 4\pi {{r}^{2}}=616 $$

$$ \pi {{r}^{2}}=154 $$

$$ {{r}^{2}}=49 $$

$$ r=7\,cm $$

Hence, the radius of the sphere is $$7\,cm$$.

$${\textbf{Step  -  1 : Assume the one arm of cube to calculate volume }}$$

                      $${\text{Let's assume the length of one arm of cube is }}a{\text{ unit}}{\text{.}}$$

                      $${\text{We know a cube has arms of same length}}{\text{.}}$$

                      $${\text{So , the volume is  =  }}a \times a \times a = {a^3}.$$

$${\textbf{Step  -  2 : Find the volume of sphere which exactly fit inside the cube}}{\text{.}}$$

                      $${\text{As given the sphere exactly fit inside the cube}}{\text{.So the radius will r}} = \dfrac{a}{2}.$$

                      $${\text{As per rule radius of a sphere is }}\dfrac{4}{3}\pi {r^3}.$$

                      $${\text{So the given sphere's radius will be  =  }}\dfrac{4}{3} \times \pi  \times {\left( {\dfrac{a}{2}} \right)^3},$$

                      $$ = \dfrac{4}{3} \times \dfrac{{22}}{7} \times \dfrac{{{a^3}}}{8}.$$

                      $$ = \dfrac{{11}}{{21}}.{a^3}.$$

$${\textbf{Step  -  3 : Calculate their ratio}}{\text{.}}$$

                      $${\text{So the ratio of volume of cube and volume of sphere is   }}$$

                      $$ = {a^3}:\dfrac{{11{a^3}}}{{21}} = 21:11.$$

$${\textbf{Thus, the ratio of the volume of a cube to that of a sphere which }}$$$${\textbf{exactly fit inside the cube is 21:11}}{\text{.}}$$

We have

Volume of Sphere $$=\ 4851\ c{{m}^{3}}$$

We know that, the volume of Sphere

$$ =\ \dfrac{4}{3}\pi {{R}^{3}} $$

$$ \text{So,} $$

$$ \dfrac{4}{3}\pi {{R}^{3}}\ =\ 4851 $$

$$ {{R}^{3}}=\ \dfrac{3\times 485\times 7}{4\times 22} $$

$$ R=\ \sqrt[3]{1157.625} $$

$$ R=\ 10.5\ cm. $$

Now, for volume of the sphere $$=\dfrac{4312}{3} cm^3$$

$$ \dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4312}{3} $$

$$ \pi {{r}^{3}}=\dfrac{4312}{4} $$

$$ {{r}^{3}}=\ 1078\times \dfrac{7}{22} $$

$$ r=\ \sqrt[3]{343} $$

$$ r\ =7 $$

Now, Change in radius

$$R-r=\ 10.5-7=3.5\ cm$$

Hence, this is the answer.

$${\textbf{Step  - 1: Find the new radius of the sphere}}$$

                    $${\text{Given, initial diameter  =  d, initial radius  =  }}\dfrac{{\text{d}}}{{\text{2}}}$$

                    $${\text{Diameter is decreased by 25% ,}}$$

                    $$ \Rightarrow {\text{ }}{{\text{d}}_{\text{N}}}{\text{  =  d  -  (25%  of d)}}$$

                    $$ \Rightarrow {\text{ }}{{\text{d}}_{\text{N}}}{\text{  =  d  -  }}\dfrac{{{\text{25d}}}}{{{\text{100}}}}$$

                    $$ \Rightarrow {\text{ }}{{\text{d}}_{\text{N}}}{\text{  =  d  -  }}\dfrac{{\text{d}}}{{\text{4}}}$$

                    $$ \Rightarrow {\text{ }}{{\text{d}}_{\text{N}}}{\text{  =  }}\dfrac{{{\text{3d}}}}{{\text{4}}}$$

                    $$\therefore {\text{ New radius, }}{{\text{r}}_{\text{N}}}{\text{  =  }}\dfrac{{{{\text{d}}_{\text{N}}}}}{{\text{2}}}{\text{  =  }}\dfrac{{{\text{3d}}}}{{\text{8}}}{\text{  =  }}\dfrac{{{\text{3r}}}}{{\text{4}}}$$

$${\textbf{Step  - 2: Calculate surface areas}}$$

                    $${\text{We know that, surface area of sphere  =  4}}\pi {{\text{r}}^{\text{2}}}$$

                    $$\therefore {\text{ Initial surface area, }}{{\text{S}}_{\text{I}}}{\text{  =  4}}\pi {{\text{r}}^{\text{2}}}$$

                    $${\text{Surface area of new sphere, }}{{\text{S}}_{\text{N}}}{\text{  =  4}}\pi {{\text{r}}_{\text{N}}}^{\text{2}}$$

                    $$ \Rightarrow {\text{ }}{{\text{S}}_{\text{N}}}{\text{  =  4}}\pi {\left( {\dfrac{{{\text{3r}}}}{{\text{4}}}} \right)^{\text{2}}}$$

                    $$ \Rightarrow {\text{ }}{{\text{S}}_{\text{N}}}{\text{  =  }}\dfrac{{{\text{9}}\pi {{\text{r}}^{\text{2}}}}}{{\text{4}}}$$

$${\textbf{Step  - 3: Calculate the percentage decrease in the surface area}}$$

                    $${\text{Percentage decrease, P  =  }}\dfrac{{{\text{Change}}}}{{{\text{Initial}}}}{\text{ }} \times {\text{ 100}}$$

                    $$ \Rightarrow {\text{ P  =  }}\dfrac{{{\text{4}}\pi {{\text{r}}^{\text{2}}}{\text{  -  }}\dfrac{{{\text{9}}\pi {{\text{r}}^{\text{2}}}}}{{\text{4}}}}}{{{\text{4}}\pi {{\text{r}}^{\text{2}}}}}{\text{ }} \times {\text{ 100}}$$

                    $$ \Rightarrow {\text{ P  =  }}\dfrac{{{\text{7}}\pi {{\text{r}}^{\text{2}}}}}{{{\text{4(4}}\pi {{\text{r}}^{\text{2}}}{\text{)}}}}{\text{ }} \times {\text{ 100}}$$

                    $$ \Rightarrow {\text{ P  =  }}\dfrac{{\text{7}}}{{{\text{16}}}}{\text{ }} \times {\text{ 100}}$$

                    $$ \Rightarrow {\text{ P  =  43}}{\text{.75% }}$$

$${\textbf{The percentage decrease in the surface area of sphere is 43}}{\textbf{.75%.  }}$$