Mensuration - Class 8 Maths - Extra Questions

Let the lengths of the parallel sides of the trapezium be $$ a\ cm $$ and $$ a + 5  \ cm $$

Area of a trapezium $$ = \dfrac { 1 }{ 2 } \times (Sum\quad of\quad parallel\quad sides)\times height  $$

$$ \Rightarrow \dfrac { 1 }{ 2 } \times (a

+ a + 5) \times 18  cm  = 279 {cm}^{2} $$

$$ \Rightarrow \dfrac { 1 }{ 2 } \times (2a + 5)

\times 18 = 279 $$

$$2a+5=31$$

$$ \Rightarrow a = 13 $$

Hence, the lengths of the parallel sides of the trapezium are $$ a  = 13  cm $$ and $$ a + 5  = 18 cm $$

Thus length of smallest parallel side is $$13$$ cm

Area of a trapezium $$ = \frac { 1 }{ 2 } \times (Sum\quad of\quad
parallel\quad sides)\times height  $$

$$ => \frac { 1 }{ 2 } \times (3a + 4a) \times 9  dm  = 126 {dm}^{2} $$

$$ =>\frac { 1 }{ 2 } \times 7a \times 9 = 126 $$

$$ => a = 4 $$

Hence, the lengths of the parallel sides of the trapezium are $$ 3a  = 12  dm $$ and $$ 4a  = 16 dm $$

Area of a trapezium $$ = \frac { 1 }{ 2 } \times (Sum\quad of\quad

parallel\quad sides)\times height  $$

$$ => \frac { 1 }{ 2 } \times (3a + 4a) \times 2a  = 175 {cm}^{2} $$

$$ =>\frac { 1 }{ 2 } \times 7a \times 2a = 175 $$

$$ {a}^{2} = 25 $$

$$ => a = 5 $$

Hence, the height of the trapezium $$ = 2a = 10  cm $$

$$TSA = (2\pi r^2+2\pi r h ) \>and\>CSA = 2\pi r h \\now (\dfrac{2}{3})(2\pi r^2+2\pi r h )=2\pi r h \\(4\pi r^2+4\pi r h )=6\pi r h \\or\>4\pi r^2=2\pi r h\\2r=h\\now$$

$$=2\pi rh = (\dfrac{2}{3})231\\$$

$$=2\pi r \cdot2r = (\dfrac{2}{3})231\\$$

$$=r^2=(\dfrac{2\cdot 231}{3\cdot 4\pi })=(\dfrac{2\cdot 231 \cdot 7}{4\cdot 22\cdot 3})=(\dfrac{49}{4})\\\therefore r = (\dfrac{7}{2})cm\\and\>hence\>h =2r = 7cm$$

$${\textbf{Step -1: Calculating the area covered in one rotation}}{\text{.}}$$

                $${\text{To find how much area will be covered in one rotation,}}$$

                $${\text{we have to find the curved surface area of cylindrical roller}}{\text{.}}$$ 

                $${\text{Given that Diameter, D}} = 1.4{\text{ m}}{\text{.}}$$

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

                $${\text{so, radius r}} = \dfrac{{\text{D}}}{2} = \dfrac{{1.4}}{2} = 0.7{\text{m}}{\text{.}}$$

                $${\text{As per formula, curved surface area}} = 2\pi {\text{rh}}{\text{}}$$

                $$ = 2 \times \dfrac{{22}}{7} \times 0.7 \times 2$$

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

$${\textbf{Step -2: Calculating the area covered in 5 revolutions}}{\text{.}}$$

                $${\text{So in 1 revolution total 8}}{\text{.8 }}{{\text{m}}^2}{\text{ area is covered}}{\text{.}}$$

                $$\therefore {\text{ in 5 revolutions total area covered is 8}}{\text{.8}} \times {\text{5 }}{{\text{m}}^2}.$$

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

$${\textbf{Hence, It will cover total 44 }}{{\textbf{m}}^2}{\textbf{ in 5 revolutions}}{\text{.}}$$

$$let\>side\>of\>cube\>be\>x\>\\then\>x^3=343\\x^3=7^3\\\therefore\>x=7cm\\now\>surface\>area\>=6x^2\\=6\cdot49\\=294$$

We have,

Curved surface area of cylinder $$2\pi rh=88$$

And height $$h=14\,cm.$$

Then,

$$ 2\pi rh=88 $$

$$ 2\pi r\times 14=88 $$

$$ r=\dfrac{88\times 7}{28\times 22} $$

$$ r=1 $$

Then, Diameter

$$ d=2r $$

$$ d=2\times 1=2\,cm. $$

Hence, this is the answer.

$$\textbf{Step -1: Find the total surface area of first cylinder.}$$

                

                

                  $$\text{Diameter, }d=2\text{ m}$$

                  $$\text{Radius, }r=1\text{ m}$$

                  $$\text{Height, }h=2\text{ m}$$

                  $$\therefore \text{Total surface area of first cylinder}=\pi (1)(2\times2+1)$$

                  $$=\pi\times5$$

                  $$=\dfrac{22}{7}\times5$$

                  $$=\dfrac{110}{7}\mathrm{m^2}$$

$$\textbf{Hence, the total surface area of first and second cylinders are }\mathbf{1320 cm^2\text{ and }\dfrac{110}{7}m^2\text{ respectively.}}$$

Volume of the given cuboid $$=3456^{3}cm$$.

Length of the given cuboid $$=24 cm$$

Breadth of the given cuboid $$=18 cm$$

Here

$$Length × Breadth × Height$$ = Volume of a cuboid

Substituting the values

$$24×18× $$Height =3456

By further calculation

Height =$$\dfrac{3456}{24\times 18}$$

So we get

Height =$$\dfrac{3456}{432}$$

Height $$=8cm$$

Volume of a cuboid =$$7.68 m^{3}$$

Length of a cuboid $$=3.2 m$$

Height of a cuboid $$=1.0 m$$

Here

$$Length \times  Breadth \times Height = Volume \,\,of \,\,a \,\,cuboid$$

Substituting the values

$$3.2 \times  Breadth \times  1.0=7.68$$

By further calculation

$$\rightarrow Breadth=\dfrac{7.68}{3.2\times 1.0}$$

so we get 

$$\rightarrow Breadth=\dfrac{7.68}{3.2}$$

$$\rightarrow Breadth= 2.4 cm $$

Consider length of the cuboid $$=6x$$

Breadth of the cuboid $$=5x$$

Height of the cuboid $$=3x$$

We know that

Total surface area of the given cuboid =$$2(1\times b+b\times h+h\times l)$$

Substituting the values

=$$2(6x\times 5x+5x\times 3x+3x\times 6x)=x(30\times 2+15\times 2+18\times 2)$$

We get

$$=(60+30+36)x=126x^{2}$$

It is given that

Total surface area of the given cuboid  $$504cm^3$$

Substituting the values

$$126x^{2}=504cm^{2}$$

By further calculation

$$\Rightarrow x^{2}=\dfrac{504}{126}$$

So we get 

$$\Rightarrow x^{2}=4\Rightarrow x=\sqrt{4}$$

$$x = 2 cm$$ 

Here

Length of the cuboid $$=6x=6\times 2=12cm$$

Breadth of the cuboid $$=5x=5\times 2=10cm$$

Height of the cuboid =  $$=3x=3\times 2=6cm$$

We get

Volume of the cuboid $$=l \times  b \times  h=12 \times 10 \times  6=720cm^{3}$$

Edge of the given cube =8cm

We know that

Volume of the given cube  $$=( Edge )^{3}=(8)^{3}=8\times 8\times 8=512cm^{3}$$

Total surface area of a cube = $$6( Edge )^{2}=6 \times  (8)^{2}=384cm^{2}$$

Edge of the given cube $$=2 m 40 cm=2.40 m$$

We know that

Volume of a cube  $$=( Edge )^{3}$$

Substituting the values

Volume of the given cube $$=(2.40)3=2.40 \times 2.40 \times 2.40=13.824m^{3}$$

Total surface area of the given cube $$=6\times 2.4\times 2.4=34.56m^{2}$$