ML AGGARWAL CLASS 8 Chapter 18 Mensuration Exercise 18.4

 Exercise 18.4

Question 1

Sol :

Given surface area = $384 \mathrm{cm}^{2}$

(i) Let length of side of cube = a 

Surface area of cube = $6 a^{2}$

$6 a^{2}=384$

$a^{2}=\frac{384}{6}$

$a^{2}=64$

$a=\sqrt{64}$

a=8 cm

∴ Length of edge = 8cm

(ii) Volume of the cube 

Volume of the cube = $a^{3}$

$=8^{3}$

Volume of the cube $=512 \mathrm{~cm}^{3}$

Question 2

Sol :

Given 

Radius of Cylinder = 5 cm

Height of Cylinder = 10 cm

Surface area of Cylinder =$2 \pi r h$

$=2 \pi \times 5 \times 10$

 Surface area of Cylinder $=100 \pi$

Question 3

Sol :

Given 

Aquarium Dimensions = $70C m \times 28 \mathrm{~cm} \times 35 \mathrm{~cm}$

To cover Base , Side and back faces Total area of 

Paper needed = $2(l b+b h+lh)$

=$2(70 \times 28+28 \times 35 \times 10 \times 35)$

=$2(5390)$

=$10780 \mathrm{~cm}^{2}$

Question 4

Sol :

Given 

Internal dimension of hall = $15 \mathrm{~m} \times 12 \mathrm{~m} \times 4 \mathrm{~m}$

Area of four walls = l h+b h+l h+lh

=2(l h+ b h)

$=2(15 \times 4+12 \times 4)$

=2(60+48)

$=2 \times 108$

Area of four walls = $216 \mathrm{m}^{2}$

Given 

4 Windows of dimension = $2 m \times 1.5 m$

2 doors of dimension = $1.5 \times 2.5 \mathrm{~m}^{2}$

∴ Remaining walls area = Area of four walls - [4 $\times$ Area of window + 2 $\times$ area of door]

$=216-[4 \times 2 \times 1.5+2 \times 1.5 \times 2.5]$

=216-[12+7.5]

=216=19-5

∴Remaining walls area $=196.5 \mathrm{~m}^{2}$

Given

Cost for white washing walls = Rs $5 / m^{2}$

Total Cost of white washing walls = $5 \times 196.5$

$=Rs 982.5$

Area of Ceiling = lb = 15 $\times$ 12

= $180 \mathrm{~m}^{2}$

Total area = Area of walls +  Area of ceiling 

= 196.5 + 180

Total Area = $376.5 \mathrm{~m}^{2}$

Total Cost of white washing walls including ceiling 

$=5 \times 376.5$

=Rs 1882.5

Question 5

Sol :

Swimming Pool length = 50m

 Breadth = 30m

Height = 2.5m

Area of walls and base = lb +lh + bh+ lh+bh

$=2(1 h+b h)+l b$

$=2(50 \times 2.5+30 \times 2.5)+50 \times 30$

$=400+1500$

Area of walls and base $=1900 \mathrm{~m}^{2}$

Given Cementing rate = Rs 27/$m^{2}$

∴ Total cost for cementing = $27 \times 1900$

=Rs 51300

Question 6

Sol :

Given rectangular hall perimeter = 236m

Hall height = 4.5m

Surface area of walls = 2 h(l+b)

$4 \cdot 5 \times 236$

Surface area of walls = 1062$m^{2}$

Painting walls cost = Rs $8.4 / m^{2}$

Total Cost of painting = $8.4 \times 1062$

Total Cost of painting= Rs 8920.8

Question 7

Sol :

Dimension of fish tank= $300 \mathrm{~m} \times 20 \mathrm{~cm} \times 20 \mathrm{~cm}$

Given Only $\frac{3}{4}$ th of tank contain water 

 

∴ Volume of water = $30 \mathrm{~m} \times 20 \mathrm{~m} \times 20 \times \frac{3}{4} \mathrm{Cm}$

Volume of water $=30 \mathrm{~cm} \times 20 \mathrm{~cm} \times 15 \mathrm{Cm}$

Area of tank in contact with water = walls area up to water level + base area 

=2 h(l+b)+l b

$=2(30+20) \times 15+30 \times 20$

=1500+600

Area of tank in contact with water $=2100 \mathrm{~cm}^{2}$

Question 8

Sol :

Given 

Volume of cuboid = $448 \mathrm{~cm}^{3}$

Let side of square = a cm

Height = 7cm

$\begin{aligned} a^{2} \times 7 &=448 \\ a^{2} &=\frac{448}{7} \\ a^{\nu} &=64 \\ a &=\sqrt{64} \\ & a=8 \mathrm{~cm} \end{aligned}$

Side of square base = 8cm

(ii) Surface area cuboid = $2\left(a^{2}+2 a h\right)$

$=2\left(8^{2}+2 \times 8 \times 7\right)$

=2(176)

Surface of area of Cuboid = $352 \mathrm{~cm}^{2}$

Question 9

Sol :

Given 

Total surface area of rectangle solid = $1216 \mathrm{~cm}^{3}$

Ratio of length, breadth and height = 5:4:2

Let length , breadth and height = 5x, 4x , 2x

Total surface area = 1216

2(l b+b h+h l)=1216

$2(5 x \times 4 x+4 x \times 2 x+2 x \times 5 x)=1216$

$2\left(20 x^{2}+8 x^{2}+10 x^{2}\right)=1216$

$2 \times 38 x^{2}=1216$

$\begin{aligned} 76 x^{2} &=1216 \\ x^{2} &=\frac{1216}{76} \\ x^{2} &=16 \\ x &=\sqrt{16} \\ x &=4 \mathrm{~cm} \end{aligned}$

∴ Length 5x = $5 \times 4=20 \mathrm{~cm}$

breadth 4x = $4 \times 4=16 \mathrm{~cm}$

Height 2x = $2 \times 4=8 \mathrm{~cm}$

Volume of rectangular solid = $l \times b \times h$

$=20 \times 16 \times 8$

Volume of rectangular Solid = $2560 \mathrm{~cm}^{3}$

Question 10

Sol :

Dimension of room = $6 \times 5 \times 3.5 \mathrm{~m}^{3}$

Dimension of window = $1.5 \mathrm{~m} \times 1.4 \mathrm{~m}$

Dimension of door = $1.1 \mathrm{~m} \times 2 \mathrm{~m}$

Area of walls = $2 h(1+h)-[3 \times 1.5 \times 1.4+2 \times 1.1 \times 2]$

$=2 \times 3.5(6+5)-[6.3+4.4]$

=77-10.7

Area of walls = $66.3 \mathrm{~m}^{2}$

Area of ceiling = lb = $6 \times 5=30 \mathrm{~m}^{2}$

Total area = $66.3+30=96.3 \mathrm{~m}^{2}$

Cost of white washing = Rs $5.3 / \mathrm{m}^{2}$

Total Cost = area $\times$ cost /$m^{2}$

$=96.3 \times 5.3$

Total cost =₹ 510.39

Question 11

Sol :

Given 

Dimension of Cuboidal block = $36 \mathrm{~cm} \times 32 \mathrm{~cm} \times 0.25 \mathrm{Cm} .$

(i) Volume of Cuboidal block =

 $\begin{aligned} & 36 \times 32 \times 25 \\=& 28800 \mathrm{~cm}^{3} \end{aligned}$

Cube of edge = 4cm

Volume of Cube = $4^{3}$

$=64 \mathrm{~cm}^{3}$

No. of Cubes = $\frac{\text { Volume of Cuboidal block }}{\text { volume of cube }}$

=$\frac{28800}{64}$

No. of Cubes = 450

∴ From given Cuboid 450 Cubes of edge 4cm Can be costed.

(ii) Cost silver coating = Rs 0.75/$m^{2}$

Surface area of cube = $6 a^{2}$

$=6 \times 4^{2}$

$=6 \times 16$

Surface Area of cube= $96 \mathrm{~cm}^{2}$

Total surface area of cubes = $450 \times 96$

thrm{~cm}^{2}$

$=43200 \mathrm{~cm}^{2}$

Total Surface Area of Cubes = 4.32 $m^{2}$

Cost of silver coating fall cubes = $4.32 \times 0.75 \times 10^{4}$

=Rs 32400

∴ Total Cost for Silver coating of Cubes is Rs 32400

Question 12

Sol :

Given three cubes of edge lengths $=3 \mathrm{~cm}, 4 \mathrm{~cm}, 5 \mathrm{~cm}$

New cube edge length $=a \mathrm{~cm}$

$a^{3}=3^{3}+4^{3}+5^{3} .$

$a^{3}=216$

$a=(216)^{1 / 3}$

A surface area of cube = $6 a^{2}$

$=6 \times 6^{2}$

Surface area of cube $=216 \mathrm{~cm}^{2}$

Cost of gold coating = Rs 3.5 / $\operatorname{Cm}^{2}$

Total Cost of gold coating of cube

 

= area $\times$ cost /$\mathrm{cm}^{2}$

$=216 \times 3.5$

$=Rs 756$

∴ Total Cost of gold coating of cube = Rs 756

Question 13

Sol :

Given

Surface area of cylinder = $4375 \mathrm{~cm}^{2}$

Rectangular sheet width = 35cm

perimeter of circle = 35 cm

$\begin{aligned} 2 \pi r &=35 \\ r &=\frac{35}{2 \pi} \end{aligned}$

radius of base $(r)=5.57 \mathrm{~cm}$

Surface area= $2 \pi r h=4375$

$\begin{array}{rl}=211 \times \pi \mathrm{hh} & 24375 \\ =35 \times h & =4375 \\ h =  \frac{4375}{35}\end{array}$

h =125 cm

Height of cylinder = 125 cm

∴ Length of sheet = 125 cm

Perimeter of sheet= $\begin{aligned} & 2(l+W) \\=& 2(125+35) \Rightarrow 2(160) \\=& 320 \mathrm{~cm} \end{aligned}$

Question 14

Sol :

Road roller diameter = 0.7m

Road roller width = 1.2m

Play ground size $= 120 \mathrm{~m} \times 44 \mathrm{~m}$

Area of ground = 5280$m ^2$

Surface area of roller =$\pi d W$

$=\pi \times 0.7 \times 1.2$

Surface area of roller = 2.538$m^{2}$

No. of revolutions to cover ground = $\frac{\text { Area of ground }}{\text { surface area of roller }}$

$=\frac{5280}{2.638}$

= $2000 \cdot 8 \approx$ 2000

∴ No , minimum no. of revolutions to cover ground is 200D

Question 15

Sol :

Given 

Diameter of cylindrical container= 14cm

Height of cylindrical container= 20cm

Label Height = 20 - (2+2)

= 16cm

∴ Area of label = surface area of cylinder of height 16cm

$=\pi d h$

$=\pi \times 14 \times 16$

=$\frac{22}{7} \times 14 \times 16$

∴ Area of label = $=704 \mathrm{~cm}^{2}$

Question 16

Sol :

Given 

Sum of radius and height of cylinder = 37cm

r+h= 37---1

Total surface area = 1628$\mathrm{cm}^{2}$

$\begin{aligned} 2 \pi r h &=1628 \\ r h &=\frac{1628}{2 \times 14} \\ r h &=\frac{1628 \times 7}{2 \times 22} \\ r h &=259 \end{aligned}$

(37-h) h=25 q

$37 h-h^{2}=259$

$h^{2}-37 h+259=0$

$h_{1}=27.63$

$r_{1}=37-27.63$

$r_{1}=9.37 \mathrm{~cm}$

Volume of Cylinder 

$=\pi r_{1}^{2} h_{1}$

$=11 \times 9.37^{2} \times 27.63$

$v_{1}$ $=7620.96 \mathrm{~cm}^{3}$

$h_{2} =9.37$

$r_{2}=37-9.37$

$r_{2}=27.63 \mathrm{~cm}$

Volume of cylinder 

$=\pi r_{2}^{2} h_{2}$

$=\pi \times27.63^{2} \times 9.32$

$V_{2}=22472.5 \mathrm{~cm}^{3}$

Question 17

Sol :

Given 

Ratio between surface area and total surface area = 1:2

Total surface area = $616 \mathrm{~cm}^{2}$

$2 \pi r h: 2 \pi r(h+r)=1: 2$

$\frac{h}{h+r}=\frac{1}{2}$

2 h=h+r

h=r

∴ Height = radius 

Total surface area = $616 \mathrm{~cm}^{2}$

$2 \pi r(h+r)=616 \mathrm{~cm}^{2}$

$2 \pi r(r+r)=616 \mathrm{~cm}^{2}$

$2\left(2 \pi r^{2}\right)=616 c m^{2}$

$2 \pi r^{2}=308 c m ^{2}$

$r^{2}=\frac{308}{2 \pi}$

$r^{2}=49$

$r= \sqrt{49}$

r=7 cm

r=h=7 cm

Volume of Cylinder = $\pi r^{2} h$

$=\pi(7)^{2} \times 7$

Volume of cylinder = $1077.56 \mathrm{~cm}^{3}$

Question 18

Sol :

Length of Cylinder = 77 cm=h

Inner diameter $d_{1}=4 c m$

Outer diameter $\left(d_{2}\right)$=4.4 cm

(i)Inner curved Surface area 

$=\pi d_{1} h$

$=\pi \times 4 \times 77$

$=967.61 \mathrm{~cm}^{2}$

(ii) Outer curved surface area 

$=\pi d_{2} h$

$=\pi \times 4.4 \times 17$

$=1064.37 \mathrm{~cm}^{2}$

(iii) Total surface area

=$\pi d _{1} h+\pi d_{2} h+2 \times \pi\left(r_{2}^{2}-r_{1}^{2}\right)$

=$967.63+1064.37+2 \pi\left(2.2^{2}-2^{2}\right)$

=967+1064.32+5.27

$=2038.08 \mathrm{~cm}^{2}$