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

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

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more