Exercise 18.1
Question 1
Given
Ratio of length and breadth of rectangular field = 9.5
Area of field = 14580$m^{2}$
Cost of fence =$₹ 3.25 / \mathrm{m}$
Let length , breadth = 9x , 5x
Area = 14580
$\begin{aligned} \ell \times b=& 14580 \\ 9 x \times 5 x &=14580 \\ 45 x^{2} &=14580 \\ x^{2} &=\frac{14580}{45} \\ x^{2} &=324 \\ x &=\sqrt{324} \end{aligned}$
$x=18 \mathrm{~m} .$
$\therefore$ length $=9 x=9 \times 18=$ 162m
breadth $=5 x=5 \times 18=90 \mathrm{~m}$
Length of fence= perimeter of rectangle section
$=2(1+b)$
$=2(162+90)$
$=2(252)$
Length of fence = 504m
Cost of fence = $504 \times 3.25$
$=₹ 1638$
Question 2
Given
Dimensions of rectangle = 16m$\times9m$
let side of square $=x m$.
Perimeter of rectangle = area of square
$\begin{aligned} l\times b &=x^{2} \\ 16 \times 9 &=x^{2} \\ x &=\sqrt{16 \times 9} \\ x &=4 \times 3=12 m \\ x &=12 m \end{aligned}$
$\therefore$ Side of square $=12 \mathrm{~m}$
Perimeter of square = 4x
$=4 \times 12$
$=48 \mathrm{~m}$
∴ Perimeter rectangle exceeds perimeter of square by 50 - 48 = 2m
Question 3
Given
Length of adjacent sides = 24cm and 18cm
(diagram should be added)
Distance between longer sides = 12cm
Let ten distance between shorter sides = xcm
Area of parallelogram = Side $\times$ ten distance between the oppositer sides
$\therefore \quad 24 \times 12=18 \times x$
$x=\frac{24 \times 12}{18}$
$x=16 \mathrm{~cm}$
∴ Ten distance between shorter sides = 16cm
Question 4
Given
(diagram should be added)
Plot dimension = $24 \mathrm{~m} \times 24 \mathrm{~m}$
House dimensions $=18 \mathrm{~m} \times 12 \mathrm{~m}$.
∴ Garden Area = plot area - house area
= $24 \times 24-18 \times 12$
Garden Area $=360 \mathrm{~m}^{2}$
Given
Cost of developing garden = Rs 50/$m^{2}$
∴ Total cos of developing garden around house = 360$\times 50$
=Rs 18000
Question 5
Dimension of tiles ( Parallelogram) $=18 \mathrm{~cm} \times 6 \mathrm{~cm}$ Height
Floor area = 540$m^{2}$
Area of one tile = $18 \mathrm{~m} \times \operatorname{6cm}(b \times h)$
= $108 \mathrm{~cm}^{2}$
Area of one tile = $108 \times 10^{-4} m^{2}\left(∴-1 m=10^{-2} m\right)$
No.of tiles required = $\frac{\text { ATotal Area }}{\text { Area of one tile }}$
$=\frac{540}{108 \times 10^{-4}}$
No.of tiles required = 50000
Question 6
(a) Diameter semi circle = 2.8cm
(Diagram should be added)
perimeter of semi circle $=\frac{\pi d}{2}$
$=\frac{\pi \times 2.8}{2}$
$=\frac{3.14 \times 2.8}{2}$
$=3.14 \times 1.4$
Perimeter of semi circle = 4.398cm
(b) (Diagram should be added)
Perimeter of given shape
= $\overline{A B}+\overline{B C}+\overline{C D}+$ Semi circle perimeter
$=1.5+2.8+1.5+4.398$
= 1$10.198 \mathrm{~cm} .$
(c) (Diagram should be added)
Perimeter of given shape
= $\overline{O A}+$ Semi circle $A B+\overline{O B}$
$=2+4.398+2$
$=8.398 \mathrm{~cm}$
∴ Comparing three figure perimeter values , we can
Say in case of figure ' B' and has covered more distance
Question 7
Given
(Diagram should be added)
Area between concentric
Circle = $770 \mathrm{Cm}^{2}$
Outer circle radius = 21cm
Let inner circle radius = rcm
Outer circle area - Inner circle area = $770 \mathrm{~cm}^{2}$
$\begin{aligned} \pi(21)^{2}-\pi r^{2}=& 770 \\ \pi\left(21^{2}-r^{2}\right) &=770 \\ 21^{2}-r^{2}=& 245.098 \\ 441-r^{2} &=245.098 \\ r^{2} &=441-245.098 \\ r^{2} &=195.90 \\ r &=\sqrt{195.9} \\ r &=13.996 \approx 14 \mathrm{~cm} . \end{aligned}$
Radius of inner circle =14cm.
Question 8
Given
Area of square = $121 \mathrm{~cm}^{2}$
$s^{2}=121$
$S=\sqrt{121}$
$S=11 \mathrm{~cm}$
∴ Side of square = 11cm
∴ Length of wire perimeter of square = $4 \times 11 \mathrm{~cm}$
$=44 \mathrm{Cm}$
Now wire is bent into a form of Circle
∴Length of wire= perimeter of circle
44 =2$\pi r$ r = radius of circle
$\pi r=\frac{44}{2}$
$\pi r=22$
$r= \frac{22}{\pi}$
$r=\frac{22}{3 \cdot 14}$
$r=7 \mathrm{~cm}$
radius of circle = 7cm
Area of circle = $\pi r^{2}$
$=3.14 \times 7^{2}$
Area of circle $=153.938 \mathrm{cm}^{2}$
Question 9
(i) (diagram should be added)
Area of $\Delta^{\text {le}}$ ABC
$=\frac{1}{2} \times b \times h$
$=\frac{1}{2} \times 3 \times 4$ (∵right angle triangle )
$=\frac{1}{2} \times 3 \times 4$
$=\frac{1}{2} \times 12$
$=6 \mathrm{cm}^{2}$
(ii) $B C^{2}=A B^{2}+A C^{2} \quad(\therefore$ Pythagorus theorem $)$
$B C^{2}=3^{v}+4^{2}$
$B C^{2}=9+16$
$B C^{2}=25$
$B C=\sqrt{25}$
$B C=5 \mathrm{cm}$
(iii) Area of triangle ABC=6$c m^{2}$
By taking $\overline{B C}$ As base
Area of triangle = $\frac{1}{2} \times b \times h$
$=\frac{1}{2} \times CB\times A D$
$6=\frac{1}{2} \times 6 \times A D$
$A D=\frac{6 \times 2}{6}$
$A D=2 \mathrm{Cm}$
Question 10
Dimension of rectangular garden = 80m$\times$40m
width of path $(w)=2.5 \mathrm{~m}$
(i) Area of cross path = $l \times w+b \times w-(w \times w)$
$=80 \times 2.5+40 \times 2.5-(2.5 \times 2.5)$
$293.75 \mathrm{~m}^{2}$
(ii) (diagram shuold be added)
Area of unshaded portion
=Area of garden - area of cross path
= $80 \times 40-(293.75)$
=$2906.25 \mathrm{~m}^{2}$
Question 11
(diagram shuold be added)
Area of shaded portion = Area of square ABCD - [Area of triangle ABC + Area of triangle AFD + Triangle EFC]
=$18 \times 12-\left[\frac{1}{2} \times 1 \times 18+\frac{1}{2} \times 12 \times 10+\frac{1}{2} \times 5 \times 8\right]$
=$216-[7 \times 9+6 \times 10+5 \times 4]$
=$216-[63+60+20]$
=$216-143$
=$73 \mathrm{~cm}^{2}$
∴ Area of shaded portion 73$\mathrm{cm}^{2}$
Question 12
Given
Area of square EFGH = $729 \mathrm{~m}^{2}$
∴ side of lawn = $\sqrt{729}$
$=27 \mathrm{~m}$
Area of square ABCD = $295 \mathrm{~m}^{2}$
Side of ABCD = $\sqrt{295}$
Side of ABCD = 17.175m
(i) ∴ Length of square filed encluding lawn and path =27m
(ii) Width of the path = side of EFGH - side of ABCD
=27 - 17.175
width of the path = 9.825m
Exercise 18.2
Question 1
(diagram should be added)
Let ABCD is a rhombus
$\begin{aligned} A B=B C &=C D=A D=13 \mathrm{~cm} . \\ A C &=10 \mathrm{Cm} . \end{aligned}$
'O' intersection point of diagnals
$\overline{O A}=\overline{O C}=5 \mathrm{~cm} .$
In $\triangle^{L e}$ AOB
(i) $\overline{A B}^{2}=\overline{O A}^{2}+\overline{O B}^{2}$ (∵pythogorus theorem)
$13^{2}=5^{2}+\overline{O B}^{2}$
$169=25+\overline{O B}^{2}$
$\overline{O B}^{2}=169-25$
$\overline{O B}^{2}=144$
$\overline{O B}=\sqrt{144}$
$\overline{O B}=12 \mathrm{~cm}$
$\overline{B D}=2 \times \overline{O B}$
$=2 \times 12$
$\overline{B D}=24 \mathrm{Cm}$
(ii) Length of diagnal = 24cm
Area of rhombus =$\frac{1}{2}\times d_{1} \times d_{2}$
$=\frac{1}{2} \times 10 \times 24$
Area of rhombus $=120 \mathrm{~cm}^{2}$
Question 2
Given ABCD is a trapezium
(diagram should be added)
Area of trapezium = $\frac{1}{2} \times $ (Sum of parllel sides) $\times $ (Distance between parllel sides)
$=\frac{1}{2} \times(1.5+8) \times 14$
Area of trapezium = 66.5$m^{2}$
Question 3
Given
Area of a trapezium = $360 \mathrm{~m}^{2}$
distance between two paralle side = 20m
Length of one parallel side = 25m
Let unknow paralle sides = x
Area of a trepezium $=\frac{1}{2}$ (sum of parallel sides) $\times$ (distance between parallel side)
$360=\frac{1}{2}(25+x) \times 20$
$(25+x)=\frac{360 \times 2}{20}$
$25+x=36$
$x=36-25$
$x=11 m$
∴ Another parellel side length =11m
Question 4
Given
ABCD is a rhombus
(diagram should be added)
$\overline{B D}=13 C m$
$\overline{A B}=\overline{B C}=\overline{C D}=\overline{A D}=65 \mathrm{~cm}$
Altitude $\overline{A C}=5 \mathrm{~cm}$
(i) Area of rhombus =$\frac{1}{2} \times$ (product of diagnals)
$=\frac{1}{2} \times(13 \times 5)$
=$6.5 \times 5$
Area of rhombys = $32.5 \mathrm{~cm}^{2}$
(ii) Another diagonal $\overline{A C}=5 \mathrm{~cm} .$
Question 5
(i) (diagram should be added)
Areaof trapezium ACDE
$=\frac{1}{2}(E D+A C) \times F G$
$=\frac{1}{2}(7+13) \times 6.5$
$=\frac{1}{2} \times 20 \times 6.5$
$=65 \mathrm{~m}^{2}$
(ii) Area of parallelogram ABCE = Base $\times$ distance between parallel sides
$=7 \times 6.5$
$=45.5 \mathrm{~m}^{2}$
(iii) The area of triangle BCD= $\frac{1}{2} \times B C \times D H$
$A C=A B+B C$
$13=7+B C$
$B C=13-7$
$B C=6 m$
$D H=G F=6.5 \mathrm{~m}$
ஃ The area of triangle BCD= $\frac{1}{2} \times 6 \times 6.5$
=$3 \times 6.5$
=$19.5 \mathrm{~m}^{2}$
Question 6
(diagram should be added)
ABCD is a rhombus and EFG is triangle
Given
Area of rhombus = Area of a triangle
$\frac{1}{2} \times d_{1} \times d_{2}$ = $\frac{1}{2} \times b \times h$
$\frac{1}{2} \times 22 \times d_{1}=\frac{1}{2} \times 24.8 \times 16.5$
$\begin{aligned} 22 \times d_{1} &=24.8 \times 16.5 \\ d_{1} &=\frac{24.8 \times 16.5}{22} \\ d_{1} &=18.6 \mathrm{~cm} . \end{aligned}$
Length of diagnal = 18.6cm
Question 7
Given
Perimeter of trapezium = 52cm
(diagram should be added)
Altitude = 8cm
Length of parllel sides = perimeter - 2(parallel sides)
= 52 - 2$\times$ 10
=52- 20
Sum of parallel sides = 32cm
Area of trapezium =$\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude
$\begin{aligned}=& \frac{1}{2} \times 32 \times 8 \\=& 32 \times 4 \end{aligned}$
Area of trapezium = $128 \mathrm{~cm}^{2}$
Question 8
Given
area of trapezium = $540 \mathrm{Cm}^{2}$
Altitude $=18 \mathrm{Cm}$.
Ratio of length of parallel sides = 7:5
Let length of parallel sides = $7 x, 5 x$
∴ Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude
$540=\frac{1}{2} \times(7 x+5 x) \times 18$
$540=\frac{1}{2}(12 x) \times 18$
$540= 6 \times 18 \times x$
$x=\frac{540}{6 \times 18}$
$x=5 \mathrm{~cm}$
Length of parallel sides = $7 x=7 \times 5=35 \mathrm{~cm}$
$5 x=5 \times 5=25 \mathrm{~cm} .$
Question 9
(i) (diagram should be added)
Area enclosed by shape = Area of square AHGE + Area of triangle BCH + Area of square DCGE
$=5 \times 5+\frac{1}{2} \times 2 \times 9+4 \times 3$
$=25+9+12$
$=46 \mathrm{~cm}^{2}$
(ii) (diagram should be added)
Area enclosed by shape = Area of square ABCD - [Area of triangle EFDH+ Area of triangle HIJ]
=$9 \times 9-\left[\frac{1}{2} \times 5 \times 7+\frac{1}{2} \times 5 \times 7\right]$
$=81-(5 \times 7) \Rightarrow 81-35=46 \mathrm{~cm}^{2}$
Question 10
(i) (diagram should be added)
In $\Delta^{l e} A B D$
$A B^{2}+A D^{2}=D B^{2}$ (∵Pythogorus theorem)
$40^{2}+A D^{2}=41^{2}$
$A D^{2}=41^{2}-40^{2}$
$=1681-1600$
$A D^{2}=81$
$A D=\sqrt{81}$
$A D=9 \mathrm{Cm} .$
(ii) Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude
$=\frac{1}{2}(15+40) \times 9$
=$\frac{1}{2} \times 55 \times 9$
Area of trapezium = $247.5 \mathrm{~cm}^{2}$
(iii) Area of triangle BCD= Area of trapezium ABCD -[area of triangle ADB]
$=247.5-\left[\frac{1}{2} \times A B \times A D\right]$
$=247.5-\left[\frac{1}{2} \times 40 \times 9\right]$
$=247.5-[180]$
Areaof triangle BCD = 67.5$\mathrm{~cm}^{2}$
Question 11
(Diagram should be added)
Area of section ①
Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude
$=\frac{1}{2} \times(28+20) \times 4$
$=96 \mathrm{~cm}^{2}$
$\therefore$ Area section (1) $=96 \mathrm{~cm}^{2}$
Ares of section ②
(diagram should be added)
Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude
$=\frac{1}{2} \times(24+32) \times 4$
Ares of section ② $=112 \mathrm{~cm}^{2}$
Section ③ Dimension are same as section ①
ஃ Area of section ③$=96 \mathrm{~cm}^{2}$
Section ④ Dimension are same as section ②
∴ Area of section ④ = 112$\mathrm{~cm}^{2}$
Question 12
(diagram should be added)
From $\Delta^{\operatorname{le}} A B D$
$B D^{2}=A B^{2}+A D^{2}$
$B D^{2}=6^{2}+88^{2}$
$B D^{2}=36+64$
$B D^{2}=100$
$B D=10 C m$
From $\Delta l e$ BDC
$B C^{2}=B D^{N}+D C^{2}$
$26^{2}=10^{2}+D C^{2}$
$676=100+D C^{2}$
$D C^{2}=676-100$
$D C^{2}=576$
$D C=\sqrt{576}$
$D C=24 \mathrm{Cm}$
Area of quadrilateral ABCD = Area of $\Delta^{l e} BAD$ + Area of $\Delta l e BDC$
$=\frac{1}{2}(A B \times A D)+\frac{1}{2}(B D \times D C)$
$=\frac{1}{2}(6 \times 8)+\frac{1}{2}(10 \times 24)$
$=\frac{1}{2}(48)+\frac{1}{2}(240)$
$=24+120$
Area of quadrilateral ABCD = $144 \mathrm{~cm}^{2}$
Question 13
(diagram should be added)
Given ABCDEFGH a regular octagon
Area of octagon ABCDEFGH = Area of square ABCH + Area of square HCDG + Area of square GDEF
= $2 \times$ Area of square ABCH + Area of square HCDG
$=2 \times\left(\frac{1}{2} \times(8+15) \times 6\right]+(8 \times 15)$
$=\left(2 \times \frac{1}{2} \times 23 \times 6\right)+(8 \times 15)$
$=\quad 23 \times 6+8 \times 15$
$=\quad 138+120$
$=258 \mathrm{~m}^{2} \mathrm{2}$
Question 14
(diagram should be added)
Jaspreet's diagram
Area of ABCD = Area square ABCF+$ Areaof square FCDE
$=2 \times($ Areaof $\square A B C F)$ (∵ Both are symmetric)
$=2 \times\left(\frac{1}{2} \times(A B+C F) \times A F\right)$
=$2 \times \frac{1}{2} \times(18+32) \times \frac{18}{2}$
= $50 \times 9$
Area of ABCDE = $450 \mathrm{~cm}^{2}$
Rahul's diagram
-----------------------
(diagram should be added)
Area of pentagon ABCDE = area of triangle DEC + area of square ECBA
$=\frac{1}{2}(E C \times D F)+B C \times A B$
$=\frac{1}{2} \times 18 \times 14+18 \times 18$
$=126+324$
$=450 \mathrm{~cm}^{2}$
We can find area of pentagon in this way
(diagram should be added)
Mahesh 's diagram
Question 15
(diagram should be added)
Area of shaded pentagon ABECD = Area of square ABCD - [Area of triangle BEC]
$=18 \times 10-\left[\frac{1}{2} \times 8 \times E B\right] \rightarrow(1)$
From triangle BEC
$B C^{2}=E C^{2}+E B^{2}$
$10^{2}=8^{2}+E B^{2}$
$E B^{2}=100-64$
$E B^{2}=36$
$E B=\sqrt{36}$
$E B=6 \mathrm{Cm}$
Sub EB value is eq ①
∴ Area of shaded pentagon ABECD = $180-\left[\frac{1}{2} \times 8 \times 6\right]$
=$180-[4 \times 6]$
=$180-24$
∴ Area of shade pentagon ABECD = $156 \mathrm{~cm}^{2}$
Question 16
(diagram should be added)
Given
ABCDE is a polygon
$\begin{aligned} A D &=8 \mathrm{~cm} \\ A H &=6 \mathrm{~m} \\ A G &=4 \mathrm{~cm} \\ A F &=3 \mathrm{Cm} . \\ B F &=2 \mathrm{~cm}, \mathrm{CH}=3 \mathrm{~cm}, \quad E G=2.5 \mathrm{Cm} . \end{aligned}$
Area of polygon ABCDE = area of triangle ABF + Area of square BCHF + Area of triangle CHD+ Area of triangle AGE
$=\frac{1}{2}(A F \times B F)+\frac{1}{2}(B F+C H) \times F H+\frac{1}{2}(D H \times C H)+$\frac{1}{2}(A D \times E G)$
$\begin{aligned} A D &=A H+H D \\ 8 &=6+H D \\ & H D=8-6 \\ H D &=2 C m \end{aligned}$
$A H=A F+F H$
$6=3+F H$
$F H=6-3$
$F H=3 \mathrm{Cm}$
∴ Area of polygon ABCDE = $\frac{1}{2}(3 \times 2)+\frac{1}{2}(5) \times 3+\frac{1}{2} \times 2 \times 3+\frac{1}{2}(8 \times 2.1)$
$=3+7 \cdot 5+3+10$
Area of PolygonABCDE $=23.5 \mathrm{Cm}^{2}$
Question 17
Given PQRSTU is a polygon
(diagram should be added)
$P S=11 \mathrm{~cm}$
$P Y=9 \mathrm{~cm}$
$PX=8 \mathrm{~cm}$
$PW=5 \mathrm{~cm}$
$P V=3 \mathrm{~cm}$
$QV=5 \mathrm{~cm}$
$U w=4 \mathrm{~cm}$
$Rx=6 \mathrm{~cm}$
$T y=2 \mathrm{~cm}$
$\begin{aligned} VX&=P x-P y \\ &=8-3 \\ Vx &=5 \mathrm{~cm} \end{aligned}$
$W Y=P y-P \omega=9-5=4 \mathrm{~cm}$
$\begin{aligned} X S &=P S-P x \\ &=11-8 \\ XS &=3Cm \end{aligned}$
$\begin{aligned} Y S &=P S-P Y \\ &=11-9 \\ Y S &=2 a m \end{aligned}$
=Area of polygon PQRSTU = Area of triangle PQN + Area of square QRXV + Area of triangle XRS + Area of triangle PUW + Area of square UMNT + Area of triangle YST
=$\left(\frac{1}{2} \times PV \times QV\right)+$ $\frac{1}{2}(Q V+R X) \times(V X)$ +$\frac{1}{2}(R X)(XS)$ + $\frac{1}{2} \times$ PW $\times$ VW +$\frac{1}{2}(V W+XT) \times(YW)$ +$\frac{1}{2}(YS) \times(Y T)$
= $\frac{1}{2} \times 3 \times 5+\frac{1}{2}(5+6) \times 5+\frac{1}{2}(6 \times 3)+\frac{1}{2}(5 \times 4)+\frac{1}{2}(4 \times 2 ) \times 4$ + $\frac{1}{2}(2 \times 2)$
= $\frac{1}{2}(15+55+18+20+24+4)$
= $\frac{1}{2}(136)$
= $68 \mathrm{~cm}^{2}$
Exercise 18.3
Question 1
Given volume of cube $=343 \mathrm{~cm}^{3}$
Let 'S' be edge of cube
Volume of cube = $S^{3}$
$\begin{aligned} s^{3} &=343 \\ S &=\sqrt[3]{343} \\ S &=7 \mathrm{~cm} . \end{aligned}$
ஃ Length of an edge of cube = 7cm
Question 2
(i) Volume of Cuboid $90 \mathrm{~cm}^{3}$
Length 6cm , Breadth 5cm , Height 3cm
(ii) Volume of cuboid $840 \mathrm{~cm}^{3}$
Length 15cm , Breadth 15cm , Height 7cm
(iii) Volume of Cuboid $62.5 \mathrm{~m}^{3}$
Length 10m , Breadth 5m , Height 1.25m
Question 3
Given
Volume of cuboid $=312 \mathrm{~cm}^{3}$
Base $A r e a=26 \mathrm{~cm}^{2}$
volume $=312 \mathrm{~cm}^{\circ}$
Area $\times$ height $=312$
$26 \times h=312$
$h=\frac{312}{26}$
h = 12cm
Question 4
Given
Godown dimension $(1 \times b \times h)=$ $55 \mathrm{~m} \times 45 \mathrm{~m} \times 30 \mathrm{~m}$
Cuboidal box volume =1.2$5 m^{3}$
$\begin{aligned} \text { godown volume } &=\operatorname{l\times b} \times h \\ &=55 \times 45 \times 30 \\ &=74250 \mathrm{~m}^{3} \end{aligned}$
No. of Cuboidal boxes = $\frac{\text { godown Volume }}{\text { box volune }}$
$=\frac{74250}{1.25}$
No. of cuboidal boxes $=59400$
Question 5
Given Dimension of rectangular pit = $1.4 \mathrm{~m} \times 90 \mathrm{~cm} \times 70 \mathrm{~cm}$
Volume of pit = $l \times b \times h$
=$140 \times 90 \times 70 \mathrm{~cm}^{3}$
Volume of pit = $882000 \mathrm{~cm}^{3}$
Given
Brick dimension (l $\times$ b) =$2.1 \mathrm{~cm} \times 10.5 \mathrm{~cm}$
Let 'h' height of brick
$1000 \times$ Brick volume $=$ pit volume
$1000 \times$ $21 \times 10.5 \times h=882000$
$h=\frac{882000}{21 \times 10.5 \times 1000}$
$h=4 \mathrm{~cm}$
$\therefore$ Height of brick $=4 \mathrm{~cm}$
Question 6
Let 'a' be edge of cube
Volume of cube = $a^{3}$
If each edge of cube is tripled = $a^{1}=3 a$
Volume of new Cube = $a^{1^{3}}$
$=(3 a)^{3}$
$=27 a^{3}$
The Volume becomes 27 times the original Volume of Cube.
Question 7
Given
Milk tank is in the from of cylinder
Diameter of tank = 1.4m $\times$ 2
Height of tank = 8m
Volume of tank = $\frac{\pi}{} d^{2} \times h$
$\frac{\pi}{} \times(1-4)^{2} \times 8$
Volume of tank = 49 .26$0 m^{3}$
∴ Volume of tank = 49.260 lit
Question 8
Given
Extened dimension of box = $84 \mathrm{~cm} \times 75 \mathrm{~cm} \times 64 \mathrm{~cm}$
Thickness of box = 2cm
∴ Internal Dimension of box = ($54 -2\times 2$)cm , (75 - 2 $\times$ 2) cm ,(64- 2$\times2$) cm
$=80 \mathrm{~cm} \times 71 \mathrm{~cm} \times 60 \mathrm{Cm}$
Volume of wood = External Volume - Internal Volume
= $(84 \times 75 \times 64)-(80 \times 71 \times 60)$
$403200-340800$
Volume of wood $=62400 \mathrm{~cm}^{3}$
Question 9
Given
Two Cylinder jar has same volume
Let
$d_{1}, d_{2}$ are diameter jar
$h_{1,} h_{2}$ are heights of jar
Given
$d_{1}: d_{2} \cdot 3: 4$
Volume of cylinder equal
∴ $\frac{\pi}{4} d_{1}^{2} \times h_{1}=\frac{\pi}{4} d_{2}^{2} \times h_{2}$
$d_{1}^{2} \times h_{1}=d_{2}^{2} \times h_{2}$
$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{h_{2}}{h_{1}}$
$\left(\frac{3}{4}\right)^{2}=\frac{h_{2}}{h_{1}}$
$\frac{h_{1}}{h_{2}}=\frac{16}{9}$
$h_{1}: h_{2}=16: 9$
∴ Height of cylinders are in the ratio = 16: 9
Question 10
Let 'r' be the ratio's of cylinder
h be the height of cylinder
Volume V = $\pi r^{2} \times h$
Now radius is $=R^{1}=\frac{r}{2}$
Height is doubled = $h^{1}=2 h$
New Volume $V^{1}=\pi r_{1}^{2} \times h^{1}$
$=\pi\left(\frac{r}{2}\right)^{2} \times(2 h)$
$=\frac{\pi r^{2}}{4} \times 2 h$
$v^{1}=\frac{\pi r^{2} \times h}{2}$
$v^{1}=\frac{v}{2}$
ஃ New Volume is half of original Volume
Question 11
Dimensions of tin sheet $=30 \mathrm{~cm} \times 18 \mathrm{~cm}$
When rolled along its length (30cm)
-----------------------------------------------
$\begin{aligned} 2 \pi r &=30, \quad h=18 C m \\ r &=\frac{30}{2 \pi} \\ r &=4.77 \mathrm{~cm} \end{aligned}$
$\begin{aligned} \text { Volume } &=\pi r^{2} \times h \\ &=\pi \times 4.77^{2} \times 18 \\ \text { Volume } &=1289.155 \mathrm{~cm} 3 \end{aligned}$
When rolled along breadth (18cm)
----------------------------------------------
$2 \pi r=18, \quad h=30 \mathrm{~lm}$
$r=\frac{18}{2 \pi}$
$r= 2.86 \mathrm{~cm}$
$\begin{aligned} \text { Volume } &=\pi r^{2} \times h \\ &=\pi \times 2.86^{2} \times 30 \\ &=773.493 \mathrm{~cm} 3 \end{aligned}$
Question 12
(i) Given dis of pipe = 7cm = 0.07m
Velocity = 5m\sec
Discharge = Area $\times$ Velocity
$=\frac{\pi d^{2}}{4} \times V$
$=\frac{\pi \times(0.07)^{2}}{4} \times 5$
Discharge $=0.0192 \mathrm{~m}^{3} / \mathrm{sec}$
$\begin{aligned} \therefore \text { Discharge } &=19.2 \text { lits } / \text { see } \\ &=19.2 \times 60 \text { lits } / \mathrm{min} \end{aligned}$
Discharge = 1154 .53 lits\min
Discharge $\approx 1155 \operatorname{lit} / \min$
(ii) Dimension of tank = $4 m \times 3 m \times 2.31 m$
Discharge $=0.0192$ $m^{3}/sec$
$1.154 \mathrm{~m}^{3} / \mathrm{min}$
Timetaken to fill the tank $=\frac{\text { Volume of tank }}{\text { Discharge }}$
$=\frac{4 \times 3 \times 2.31}{1.154}$
Time taken to fill the tank = 24min
Question 13
Given
Vessel 1
radius 15cm
Height 40cm
Volume $\pi x r^{2} \times h$
$\pi \times(15)^{2}+40$
Volume $\quad 28274.33 \mathrm{~cm}^{3}$
Vessel 2
Radius 20cm
Height 40cm
Volume $\pi x r^{2} \times h$
$\pi \times(20)^{2}+45$
Volume $56548.667 \mathrm{~cm}^{3}$
Given
Another vessel with capcity equal to sum of vessel 1 and vessel 2
Let radius of vessel's = R
Height of vessel = 30cm
$\left(\pi R^{2}\right) \times 30=28274.33+56548.667$
$\begin{aligned} 30 \times\left(11 R^{2}\right) &=84823 \\ R^{2} &=\frac{84823}{\pi \times 30} \end{aligned}$
$R^{2}=900$
$R=\sqrt{900}$
$R=30 \mathrm{Cm}$
∴ Radius of vessel = 30cm
Question 14
Given
Pole height = 7m
Pole diameter = 20cm = 0.2m
Density = $225 \mathrm{~kg} / \mathrm{m}^{3}$
Volume of wood = $\frac{\pi d^{2}}{4} \times h$
$=\frac{\pi}{4} \frac{(20)^{2}}{10^{4}} \times 7$
Volume of wood $=0.219 \mathrm{~m}^{3}$
Weight of wood = 49.48kg
Question 15
(diagram should be added)
A cylinder with diameter of 14cm and height of 30cm is the maximum Volume
That can be cut from given Cuboid
Volume of cylinder = $\frac{\pi}{4} d^{2} \times h$
$=\frac{\pi}{4}(14)^{2} \times 30$
Volume of cylinder = 4618.14 $\mathrm{cm}^{3}$
Volume of wood wasted = Volume of cuboid - Volume of cylinder
= $14 \times 14 \times 30-(4618.14)$
Volume of Wood wasted = $1261.85 \mathrm{~cm}^{3}$
Exercise 18.4
Question 1
Given surface area = $384 \mathrm{cm}^{2}$
(i) Let length of side of cube = a
Surface area of cube = $6 a^{2}$
$6 a^{2}=384$
$a^{2}=\frac{384}{6}$
$a^{2}=64$
$a=\sqrt{64}$
$a=8 \mathrm{~cm}$
∴ 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
Given
Radius of Cylinder = 5cm
Height of Cylinder = 10cm
Surface area of Cylinder =$2 \pi r h$
$=2 \pi \times 5 \times 10$
Surface area of Cylinder $=100 \pi$
Question 3
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
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
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
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
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
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
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
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
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 = $4.32 \times 0.75 \times 10^{4}$
$=Rs32400$
∴ Total Cost for Silver coating of Cubes is Rs 32400
Question 12
Givew 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
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 = $125 \mathrm{~cm}$
Height of cylinder = 125cm
∴ Lenght 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
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
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
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
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 = $616 \mathrm{~cm}^{2}$
$2 \pi r(h+r)=616 \mathrm{~cm}^{2}$
$2 \pi r(r+r)=616 \mathrm{~cm}^{2}$
$2\left(2 \pi r^{2}\right)=616 c m^{2}$
$2 \pi r^{2}=308 c m ^{2}$
$r^{2}=\frac{308}{2 \pi}$
$r^{2}=49$
$r= \sqrt{49}$
$r=7 \mathrm{~cm}$
$r=h=7 \mathrm{~cm}$
Volume of Cylinder = $\pi r^{2} h$
$=\pi(7)^{2} \times 7$
Volume of cylinder = $1077.56 \mathrm{~cm}^{3}$
Question 18
Length of Cylinder = $77 \mathrm{~cm}=\mathrm{h}$
Inner diameter $d_{1}=4 c m$
Outer diameter $\left(d_{2}\right)=4.4 \mathrm{~cm}$
(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
=$\pi d _{1} h+\pi d_{2} h+2 \times \pi\left(r_{2}^{2}-r_{1}^{2}\right)$
=$967.63+1064.37+2 \pi\left(2.2^{2}-2^{2}\right)$
$=967+1064.32+5.27$
$=2038.08 \mathrm{~cm}^{2}$
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