Exercise 18 A
Question 1
Ans: (a) length $=3 m$, Breadth $=4 m$, Height $=5 m$
In a cuboid
Total surface a area $=2(l b+bh+hl)$
$2(3 \times 4+4 \times 5+5 \times 3)$
$2(12+20+15) m^{2}$
$2 \times 47$
$=94 m^{2}$
(b) length = $4 \mathrm{~cm}$, Breadth $=1.7 \mathrm{~cm}$, Height $=2.3 \mathrm{~cm}$
In a Cuboid
To total surface area $=2(16+6 h+h 1)$
$2(4 \times 1.7+1.7 \times 2.3+2.3 \times 4) \mathrm{cm}^{2}$
$2(6.8+3.9+9.2) \mathrm{cm}^{2}$
$2 \times 19.91$
$=39.82 \mathrm{~cm}^{2}$
Question 2
Ans: (i) (a) Side of cube $=7 \mathrm{~cm}$ 2usipace ariea of cube $=6 a^{2}$ where a is $7 \mathrm{~cm}$ $6 \times(7) 2 \mathrm{~cm}^{2}$
$6 \times 7 \times 7$
$6 \times 49$
$294 \mathrm{~cm}^{2}$
(b) Side of cube $=10 \mathrm{~m}$
Surface area of cube $=6 a^{2}$
where a is 10 $6 \times(10)^{2} \mathrm{~m}^{2}$ $6 \times 10 \times 10$ $600 \mathrm{~m}^{2}$
(ii) Total area of a model of cube $=96 \mathrm{~cm}^{2}$ length ol its edge $=\sqrt{\frac{\text { Area }}{6}}$
$\begin{aligned}&=\sqrt{\frac{96}{6}} \\&=\sqrt{16} \\&=4 \mathrm{~cm}\end{aligned}$
(iii) Surface area =$726 \mathrm{~cm}^{2}$
Edge $=\sqrt{\frac{\text { Surface area }}{6}}$
$=\sqrt{\frac{726}{6}}$
$\sqrt{121 \mathrm{~cm}}$
$=110 \mathrm{~cm} .$
Question 3
Ans:
(i) Length = 4cm, breadth = 3cm, height = 5cm
Volume of cuboid = $1 \times b \times h$
$=4 \times 3 \times 5 \mathrm{~cm}^{3}$
$=60 \mathrm{~cm} 3$
[ii]
$\begin{aligned} \text { Length } &=2 \mathrm{~cm}, \text { Breadth }=6 \mathrm{~cm}, \text { Height }=8 \mathrm{~cm} \\ \text { Volume of cuboid } &=1 \times 6 \times h \\ &=2 \times 6 \times 8 \mathrm{~cm}^{3} \\ &=96 \mathrm{~cm}^{3} \end{aligned}$
(iii) length $=7 \mathrm{~cm}$, Breadth $=3 \mathrm{~cm}$, Height $=4 \mathrm{~cm}$
Volume of cuboid $=1 \times b \times h$
$7 \times 3 \times 4 \mathrm{~cm}^{3}$
$84 \mathrm{~cm}^{3}$
(iv)
$\begin{aligned} \text { Length } &=12 \mathrm{~cm}, \text { Breadth }=9 \mathrm{~cm}, \text { Height }=12 \mathrm{~cm} \\ \text { Volume of cuboid } &=1 \times 6 \times h \\ &=12 \times 9 \times 12 \mathrm{~cm}^{3} \\ &=1296 \mathrm{~cm}^{3} \end{aligned}$
(v)Length = $16 \mathrm{~cm}$, Breadth $=14 \mathrm{~cm}$, Height $=18 \mathrm{~cm}$
Volume of cuboid $=1 \times 6 \times h$
$16 \times 14 \times 18 \mathrm{~cm}^{3}$
$4032 \mathrm{~cm}^{3}$
(vi) Length $=7 \mathrm{~cm}$, Breadth $=28 \mathrm{~cm}$, Height $=26 \mathrm{~cm}$
Volume of cuboid $=1 \times 6 \times h$
$7 \times 28 \times 24 \mathrm{~cm}^{3}$
$=25984 \mathrm{~cm}^{3}$
(vii) Length = 40cm, Breadth = 24cm,
Height = $\frac{\text { Volume }}{\text { Length } \times \text { Breadth }}$
$=\frac{2400}{40 \times 24}$
$=\frac{5}{2}$
$=2.5 \mathrm{~cm}$
(vii) Length $=60 \mathrm{~cm}$, Height $=5 \mathrm{~cm}$, volume $=5400 \mathrm{~cm}^{3}$
$\begin{aligned} \text { Breadth} &=\frac{\text { Volume }}{\text { length } \times \text { Height }} \\ &=\frac{5400}{60 \times 4} \\ &=18 \mathrm{~cm} \end{aligned}$
$\begin{array}{|l|c|c|c|c|c|c|c|c|}\hline & \text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } & \text { (v) } & \text { (vi) } & \text{ (viii) 6riis } \\\hline \text { length } & 4 & 2 & 7 & 12 & 16 & 38.6 & 40 & 60 \\\hline \text { brendth } & 3 & 6 & 3 & 9 & 14 & 28 & 24 & 18 \\\hline \text { height } & 5 & 8 & 4 & 12 & 18 & 24 & 2.5 & 5 \\\hline \text { volume } & 60 & 96 & 84 & 1296 & 4032 & 25984 & 2400 & 5409 \\\hline\end{array}$
Question 4
Sol: (a) Dimension $=2,3$, $4 \mathrm{~cm}$
Diagonal of cuboid $=\sqrt{1^{2}+b^{2}+h^{2}}$
$=\sqrt{(2)^{2}+(3)^{2}+(4)^{2}}$
$=\sqrt{4+9+16}$
$=\sqrt{29} \mathrm{~cm}$
$=5.38 \mathrm{~cm}$
(b) Dimension $=3,4,5 \mathrm{~cm}$ Diagonal of cuboid $=\sqrt{1^{2}+b^{2}+h^{2}}$
$\begin{aligned}&\sqrt{(3)^{2}+(1)^{2}+(5)^{2}} \\&\sqrt{9+16+25} \\&\sqrt{50} \\&=7.07 \mathrm{~cm}\end{aligned}$
Question 5
Sol: (a) Edge =2m
length of Diagonal $=\sqrt{3} \times$ side
where side is $2 \mathrm{~m}$
$\begin{aligned}& \sqrt{3} \times 2 \mathrm{~m} \\=& 2 \sqrt{3} \mathrm{~m} \\=& 2(1.732) \\& 2 \times 1.732 \\&=3.464 \mathrm{~m}\end{aligned}$
Volume of cuboid $=a^{3}$ where a is 2
$\begin{aligned}&(2)^{3} \mathrm{~m}^{3} \\&2 \times 2 \times 2 \\&=8 \mathrm{~m}^{3}\end{aligned}$
(b) Edge $=5 \mathrm{~m}$
length of diagonal $=\sqrt{3} \times$ side
where side is $5 \mathrm{~m}$
$\sqrt{3} \times 5 \mathrm{~m}$
$=\begin{gathered}5(1.732) \\ 5 \times 1.732\end{gathered}$
=8.660
$=8.66 \mathrm{~m}$
Volume $=a^{3}$
$=(5)^{3}$
$=5 \times 5 \times 5 \mathrm{~m}^{3}$
$=125 \mathrm{~m}^{3}$
(c) (i) Edge $=8 \mathrm{~cm}$
length of Diagonal $=\sqrt{3} \times$ side where side is $8 \mathrm{~cm}$
$\sqrt{3} \times 8 \mathrm{~cm}$
$8(1.732)$
$8 \times 1.732$
$=13.858$
$=13.86 \mathrm{~cm}$
(ii) Edge $=12 \mathrm{~cm}$
Volume of cube $=a^{3}$
where a is $12 \mathrm{~cm}$
$\begin{aligned}&(12)^{3} \mathrm{~cm}^{3} \\&12 \times 12 \times 12 \mathrm{~cm}^{3} \\&=1728 \mathrm{~cm}^{2}\end{aligned}$
Total Surface area of cube
$\begin{aligned}\text { where } & \text { a ig } 12 \mathrm{~cm} \\& 6 \times(12)^{2} \\& 6 \times 12 \times 12 \\& 6 \times 144 \mathrm{~cm}^{2} \\& 864 \mathrm{~cm}^{2}\end{aligned}$
Question 6
Ans: (a) Volume $=216 \mathrm{~m}^{3}$
Edge $=3 \sqrt{\text { volume }}$
$=3 \sqrt{216}$
$=(6 \times 6 \times 6)^{\frac{1}{3}}$
$=\left(6^{3}\right)^{\frac{1}{3}}$
$=6^{3 / \frac{1}{3}}$
$=6 \mathrm{~m}$
(b) Volume $=2197 \mathrm{~m}^{3}$
$\begin{aligned} \text { Edge } &=\sqrt[3]{\text { volume }} \\=& 3 \sqrt{2197} \\=&\left(13^{3}\right)^{\frac{1}{3}} \\ & 13^{3 \times \frac{1}{3}} \\ &=13 \mathrm{~m} \end{aligned}$
(ii)
$\begin{aligned} \text { Length of cuboid } &=1 \mathrm{~m} \\ &=100 \mathrm{~cm} \\ \text { Breadth } &=50 \mathrm{~cm} \\ \text { Height } &=0.5 \mathrm{~m} \\ &=\frac{5}{10} \times 100 \\ &=50 \mathrm{~cm} \end{aligned}$
$\begin{aligned} \text { Volume of cuboid } &=1 \times 6 \times h \\ &=100 \times 50 \times 50 \\ &=250000 \mathrm{~cm}^{3} \end{aligned}$
This is given
It is true
(iii) Diagonal of a rectangular solid $=5 \sqrt{2} \mathrm{dm}$
Let the height be $=h$
Length $=5 \mathrm{dm}$
Breadth $=4 \mathrm{dm}$
Diagonal of cuboid $=\sqrt{1^{2}+b^{2}+h^{2}}$
$5 \sqrt{2}=\sqrt{(5)^{2}+(4)^{2}+h^{2}}$
squaring both Side
$\begin{aligned}50 &=25+16+n^{2} \\&=50 \\&=41+n^{2} \\h^{2} &=50-41 \\&=9 \\&=(3)^{2} \\&=3 \\\text { Height } &=30 \mathrm{~m}\end{aligned}$
Question 7
Ans: Volume of Rectangular solid $=3600 \mathrm{~cm}^{3}$
Length $=200 \mathrm{~m}$
Height $=90 \mathrm{~m}$
$\begin{aligned} \text { Breadth } &=\frac{\text { Volume }}{\text { Length x Breadth }} \\ &=\frac{3600}{20 \times 9} \\ &=20 \mathrm{~cm} \end{aligned}$
Question 8
Sol: Length of rectangular solid $=25 \mathrm{~cm}$
Breadth $=20 \mathrm{~cm}$
Volume $=7000 \mathrm{~cm}^{3}$
$\begin{aligned} \text { Height } &=\frac{\text { Volume }}{\text { Length } \times \text { Breadth }} \\ & \frac{7000}{25 \times 20} \\ &=14 \mathrm{~cm} \end{aligned}$
Question 9
Ans: (9) Perimete $r$ of one diace $=20 \mathrm{~cm}$
$\begin{aligned}\text { Side } &=\frac{20}{4} \\&=5 \mathrm{~cm}\end{aligned}$
(i) Total surface area of cube = $6 a^{2}$
Where a is 5cm
$6 \times(5)^{2}$
$6 \times 5 \times 5 \mathrm{~cm}^{2}$
$=150 \mathrm{~cm}^{2}$
(ii)Volume $=a^{3}$
where a is 5 $(5)^{3}$
$5 \times 5 \times 5 \mathrm{~cm}^{3}$
$125 \mathrm{~cm}^{3}$
Question 10
Sol: Area of playground = $=4800 \mathrm{~m}^{2}$
$\begin{aligned} Depth &=10 m \\ &=\frac{1}{100} m \end{aligned}$
Volume = Area of the Base height
$\begin{aligned}&=4800 \times \frac{1}{100} \\&=48 \mathrm{~m}^{3}\end{aligned}$
Rate of growing = Rs 4.80 per cubic meter
Total cost = Rs $48 \times 4.80$
$=\operatorname{Rs} \quad 230.40$
Question 11
Sol: Base of the tank $=7 \mathrm{~m} \times 6 \mathrm{~m}$
Area of the base $=7 \times 6$
$=42 \mathrm{~m}^{2}$
Depth $=5 \mathrm{~m}$
Volume of the water in the tanc
$=$ Area \times $ Depth
$=42 \times 5$
$=210 \mathrm{~m}^{2}$
Question 12
Ans: Internal length of Box $=20 \mathrm{~cm}$
Breadth $=16 \mathrm{~cm}$
Height $=24 \mathrm{~cm}$
Volume of cuboid $=1 \times 6 \times h$
$=20 \times 16 \times 24$
$=7680 \mathrm{~cm}^{3}$
Edge $=4 \mathrm{~cm}$
Volume of one cube $=a^{3}$
Where a. is 4
4 $^{3} \mathrm{~cm}^{3}$
$64\mathrm{~cm}^{3}$
No. of cubes kept in the Box
$=\frac{\text { volume of box }}{\text { Volume of ore cube }}$
$\frac{7680}{64}$
$=120$
Question 13
Sol: Ratio in dimension=5:4:9
Surface area = 1216$\mathrm{cm}^{3}$
Let length of rectangular solid = 5x
Breadth = 4x
Height = 2x
Surface area of cuboid = 2(lb +bh + hl)
=2(5x $\times$ 4x + 4x $\times$ 2x $\times$ 5x )=216
=2($20 x^{2}+$$8 x^{2}+10 x^{2}$)= 1216
$=2 \times 38 x^{2}=1216$
$=x^{2}=\frac{1216}{2 \times 38}$
$=x^{2}=16=(4)^{2}$
$x=4$
$\begin{aligned} \text { Length } &=5 x \\ \text { whers } x \text { is } 4 & \\ & 5 \times 4 \\ &=20 \mathrm{~cm} \\ \text { Buleadth } &=4 \times \\ & 4 \times 4 \\ &=16 \mathrm{~cm} \\ \text { Height } &=21 \\ 2 & \times 4 \\ & 8 c m \end{aligned}$
Question 14
Ans : Length of lock of canal $=40 \mathrm{~m}$
Breadth $=7 \mathrm{~m}$
level of water decreased $=5-3.80$
$=1.20 \mathrm{~m}$
Volume of water runs but $=1 \times 6 \times h$
$\begin{aligned}&=40 \times 7 \times 1.20 \mathrm{~m}^{3} \\&=336 \mathrm{~m}^{3}\end{aligned}$
Question 15
Ans: (15) Length of $60 x=90 \mathrm{~cm}$
Breadth $=780 \mathrm{~m}$
Height $=42 . \mathrm{cm}$
Volume of cuboid $=1 \times 6 \times h$
$=90 \times 78 \times 48 \mathrm{~cm}^{3}$
(i) Length of rectangular block = $2 \frac{1}{2} \mathrm{~cm}$ or $\frac{5}{2}$
$\begin{aligned} \text { Breadth } &=2 \mathrm{~cm} \\ \text { Height } &=1 \frac{1}{2} \mathrm{~cm}=\frac{3}{2} \end{aligned}$
Volume of one block
$=l \times b \times h$
$=\frac{5}{2} \times 2 \times \frac{3}{2}$
$=\frac{15}{2} \mathrm{~cm}^{3}$
No. of block pocked in the box
$\begin{aligned}&=\frac{V \cdot v 1 \text { box }}{V \cdot \text { ol one box }} \\&=\frac{98 \times 78 \times 42 \times 2}{15} \\&=39312\end{aligned}$
(ii)Edge op one cube $=4 \mathrm{~m}$
Volume of cube $=a^{3}$
where a is $4 \mathrm{~cm}$ $=(4)^{3}$.
$\begin{aligned}&=4 \times 4 \times 4 \\&=64 \mathrm{~cm}^{3}\end{aligned}$
$\therefore$ No. of cubes to be packed
$=\frac{90 \times 78 \times 42}{4}$
= Along length wise, no of complete cubes $=\frac{90}{4}=22$
And along breadth wise, no complete 4 cubes = $\frac{78}{4}=19$
And along height wise = No. of complete cubes = $\frac{42}{4}=10$
Total number of cubes $=22 \times 19 \times 10$$=4180$
Question 16
Ans: Length of tank(l)=72 \mathrm{~cm}$
Breath $(b)=60 \mathrm{~cm}$.
Height $(h)=36 \mathrm{~cm}$
Depth of water $=18 \mathrm{~cm}$
Volume of water in the tank
$=72 \times 60 \times 18 \mathrm{~cm}^{3}=77760 \mathrm{~cm}^{3}$
Length of block
$=48 \mathrm{~cm}$
Breath $=36 \mathrm{~cm}$
Height $=15 \mathrm{~cm}$
Volume cy block
$=48 \times 36 \times 15 \mathrm{~cm}^{2}$
$=25920 \mathrm{~cm}^{3}$
$\therefore$ Height of water level rose up in the tank
$=\frac{25920}{72 \times 66}=6 \mathrm{~cm}$
Question 17
Ans:
$\begin{aligned} & \text { Internal length of the closed box } \\ & \text { length (l) }=20 \mathrm{~cm} \\ \text { Areath (b) }=& 12.5 \mathrm{~cm} \\ & \text { Height }(\mathrm{H})=9.5 \mathrm{~cm} \\ \therefore & \text { Inner volume }=l \mathrm{bh} \\=& 20 \times 12.5 \times 9.5 \mathrm{~cm}^{3}=2375 \mathrm{~cm}^{3} \\ \text { Thickness of wood }=1.25 \mathrm{~cm} \\ \therefore \text { Outer length }(l)=\\=& 20+2 \times 1.25 \mathrm{~cm} \\=& 20+2.5=22.5 \mathrm{~cm} \\ \text { Outer breadth }(B)=& 12.5+2 \times 1.25 \mathrm{~cm} \\=& 12.5+2.5=15 \end{aligned}$
and outer height (H) 2.5+$12 \times 1.25 \mathrm{~cm}$
$=9.5+2.5=12 \mathrm{~cm}$
Outer volume $=22.5 \times 15 \times 12 \mathrm{~cm}^{2}$
$=4050 \mathrm{~cm}^{3}$
$\therefore$ Volume of wood used
Outer volume - Inner volume
$=1050-2375=1675 \mathrm{~cm}^{3}$
Question 18
Ans : In an open box
External Length $(L)=17.5 \mathrm{~cm}$
Breadth (B) $=14 \mathrm{~cm}$
Height (H) $=10 \mathrm{~cm}$
$\begin{aligned} \therefore \text { valume } &=l \mathrm{BH}=17.5 \times 14 \times 10 \mathrm{~cm}^{3} \\ &=2450 \mathrm{~cm}^{3} \end{aligned}$
Thickness of word $=7.5 \mathrm{~mm}=0.75 \mathrm{~cm}$
$\therefore$ Inner length $(l)=17.5-2 \times 7.5$
$=17.5-1.5=16 \mathrm{~cm}$
Breadth (b) $=14-2 \times 75$
$=14-1.5=12.5 \mathrm{~cm}$
height $=10-0.75=9.25 \mathrm{~m}$
Hence box is open
$\therefore$ Inner volume
$=16 \times 12.5 \times 9.25 \mathrm{~cm}^{3}$
$=1850 \mathrm{~cm}^{3}$
∴ Volume of wood used = outer volume - inner volume
=2350-1850
= $600 \mathrm{~cm}^{3}$
Question 19
Ans: Length of field(L) $=30 \mathrm{~m}$
$\text { Breadth }=18 \mathrm{~m}$
Area $=L \times B$
$=30 \times 18=540 \mathrm{~m}^{2}$
Length of the pit (l) =6m
Breadth (b) $=4 \mathrm{~m}$
Depth $(h)=3 m$
$\therefore$ Area of the face of the pit $=l \times b$
$=6 \times 4=24 \mathrm{~m}^{2}$
Volume of the earth
$=l b h=6 \times 4 \times 3=72 \mathrm{~m}^{3}$
So , Area of the remaining field leaving the pit =$540-24=516 \mathrm{~m}^{2}$
$\therefore$ Height of the earth level
$=\frac{\text { volume af earth }}{\text { Area ay the remaining field}}$
$\begin{aligned}&=\frac{72}{516} \mathrm{~m} \\&=\frac{72}{516} \times 100 \mathrm{~cm}=13.9 \mathrm{~cm} \text { Approx }\end{aligned}$
Question 20
Ans : Length of field(L) $=2 \mathrm{om}$
$\text { Breadth }=14 \mathrm{~m}$
$\therefore$ Area of the field $=L \times B$
$=20 \times 14=280 \mathrm{~m}^{2}$
Length of the pit (l) = 6m
Breadth (b) = 3m
Depth (h) = 2.5m
$\therefore$ Area of the face cof the pit $=$l \times b$
$=6 \times 3=18 \mathrm{~m}^{2}$
volume cy earth dug out $=l \cdot b \cdot h$
$=6 \times 3 \times 2.5=45 \mathrm{~m}^{3}$
∴Area of the remaining field excluding the pit = 280 -18 = $262 \mathrm{~m}^{2}$
∴Height of the earth level
$=\frac{\text { volume of the earth }}{\text { Area of the remaining field }}$
$=\frac{45}{262} \mathrm{~m}=\frac{45 \times 100 \mathrm{~cm}}{262}$
$=17.175=17.18 \mathrm{~cm} .$
Question 21
Ans: Total cost of wood = Rs 182.25
Rate of wood = Rs 250 per m $^{3}$
$\therefore$ volume af the cubical block$=\frac{182.25}{250}$
$\Rightarrow \frac{18225}{100 \times 250}=0.729 \mathrm{~m}^{3}$
$=729 \times 100 \times 100 \times 100$
$=729000 \mathrm{~cm}^{3}$
$\therefore$ Edge ef cubical block
$\begin{aligned}&=\sqrt{\text { volume }} \\&=\sqrt[3]{0.729} \\&=\left[(0.9 \times 0.9 \times 0.9)^{3}\right]^{\frac{1}{3}} \\&=0.9 \mathrm{~m}=90 \mathrm{~cm} .\end{aligned}$
Question 22
Ans: A rectangular container,
Side of the square base the container, $=6 \mathrm{~cm}$
$=$ water level $=1 \mathrm{~cm}$ from the top.
- un Placing a cube in it, volume of water
$\begin{aligned}&=6 \times 6 \times 1 \mathrm{~cm}^{3}+2 \mathrm{~cm}^{2} \\&=36+2=38 \mathrm{~cm}^{3}\end{aligned}$
$\therefore$ volume of cube $=38 \mathrm{~cm}^{3}$
Question 23
Sol: By joining two cubes of 8 cm edge the resulting cuboid wall have
Length ( l) 8+ 8=16
Breadth (b) =8cm
Height (h) = 8cm
$\therefore$ Surface area $=2(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})$
$=2(16 \times 8+8 \times 8+8 \times 16) \mathrm{cm}^{2}$
$=2(128+69+128)$
$=2 \times 320 \mathrm{~cm}^{2}$
$=640 \mathrm{~cm}^{2} .$
Question 24
Sol: Let the length be $=l$
$B$ breadth $=B$
Height $\quad=H$
x= lh
y=bh
z=lb
And volume = $1 \times b \times h$
Now $x y z=16+6 h+h l$
$1^{2} b^{2} h^{2}=(lb h)^{2}$
$=v^{2}$
Proved
Question 25
Ans: Edge of metal cube $=12 \mathrm{~cm}$
Volume of cube $=(a)^{3}$
Where $a$ is 12
$\begin{aligned}&(12)^{3} \mathrm{~cm}^{3} \\&12 \times 12 \times 12 \mathrm{~cm}^{3} \\&1728 \mathrm{~cm}^{3}\end{aligned}$
On meeting it three smaller cubes are formed
Edge of first smaller cube $\left(a_{2}\right)$= 6cm
Volume =$\left(a_{2}\right)^{3}$
$=(6)^{3} \mathrm{~cm}^{3}$
$=6 \times 6 \times 6 \mathrm{~cm}^{3}$
$=216 \mathrm{~cm}^{3}$
Edge of second smaller cube $\left(a_{2}\right)=8 \mathrm{~cm}$
$\begin{aligned}\text { Volume } &=\left(a_{2}\right)^{3}=(8)^{8} \mathrm{~cm}^{3} \\&=8 \times 8 \times 8 \mathrm{~cm}^{3} \\&=512 \mathrm{~cm}^{3}\end{aligned}$
Volume of third smaller cube
$\begin{gathered}=1728-(216+512) \\=1728-728 \\=1000 \mathrm{~cm}^{3}\end{gathered}$
Edge of the third smaller cube
$\begin{aligned}&=\sqrt[3]{\text { volume }} \\=& \sqrt[3]{1000} \\=&\left(10^{3}\right)^{\frac{1}{3}} \\& 10^{3 \times \frac{1}{3}} \\&=10 \mathrm{~cm}\end{aligned}$
Question 26
Ans: No. of students $=50$
Area required for each student $=9 \mathrm{~m}^{2}$
Total area of the floor=50 \times 9 \mathrm{~m}^{2}$ $=450 \mathrm{~m}^{\circ}$
Volume required for each student =$=108 \mathrm{~m}^{3}$
Total volume of air of the room $\begin{aligned}=& 108 \times 50 \mathrm{~m}^{3} \\ &=5400 \mathrm{~m}^{3} \end{aligned}$
Length of $100 \mathrm{~m}=25 \mathrm{~m}$
$\begin{aligned}\text { Breadth } &=\frac{\text { Area of floor }}{\text { length }} \\&=\frac{450}{25} \\&=18 \mathrm{~m}\end{aligned}$
$\begin{aligned}\text { height } &=\frac{\text { volume }}{1 \times b} \\&=\frac{5400}{25 \times 18} \\&=12 m\end{aligned}$
Now breadth = 18 m and height = 12m
Question 27
Ans: Length of rectangular cardboard sheet = 42cm
Breadth = 36cm
By cutting squares from each corner , a box
is formed whose length = $42-(2 \times 6)$
$=42-12$
$=30 \mathrm{~cm}$
$\begin{aligned} \text { Breadth } &=36-2 \times 6 \\ &=36-12 \\ &=24 \mathrm{~cm} \\ \text { height } &=6 \mathrm{~cm} \\ \text { Volume } &=1 \times 10 \times \mathrm{h} \\=30 \times 24 \times 6 \mathrm{~cm}^{3} & \\=& 4320 \mathrm{~m}^{3} \end{aligned}$
Question 28
Sol: Volume of each of two cubes = $343 \mathrm{~cm}^{3}$
Edge of each cube = $\sqrt[3]{243}$
$=\left(7^{3}\right)^{\frac{1}{3}}$
$7^{3 \times \frac{1}{3}}$
$=7 \mathrm{~cm}$
Now by joining two cubes a cuboid is formed then '
Length of cuboid = 7+7=14cm
Breadth $=7 \mathrm{~cm}$
Height $=7 \mathrm{~cm}$
Surface area of the cuboid
$=2(1 b+b h+h 1)$
$=2(14 \times 7+7 \times 7+7 \times 14) \mathrm{cm}^{2}$
$=2(98+49+98) \mathrm{cm}^{2}$
$=2 \times 245$
$=490 \mathrm{~cm}^{2}$
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