SChand CLASS 9 Chapter 18 Surface Area and Volume of 3D Solids Exercise 18(A)

 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

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

$=2(14 \times 7+7 \times 7+7 \times 14) \mathrm{cm}^{2}$

$=2(98+49+98) \mathrm{cm}^{2}$

$=2 \times 245$

$=490 \mathrm{~cm}^{2}$