Verified answer

Answer:

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: Volume_{(Cuboid)}=81 \: {m}^{3} \qquad \: \\ \\& \qquad \:\sf \: CSA_{(Cuboid)}=37.8 \: {m}^{2} \\ \\& \qquad \:\sf \: TSA_{(Cuboid)}= 217.8\end{aligned}} \qquad \: \\ \\ [/tex]

and

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: Volume_{(Cuboid)}=0.288 \: {m}^{3} \qquad \: \\ \\& \qquad \:\sf \: CSA_{(Cuboid)}=2.16 \: {m}^{2} \\ \\& \qquad \:\sf \: TSA_{(Cuboid)}= 2.736\end{aligned}} \qquad \: \\ \\ [/tex]

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-1}}[/tex]

Given that,

Dimensions of cuboid are

Length of cuboid, l = 15 m

Breadth of cuboid, b = 6 m

Height of cuboid, h = 9 dm = 0.9 m

Now,

[tex]\sf \: Volume_{(Cuboid)} = lbh \\ \\ [/tex]

[tex]\sf \: \qquad\qquad = 15 \times 6 \times 0.9 \\ \\ [/tex]

[tex]\sf \: \qquad\qquad = 81 \: {m}^{3} \\ \\ [/tex]

Now,

[tex]\sf \: CSA_{(Cuboid)} = 2(l + b) \times h \\ \\ [/tex]

[tex]\sf \: \qquad\qquad= 2(15 + 6) \times 0.9 \\ \\ [/tex]

[tex]\sf \: \qquad\qquad= 42 \times 0.9 \\ \\ [/tex]

[tex]\sf \: \qquad\qquad= 37.8 \: {m}^{2} \\ \\ [/tex]

Now,

[tex]\sf \: TSA_{(Cuboid)} = 2(lb + bh + hl) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2(15 \times 6 + 6 \times 0.9 + 0.9 \times 15) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2(90 + 5.4 + 13.5) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2 \times 108.9 \\ \\ [/tex]

[tex]\qquad\sf \: = \: 217.8 \: {m}^{2} \\ \\ [/tex]

[tex] \\ \large\underline{\sf{Solution-2}}[/tex]

Given that,

Dimensions of cuboid are

Length of cuboid, l = 48 cm = 0.48 m

Breadth of cuboid, b = 6 dm = 0.6 m

Height of cuboid, h = 1 m

Now,

[tex]\sf \: Volume_{(Cuboid)} = lbh \\ \\ [/tex]

[tex]\sf \: \qquad\qquad = 0.48 \times 0.6 \times 1 \\ \\ [/tex]

[tex]\sf \: \qquad\qquad = 0.288 \: {m}^{3} \\ \\ [/tex]

Now,

[tex]\sf \: CSA_{(Cuboid)} = 2(l + b) \times h \\ \\ [/tex]

[tex]\sf \: \qquad\qquad= 2(0.48 + 0.6) \times 1 \\ \\ [/tex]

[tex]\sf \: \qquad\qquad= 2.16 \: {m}^{2} \\ \\ [/tex]

Now,

[tex]\sf \: TSA_{(Cuboid)} = 2(lb + bh + hl) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2(0.48 \times 0.6 + 0.6 \times 1 + 1 \times 0.48) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2(0.288 + 0.6 + 0.48) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2.736 \: {m}^{2} \\ \\ [/tex]

1)

according to ur question,

length (1)=15m

breadth (b) = 6m height (h)=9dm=(9/10) 1 = 0.9m

therefore, volume = l×b×h

= 15 × 6 × 0.9

= 81m³

2)

Given Length = 48cm = 0.48m breadth = 6dm = 0.6m and height = 1n

We know that volume of cuboid × length × breadthheight

So,

V = (0.48 × 0.6 × 1)

V = 0.288m³

We know that total surface area of cuboid = 2(lb + bh + hl)

So,

Surface area =2( 0.48 × 0.6 + 0.6 × 1 + 1 × 0.48)

Surface area = 2(0.288 + 0.6 + 0.48)

Total Surface area = 2.736 m²

So,

Lateral surface area = 2[(0.48 + 0.6) × 1] Lateral surface area = 261m²