Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:

Solution:

Volume of a cuboid $=($ Length $\times$ Breadth $\times$ Height $)$ cubic units

Total surface area $=2(l b+b h+l h)$ sq units

Lateral surface area $=[2(l+b) \times h]$ sq units

(i) Length $=22 \mathrm{~cm}$, breadth $=12 \mathrm{~cm}$, height $=7.5 \mathrm{~cm}$

Volume $=($ Length $\times$ Breadth $\times$ Height $)=(22 \times 12 \times 7.5)=1980 \mathrm{~cm}^{3}$

Total surface area $=2(l b+b h+l h)=2[(22 \times 12)+(22 \times 7.5)+(12 \times 7.5)]=2[264+165+90]=1038 \mathrm{~cm}^{2}$

Lateral surface area $=[2(l+b) \times h]=2(22+12) \times 7.5=510 \mathrm{~cm}^{2}$

(ii) Length $=15 \mathrm{~m}$, breadth $=6 \mathrm{~m}$, height $=9 \mathrm{dm}=0.9 \mathrm{~m}$

Volume $=($ Length $\times$ Breadth $\times$ Height $)=(15 \times 6 \times 0.9)=81 \mathrm{~m}^{3}$

Total surface area $=2(l b+b h+l h)=2[(15 \times 6)+(15 \times 0.9)+(6 \times 0.9)]=2[90+13.5+5.4]=217.8 \mathrm{~m}^{2}$

Lateral surface area $=[2(l+b) \times h]=2(15+6) \times 0.9=37.8 \mathrm{~m}^{2}$

(iii) Length $=24 \mathrm{~m}$, breadth $=25 \mathrm{~cm}=0.25 \mathrm{~m}$, height $=6 \mathrm{~m}$

Volume $=($ Length $\times$ Breadth $\times$ Height $)=(24 \times 0.25 \times 6)=36 \mathrm{~m}^{3}$

Total surface area $=2(l b+b h+l h)=2[(24 \times 0.25)+(24 \times 6)+(0.25 \times 6)]=2[6+144+1.5]=303 \mathrm{~m}^{2}$

Lateral surface area $=[2(l+b) \times h]=2(24+0.25) \times 6=291 \mathrm{~m}^{2}$

(iv) Length $=48 \mathrm{~cm}=0.48 \mathrm{~m}$, breadth $=6 \mathrm{dm}=0.6 \mathrm{~m}$, height $=1 \mathrm{~m}$

Volume $=($ Length $\times$ Breadth $\times$ Height $)=(0.48 \times 0.6 \times 1)=0.288 \mathrm{~m}^{3}$

Total surface area $=2(l b+b h+l h)=2[(0.48 \times 0.6)+(0.48 \times 1)+(0.6 \times 1)]=2[0.288+0.48+0.6]=2.736 \mathrm{~m}^{2}$

Lateral surface area $=[2(l+b) \times h]=2(0.48+0.6) \times 1=2.16 \mathrm{~m}^{2}$