Question

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

(i) length = 22 cm, breadth = 12 cm and height = 7.5 cm

(ii) length = 15 in, breadth = 6 in and height = 9 dm

(ill) length = 24 m, breadth = 25 cm and height = 6 m

(iv) length tr- 48 cm. breadth = 6 dm and height = 1 In

Answer 1

Length of cuboid (l)=22 cm.

Breadth (b)=12 cm.

and height (h)=7.5 cm.

(a)Volume =lb $=22\times 12\times 7.5c{m}^3$

$=1980c{m}^3$

(b)Lateral surface area $=2(l+b)\times h$

$=2(22+12)\times 7.5c{m}^2$

$=2\times 34\times 7.5=510c{m}^2$

(c)Total surface area =2[lb+bh+hl]

$=2[22\times 12+12\times 7.5+7.5\times 22]c{m}^2$

$=2[264+90+165]c{m}^2$

$=2\times 519=1038c{m}^2$

(ii)Length of cuboid (l)=15 m

Breadth (b)=6 m

Height (h)=9 dm =0.9 m

(a)Volume =lbh $=15\times 6\times 0.9=81m^3$

(b)Lateral surface area $=2(l+h)\times h$

$=2(15+6)\times 0.9m^2$

$=2\times 21\times 0.9=37.8m^2$

(c)Total surface area =2[lb+bh+hl]

$=2(15\times 6+6\times 0.9+0.9\times 15)m^2$

$=2[90+5.4+13.5]m^2$

$=2\times 108.9=217.8m^2$

(iii)Length of cuboid (l)=24 m

Breadth (b)=25 cm $=\frac{1}{4}m$

Height (h)=6 m

(a)Volume =lbh $=24\times \frac{1}{4}\times 6=36m^3$

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

$=2(24+\frac{1}{4})\times 6m^2$

$=2\times \frac{97}{4}\times 6=291m^2$

(c)Total surface area =2(lb+bh+hb)

$=2(24\times \frac{1}{4}+\frac{1}{4}\times 6+6\times 24)m^2$

$=2(6+\frac{3}{2}+144)m^2$

$=2\left(\frac{12+3+288}{2}\right)m^2$

$=2\times \frac{303}{2}=303m^2$

(iv)Length of cuboid (l)=48 cm.

Breadth (b)=6 dm =60 cm.

Height (h)=1 m =100 cm.

(a)Volume =lbh $=48\times 60\times 100$

$=288000c{m}^3$

(b)Lateral surface area $=2(l+b)\times h$

$=2(48+60)\times 100c{m}^2$

$=2\times 108\times 100=21600c{m}^2$

(c)Total surface area =2[lb+bh+hl]

$=2[48\times 60+60\times 100+100\times 48]c{m}^2$

$=2[2800+6000+4800]c{m}^2$

$=2\left[13680\right]=27360c{m}^2$