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A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket

[Use $\pi = 3.14$ ].

Answers (1)

Diameter of cylinder = 6 cm

Radius = 3 cm

Height = 12 cm

$\text{ Surface are}$ $=2 \pi r h = 2\pi \times 3 \times 12 = 72 \pi cm^2$

Similarly radius of circle = 3 cm

$\text{ Area}$ $=\pi r^2=\pi 3 \times 3 = 9 \pi cm^2$

Slant height of cone (l)=5cm

Radius = 3 cm

$\text{ Surface area}=\pi r l =\pi \times 5 \times 3 = 15\pi cm^2$

Total surface area = area of cylinder + area of circle + area of the cone

$=72\pi + 9 \pi + 15 \pi$

$=96\pi$

$=96\times 3.14=301.7 cm^2$

Slant height (l) = 5cm

We know that,

$h^2+r^2=l^2$

$h^2+3^2=5^2$

$\\h^2= 25 -9 \\ h^2=16 \\\ h = 4$

$\text{ Volume of cylinder= }\pi r^2 h \\ = \pi \times 3 \times 3 \times 12 = 108 \pi cm^3$

$\\\text{ Volume of cone}=\frac{1}{3}\pi r^2 h \\ = \frac {1}{3} \pi \times 3 \times 3 \times 12 = 12 \pi cm^3$

$\text{ Volume of rocket }$ $=180 \pi + 12 \pi$

$=120 \pi$

$=120 \times 3.14 =377.14 cm^3$

Posted by

infoexpert21

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