Question

An option metallic bucket is in the shape of a frustum of a cone. If the diameters of the two circular ends of the bucket are 45 cm and 25 cm and the vertical height of the bucket is 24 cm, find the area of the metallic sheet used to make the bucket. Also find the volume of the water it can hold.
（Use π = 22/7)

Answer 1

$\boxed{\boxed{ \overline { \underline{\bf \red{ SOLUTION ☻ }}}}}$

$\rm \: Given, \: radii \: of \: both \: circular \: ends \: are \:$

$\rm \: r_1 \: = \frac{45}{2} = 22.5 \: cm,$

$\rm \: r_2 \: = \frac{25}{2} = 12.5 \: cm \: and \: h = 24 \: cm$

$\rm \: Now, \: slant \: height,$

$l = \rm \: \sqrt{h {}^{2} + (r_1 \: - \: r_2) {}^{2} }$

$= \rm \: \sqrt{(24) {}^{2} + (22.5 - 12.5 ) {}^{2} }$

$= \rm \: \sqrt{(24) {}^{2} + (10) {}^{2} }$

$= \rm \: \sqrt{576 - 100}$

$= \rm \: \sqrt{676} = 26 \: cm$

$\rm \red{• Area \: of \: metallic \: sheet \: used} \\ \rm = (curved \: surface \: area \: of \: the \: frustum \: of \: the \: cone) + (area \: of \: base)$

$= \pi \: l \: (r_1 \: + \: r_2) + \pi \: r_2 \: {}^{2}$

$= \pi \: [l( \rm \: r_1 \: + r_1) + r_2 \: {}^{2}]$

$= \pi \: [26(22.5 + 12.5) + (12.5) {}^{2}]$

$= \frac{22}{7} \: [26 \times 35 + (12.5) {}^{2}]$

$= \frac{22}{7}(910 + 156.25)$

$= \frac{22}{7} \times 1066.25$

$= \rm \frac{23457.5}{7} = 3351.07 \: cm {}^{2}$

$\rm \: Now, \: the \: volume \: of \: the \: water \: hold \: bucket$

$\rm = \frac{1}{3}\pi \: h(r_1 \: + \: r_2 \: + r_1 \: r_2)$

$= \rm \frac{1}{3} \times \frac{22}{7} \times 24(22.5) {}^{2} + (12.5) {}^{2} + 22.5 \times 12.5$

$= \rm \frac{22}{7} \times 8(506.25 + 156.25 + 281.25)$

$= \rm \frac{22}{7} \times 8(943.75) = \frac{166100}{7}$

$\boxed{ \bf \red{ = 23728.57 \: cm {}^{3}. }}$

Answer 2

Thanks for free points