Question

A solid is made up of a cube and a hemisphere is attached on its top, as shown in the figure. Each edge of the cube measures 5 cm and the hemisphere has a diameter of 4.2 cm. Find the total area to be painted.
Or, the diameters of the lower and upper ends of a bucket, in the form of a frustum of a cone, are 10 cm and 30 cm respectively. If its heights is 24 cm, find
(i) the capacity of the bucket
(ii) the area of the metal sheet used to make the bucket

Answer 1

Total surface area of the cube = 6 × (edge)²
= (6 × 5 × 5) cm²
Area to be painted = (Total surface area of the cube) - (Base area of the hemisphere) + (CSA of the hemisphere)
= $\left(150-{\mathrm{\pi r}}^2+2{\mathrm{\pi r}}^2\right)$ cm²
$=\left(150+{\mathrm{\pi r}}^2\right)$cm²
$=\left(150+\frac{22}{7}\times 2.1\times 2.1\right)$ cm²
$=\left(150+\frac{693}{50}\right)$ cm²
= (150 + 13.86) cm² = 163.86 cm²
Hence, required area to be painted is 163.86 cm².

OR

We have:
R = 15 cm, r = 5 cm and h = 24 cm
l² = h² + (R - r)²
= ((24)² + (15 - 5)²} = (576 + 100) = 676
⇒ l = 26 cm

(i)
$\mathrm{Capacity}\mathrm{of}\mathrm{the}\mathrm{bucket}=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{R}}^2+{\mathrm{r}}^2+\mathrm{Rr}\right){\mathrm{cm}}^3$
$=\frac{1}{3}\left\{3.14\times 24\left({15}^2+5^2+15\times 5\right)\right\}{\mathrm{cm}}^3$
$=\frac{1}{3}\left\{3.14\times 24\left(225+25+75\right)\right\}{\mathrm{cm}}^3=\frac{1}{3}\left\{3.14\times 24\times 325\right\}{\mathrm{cm}}^3$
= 8164 cm³

(ii)
$\mathrm{Required}\mathrm{area}\mathrm{of}\mathrm{t}\mathrm{he}\mathrm{metal}\mathrm{sheet}=\mathrm{\pi }\left[{\mathrm{r}}^2+\mathrm{l}\left(\mathrm{R}+\mathrm{r}\right)\right]\mathrm{sq}.\mathrm{cm}$
$=3.14\left[5^2+26\left(15+5\right)\right]{\mathrm{cm}}^2$
$=3.14\times \left(25+520\right){\mathrm{cm}}^2$
$=3.14\times 545{\mathrm{cm}}^2$
$=1711.3{\mathrm{cm}}^2$