Question

a bucket open at the top and made up of a metal sheet is in the form of a frustum of a cone.The depth of bucket is 24cm and the diameters of its upper and lower circular ends are 30cm and 10cm.Find the cost of metal sheet used in it at the rate of Rs10 per 100cm2.(pi=3.14)

Answer 1

Solution:

Total Surface Area of Frustum of cone:

$\pi \: l(r + R) + \pi {r}^{2} + \pi {R}^{2}$
Where

$l = \sqrt{ {h}^{2} + ( {R - r)}^{2} }$
As we know that bucket is open from top,So to calculate total surface area of Bucket we must subtract area of top

TSA of Bucket
$= \pi \: l(r + R) + \pi {r}^{2}$

h =24 cm

R = 15 cm

r = 5 cm

$l = \sqrt{ ({24})^{2} + {(15 - 5)}^{2} } \\ \\ = \sqrt{576 + 100} \\ \\ = \sqrt{676} \\ \\ l = 26 \: cm$
Now put all these values in Formula

$\pi \: l(r + R) + \pi {r}^{2} \\ \\ = 3.14 \times 26(5 + 15) + 3.14 {(5)}^{2} \\ \\ = 3.14(26 \times 20 + 25) \\ \\ = 3.14(520 + 25) \\ \\ = 3.14 \times 545 \\ \\ T.S.A.= 1711.3 \: {cm}^{2}$

To find the total cost of sheet metal:

$= \frac{10 \times 1711.3}{100} \\ \\ = 171.13 \: Rs$
Hope it helps you.

Answer 2

Answer:

this is the correct answer....