Question

A bucket open at the top is in the form of a frustum of a cone the diameter of its upper and lower circular endsare 40cm and 20cm respectively if total 17600 cm^3 of water can be filled in bucket find its total surface area of the b
ucket

Answer 1

here total surface area is CSA + area of base as it is opened at top

Answer 2

Its total surface area of the  bucket is 2547.31 sq.cm.

Step-by-step explanation:

Bucket is in the form of frustum

Upper diameter = 40 cm

Upper radius R= $\frac{40}{2}=20 cm$

Lower diameter = 20 cm

Lower radius r= $\frac{20}{2} = 10 cm$

Volume of bucket= $\frac{\pi}{3}h(R^2+r^2+Rr)=\frac{\pi}{3} h(20^2+10^2+(20)(10))=\frac{\pi}{3} h(700)$

We are given that  total 17600 cm^3 of water can be filled in bucket

So,$\frac{\pi}{3}h(700)=17600\\\\\frac{\frac{22}{7}}{3}(700)h=17600\\\\h = \frac{17600}{\frac{22}{7} \times 3 \times 700}$

h = 2.67

Total Surface area of bucket =$\pi (R+r)\sqrt{(R-r)^2+h^2}+\pi r^2 + \pi R^2 = \frac{22}{7}(20+10)\sqrt{(20-10)^2+2.67^2}+\frac{22}{7} (20)^2+\frac{22}{7}(10)^2=2547.31 cm^2$

Hence its total surface area of the  bucket is 2547.31 sq.cm.

#Learn more:

The diameter of the lower and upper hands of a bucket is in the form of frustum of a cone are 10 cm and 30 cm respectively if its height is 24 cm find the area of the metal sheet used to make the bucket

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