Math, asked by ravneesh1, 1 year ago

an open metal bucket is in shape of a frustum mounted on a cylindrical base made of same metallic sheet the dia of two circular ends of bucket are 45 and 25 cm the total height of bucket is 40 cm and base of cylinder is 6cm find area of metallic sheet

Answers

Answered by Golda
197
Solution :-

Area of metallic sheet used = CSA of Frustum + CSA of Cylinder + CSA of Base

CSA of Frustum -

Diameter of the bigger circular end = 45 cm

Radius = 45/2 = 22.5 cm

Diameter of the smaller circular end = 25 cm

Radius = 25/2 = 12.5 cm

Height of the frustum = Total height of the bucket - Height of the circular base 

⇒ 40 - 6 = 34 cm

Slant Height = l √h² + (r1² - r2²)²

⇒ √34² + (22.5 - 12.5)²

⇒ √1156 + (10)²

⇒ √1156 + 100

⇒ √1256

⇒ Slant Height = 35.44 cm

CSA of Frustum = π(r1 + r2)l

⇒ 22/7*(22.5 + 12.5)*35.44

⇒ 22/7*35*35.44

= 3898.4 cm²

Area of Circular Base -

Base is a circular part with radius 25/2 = 12.5 cm

Area of circular base = πr²

⇒ 22/7*12.5*12.5

491.07 cm²

CSA of Cylinder = 2πrh

⇒ 2*22/7*12.5*6

⇒ 471.428

Area of metallic sheet used = 3898.4 cm² + 491.07 cm² + 471.428

= 4860.898 cm²

Answer.
Answered by psupriya789
1

Area of metallic sheet used = curved surface area of Frustum + curved surface area of Cylinder + area of circular Base.

Diameter of the bigger circular end = 45 cm

Radius, r1 = 452= 22.5 cm  

Diameter of the smaller circular end = 25 cm

Radius, r2 = 252=12.5 cm

Height of the frustum, h = Total height of the bucket (H) - Height of the circular base (h1)

= 40 - 6 = 34 cm

Slant Height of frustum (l) =h2+(r1−r2)2−−−−−−−−−−−−√

⇒l=342+(22.5−12.5)2−−−−−−−−−−−−−−−−√

⇒l=1156+(10)2−−−−−−−−−−√

⇒l=1156+100−−−−−−−−−√

⇒l=1256−−−−√

⇒ Slant Height, l = 35.44 cm

Curved surface area of Frustum = π(r1+r2)l

= 227×(22.5+12.5)×35.44

⇒227×35×35.44

= 3898.4 cm2

Area of Circular Base with radius 252=12.5 is given by,

Area of circular base = πr

=227×12.5×12.5

= 491.07 cm²

Curved surface area of Cylinder = 2πr2h1

= 2×227×12.5×6

= 471.428 cm²

Area of metallic sheet used = 3898.4 cm² + 491.07 cm² + 471.428 cm²

= 4860.898 cm²

Volume of water in bucket = Volume of Frustum

=  13πh(r21+r22+r1r2)  

= 13×227×34(22.52+12.52+22.5×12.5) = 33615.48 cm³

Now, 1 litre = 1000 cm

Thus, Volume of water in bucket in litres = 33.62 litres

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