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open metallic bucket is in the shape of a frustum of a cone. If themeters of the two circular ends of the bucket are 45 cm and 25 cm ande vertical height of the bucket is 24 cm, find the area of the metallicthsheet used to make the bucket. Also find the volume of the water it canhold. (Use π-_-)7 |
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Answer» 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 |
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