Math, asked by BrainlyHelper, 1 year ago

A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5cm and the total wood used in the making of toy is 166 5/6cm³. Find the height of the toy. Also,find the cost of painting the hemi-spherical part of the toy at the rate of ₹ 10 per cm².(use π = 22/7 ).

Answers

Answered by nikitasingh79
322
SOLUTION:
Given: Radius of hemisphere = radius of cone = r = 3.5 cm
Volume of  total wood used in making a toy = 166 ⅚ = 1001/6 cm³

Let h be height of a cone.
Volume of  total wood used in making a toy = volume of hemisphere + volume of cone

(⅔) × πr³ + (⅓)π²h
= (⅓)πr²(2r + h)
1001/6 = (⅓) ×( 22/7) × (3.5)² × (h +2×3.5)
1001/6 = (⅓) ×( 22/7) × 3.5 ×3.5 ×( h+7)
1001/6 = (⅓) ×  22 × .5 ×3.5 × (h +7)
1001 × 3  = 6 × 22 × .5 ×3.5 × (h+7)
h +7  = 1001 × 3 / (6 ×  22 × .5 ×3.5)
h +7 = 1001 × 3 / 132 × 1.75
h +7 = 1001 × 3 × 100 / 132 × 175
h +7 = 91 × 3 × 4 / 12 × 7
h +7 = 91 × 12 / 12×7
h +7 = 13
h + 7 = 13 -7 = 6

Height of a cone(h) = 6

Height of the toy = Height of a cone + Height of a hemisphere
Height of the toy = 6 +3.5 = 9.5 cm

Curved surface area of hemisphere = 2πr²
CSA of hemisphere = 2 ×( 22/7) × 3.5 × 3.5
= 2 × 22 × .5 × 3.5 = 44 × 1.75 = 77 cm²

Rate of painting the hemispherical part of the toy = ₹ 10 per m².
Cost of painting the hemispherical part of the toy = 77 × 10 =  ₹ 770 .

Hence, the Height of the toy is 9.5 m & Cost of painting the hemispherical part of the toy is  ₹ 770.

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Answered by rajputdeepali832
32

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