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Sagot :
Given: Radius of Cylinder= 1m
Height of Cylinder=2m
Volume of Cylinder= πr²h
=π(1m)²(2m)
2πm³
Since the radius of the base is 1m we can assume that the small cone has 1m radius also.
Volume of Small Cone=1/3πr²h
2πm³=1/3π(1m)²h (Volume of Cylinder=Volume of Small Cone)
Solving for h.
h=6m
Substituting h=6m
Volume of Small Cone=1/3π(1m)²(6m)
Volume of Small Cone=2πm³ or 6.28m³
By Pythagorean Theorem,
l²=h²+r²
l=√(6m)²+(1m)²
l=√37 m or 6.08m
Height of Cylinder=2m
Volume of Cylinder= πr²h
=π(1m)²(2m)
2πm³
Since the radius of the base is 1m we can assume that the small cone has 1m radius also.
Volume of Small Cone=1/3πr²h
2πm³=1/3π(1m)²h (Volume of Cylinder=Volume of Small Cone)
Solving for h.
h=6m
Substituting h=6m
Volume of Small Cone=1/3π(1m)²(6m)
Volume of Small Cone=2πm³ or 6.28m³
By Pythagorean Theorem,
l²=h²+r²
l=√(6m)²+(1m)²
l=√37 m or 6.08m
Please refer to the attached photo for the solution.
V = [tex] \frac{9}{32} [/tex] π m³
LAsmallcone = [tex] \frac{9 \sqrt{5} }{16} [/tex] π m²
V = [tex] \frac{9}{32} [/tex] π m³
LAsmallcone = [tex] \frac{9 \sqrt{5} }{16} [/tex] π m²

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