IDNStudy.com, kung saan ang iyong mga tanong ay natutugunan ng mabilis na sagot. Makakuha ng hakbang-hakbang na mga gabay para sa lahat ng iyong teknikal na tanong mula sa mga miyembro ng aming komunidad.
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

Salamat sa iyong pakikilahok. Huwag kalimutang magtanong at magbahagi ng iyong kaalaman. Ang iyong ambag ay napakahalaga sa aming komunidad. Ang IDNStudy.com ay nangako na sasagutin ang lahat ng iyong mga tanong. Salamat at bisitahin kami palagi.