Respuesta :
Question Q1:
Surface area of the cone:
[tex] \pi r l + \pi r^{2}=125 \pi /: \pi rl+ r^{2}=125 5l+ 5^{2}=125 5l+25=125 5l=100 l=100:5=20 [/tex]
Answer: l=20 cm.
Surface area of the cone:
[tex] \pi r l + \pi r^{2}=125 \pi /: \pi rl+ r^{2}=125 5l+ 5^{2}=125 5l+25=125 5l=100 l=100:5=20 [/tex]
Answer: l=20 cm.