A semicircular sheet of diameter 28 cm is bent into an open conical cup . Find the depth and capacity of this cup
Answers
circumference of circular part=28
2×22/7×r=28
r=28×7/44
r=49/11 cm.
semicircular sheet when fold then it's radius become height of cone.
so, height of cone=14cm (d=28cm)
now capacity =volume of cone=1/3×22/7×49/11×14
=196/3cm cubic
=68.7cm cubic.
Answer:
depth = 12 cm (approx.)
volume = 616 cm³
Step-by-step explanation:
diameter= 28 cm
radius of sphere = 14 cm = slant height of cone
curved part of the sheet = circumference of the base of cone
curved part of sheet = (2π r)/2
= π r
= 22/7 *14
= 44 cm
circumference of base of cone = 2 π r
⇒ 44 = 2*22/7*r
⇒ 7 cm = r
height of cone = √l²-r²
⇒√(14)²-(7)²
⇒√196-49
⇒√146⇒12 cm(approx.)
∴depth = 12 cm (approx.)
∴ volume = 1/3πr²h
⇒1/3×22/7×7×7×12
⇒22×7×4⇒616 cm³