5 A solid cone of base radius 10 cm is cut into two
parts through the mid-point of its height by a plane
parallel to its base. Find the ratio of the volumes of
the two parts of the cone.
1lt
Answers
Answer:
Step-by-step explanation:
SOLUTION :
Let r & R be the radius of the lower part of the frustum.
Height of a cone , AB’ = 10 cm
Height of a Smaller cone, AB = 5 cm
[Cut through the midpoint of its height]
From the figure,
AB = h = 5
AB’ = 2h = 10
BC = r
B'C = R
In ∆ABC & ∆AB’C’ ,
∠ABC = ∠AB’C’ (each 90°)
∠ACB = ∠AC’B’ (corresponding angles)
∆ABC ∼ ∆AB’C’ [By AA Similarity]
BC/B'C’ = AB/AB’
[Corresponding sides of a similar triangles are proportional]
r/R = 5 /10
r/R = ½
R = 2r
Volume of the upper part (Smaller cone) = ⅓ πr²h
Volume of solid cone = ⅓ π R²2h
= ⅓ π (2r)² 2h = ⅓ π × 4r² × 2h = 8/3πr²h
Volume of lower part (frustum) = volume of solid cone - volume of Smaller cone = 8/3πr²h - ⅓ πr²h = 7/3 πr²h
Volume of lower part (frustum) = 7/3 πr²h
Volume of the upper part (Smaller cone)/ Volume of lower part (frustum) =
⅓ πr²h / 7/3 πr²h
= 1/7
Hence, the ratio of volume of two parts of the cone is 1 : 7 .