The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.
Answers
Answer:
The ratio of volume of two parts of the cone is 1 : 7 .
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 .
{The height and radius of the given cone be H and R respectively}
{Upper part is a smaller cone and the bottom part is the frustum of the cone.}
{⇒ OC = CA = h/}
{Let the radius of smaller cone be r cm}
{In ΔOCD and ΔOAB,}
{∠OCD = ∠OAB = 90°}
{∠COD = ∠AOB (common)}
{NOTE∴ ΔOCD ∼ ΔOAB (AA similarity}
{⇒ OA/OC = AB/CD = OB/OD}
{⇒ h / h/2 = R/r}
{⇒ R = 2r}
{ Radius and height of the cone OCD are r cm and h/2 cm}
{therefore the volume of the cone OCD = 1/3 x π x r x h/2 = 1/6 πr h}
{Volume of the cone OAD = 1/3 x π x R x h = 1/3 x π x 4r x h}
{The volume of the frustum = Volume of the cone OAD - Volume of the cone OCD}
{= (1/3 x π x 4r x h) – (1/3 x π x r x h/2)}
{= 7/6 πr h}
{Ratio of the volume of the two parts = Volume of the cone OCD : volume of the frustum}
{= 1/6 πr h : 7/6 πr h}
{}= 1 : 7{}
ANSWER---(1:7)
hence proved:---
HOPE IT HELPS:---
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