A cone of radius 10cm is cut into 2parts by a plane through the midpoint of its vertical axis parallel to the base . Find the ratio of the volumes of the smaller cone and frustum of the cone?
Answers
Let’s assume from the attached figure below,
“2h” = OB = height of the large cone
“R” = BC = radius of the base of the large cone = 10 cm (given)
When the cone is cut into 2 by the plane through the midpoint of its vertical axis parallel to its base, then we will get a smaller cone and a frustum.
Then,
“r” = AD = radius of the base of the smaller cone
“h” = OA = height of the smaller = AB = height of the frustum = ½ * OB
Now, first, let’s consider ∆ OAD and ∆ OBC, we have
∠A = ∠A …… [common angle]
∠OAD = ∠OBC ……. [corresponding angles, since AD//BC]
∴ By AA similarity, ∆ OAD ~ ∆ OBC
Since the corresponding sides of two similar triangles are proportional to each other.
∴ OA/OB = AD/BC
⇒ h/2h = r/R
⇒ r/R = ½
⇒ r = ½ * 10 ….. [since R = 10 cm]
⇒ r = 5 cm
Therefore,
The volume of the smaller cone is given by,
= 1/3 * π * (r)²(h)
= 1/3 * π * (5)²(h) …… (i)
And,
The volume of the large cone is given by,
= 1/3 * π * (R)²(2h)
= 1/3 * π * (10)²(2h) …… (ii)
So, by subtracting (i) from (ii), we get
The volume of the frustum as,
= [1/3 * π * (10)² * (2h)] - [1/3 * π * (5)² * (h)]
= 1/3 * π * h * (200 – 25)
= 1/3 * π * h * 175 ……. (iii)
Thus, from (i) & (iii), we get
The ratio of the volume of the smaller cone to the volume of the frustum as,
= [volume of the smaller cone] / [volume of the frustum]
= [1/3 * π * (5)² * (h)] / [1/3 * π * h * 175]
= [1/3 * π * (5)²(h)] / [1/3 * π * h * 175]
Cancelling all the similar terms
= (25) / (175)
= 1/7
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