Question

A cone of radius 10cm is cut into two parts by a plane through the mid-point of its vertical axis parallel to the base. Find the ratio of the volumes of the smaller cone and frustum of the cone​

Answer 1

Let the height of the cone be H and radius be R.

Let the height of the cone formed by the cut will be h and radius be r.

Fig. shows the side view of the cone, on which it is clear that the triangles ABE and CDE are similar. So,

$\longrightarrow \dfrac{h}{H}=\dfrac{r}{R}\quad\quad\dots(i)$

Volume of the whole cone will be,

$\longrightarrow V=\dfrac{1}{3}\,\pi R^3H\quad\quad\dots(1)$

Volume of the cone formed by the cut will be,

$\longrightarrow v=\dfrac{1}{3}\,\pi r^3h\quad\quad\dots(2)$

Dividing (2) by (1), we get,

$\longrightarrow\dfrac{v}{V}=\left(\dfrac{r}{R}\right)^3\dfrac{h}{H}$

But by (i),

$\longrightarrow\dfrac{v}{V}=\left(\dfrac{h}{H}\right)^3\dfrac{h}{H}$

$\longrightarrow\dfrac{v}{V}=\left(\dfrac{h}{H}\right)^4$

In the question, the cone is cut at the midpoint of the vertical axis, so,

• $\dfrac{h}{H}=\dfrac{1}{2}$

Therefore,

$\longrightarrow\dfrac{v}{V}=\left(\dfrac{h}{H}\right)^4$

$\longrightarrow\dfrac{v}{V}=\left(\dfrac{1}{2}\right)^4$

$\longrightarrow\dfrac{v}{V}=\dfrac{1}{16}\quad\quad\dots(3)$

We need to find ratio of volume of smaller cone to frustum. Volume of frustum is equal to $V-v$ here.

Apply rule of dividendo in (3) as the following.

$\longrightarrow\dfrac{v}{V-v}=\dfrac{1}{16-1}$

[Here denominator is replaced by numerator subtracted from it.]

$\longrightarrow\underline{\underline{\dfrac{v}{V-v}=\dfrac{1}{15}}}$

This is the required answer.

Answer 2

☆Given Question :-

• A cone of radius 10cm is cut into two parts by a plane through the mid-point of its vertical axis parallel to the base. Find the ratio of the volumes of the smaller cone and frustum of the cone.

$\huge \orange{AηsωeR}$

☆Given :-

• A cone of radius 10cm is cut into two parts by a plane through the mid-point of its vertical axis parallel to the base.

☆To Find :-

• The ratio of the volumes of the smaller cone and frustum of the cone.

☆Formula used :-

${{ \boxed{\large{\bold\red{Volume_{(Cone)}\: = \dfrac{1}{3} \:\pi r^2 h }}}}}$

☆Solution :-

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{h \: be \: the \: height \: of \: small \: cone} \\ &\sf{H \: be \: the \: height \: of \: big \: cone} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{r \: be \: the \: radius \: of \: small \: cone} \\ &\sf{R \: be \: the \: radius \: of \: big \: cone} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{v \: be \: the \: volume \: of \: small \: cone} \\ &\sf{V \: be \: the \: volume \: of \: big \: cone} \end{cases}\end{gathered}\end{gathered}$

☆According to statement,

⟼ Radius, R of big cone = 10 cm

⟼ Height of small cone = h cm

⟼ Height of big cone, = H cm

⟼ Since, cone is cut into two parts by a plane through the mid-point of its vertical axis parallel to the base.

$\sf \: ⟼Therefore, \: H = \: 2h$

⟼ Now, Using Concept of Similarity,

$\sf \: ⟼\dfrac{r}{R} = \dfrac{h}{H}$

$\sf \: ⟼\dfrac{r}{10} = \dfrac{h}{2h}$

$\sf \: ⟼\dfrac{r}{10} = \dfrac{1}{2}$

$\bf\implies \:r = 5 \: cm$

⟼ Now, we compare the volumes of small cone with volume of big cone.

⟼ So, Let us Consider,

$\bf \:\dfrac{v}{V} = \dfrac{\dfrac{1}{3} \pi \: {r}^{2} h}{\dfrac{1}{3} \pi \: {R}^{2} H}$

☆ On substituting the values of r, R and H, we get

$\sf \: ⟼\dfrac{v}{V} = \dfrac{5 \times 5 \times h}{10 \times 10 \times 2h}$

$\bf \:  ⟼ \dfrac{v}{V} = \dfrac{1}{8}$

$\bf\implies \:V = 8v$

☆ Now,

$\sf \: ⟼Volume \: of \: Frustum = V - v$

$\sf \: ⟼Volume \: of \: Frustum = 8v - v = 7v$

$\bf\implies \:\dfrac{Volume \: of \: smaller \: cone}{Volume \: of \: Frustum = } = \dfrac{v}{7v} = \dfrac{1}{7}$

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