Find ratio of the volume of two cone

geometryratiovolume

Given two sector ABC and PQR, $\angle A=2\theta$, $\angle P=3\theta, AC=2r, PR=3r, $ both sectors are folded into a right circular cone, find the ratio of the volume of two cone.

I am having trouble doing this question, and I doubt if the result is not a simple ratio. Here is what I have got:

The ratio of the base area = $16 : 81$

The ratio of the height =
$\frac{2r}{360^\circ}\sqrt{(360^\circ)^2-4\theta^2}:\frac{3r}{360^\circ}\sqrt{(360^\circ)^2-9\theta^2}$

And it cannot be further simplified.

Any form of help will be appreciated.

Best Answer

I got same ratio of the base area. $r_a=\dfrac{4r\theta}{\pi}$, $r_b=\dfrac{9r\theta}{\pi}$

$h_a^2=4r^2-r_a^2=4r^2-\dfrac{16r^2\theta^2}{\pi^2}$

$h_b^2=9r^2-r_b^2=9r^2-\dfrac{81r^2\theta^2}{\pi^2}$

$Va=\dfrac13{\pi}r_a^2h_a=\dfrac13{\pi}(\dfrac{4r\theta}{\pi})^2\sqrt{4r^2-\dfrac{16r^2\theta^2}{\pi^2}}$

$V_a:V_b=32\sqrt{1-\dfrac{4\theta^2}{\pi^2}}:243\sqrt{1-\dfrac{9\theta^2}{\pi^2}}$

Related Solutions

[Math] How to find the volume of a cone in terms of theta

The angle of the sector not taken is $2\pi-\theta$. The length of the circular arc with central angle $2\pi-\theta$ radians and radius $r$ is $r(2\pi-\theta)=2\pi r-\theta r$ (by the definition of radian measure). When you form the cone, this will be the circumference of the circle at the base of the cone. The radius of that circle is $\frac{2\pi r-\theta r}{2\pi}=r\left(1-\frac{\theta}{2\pi}\right)$ (call it $R$).

The "slant height" of the cone is the radius of the original circle, $r$, and is the hypotenuse of a right triangle with base $R=r\left(1-\frac{\theta}{2\pi}\right)$ and height $h$ (let's say). Therefore

$$h=\sqrt{r^2-\left[ r\left(1-\frac{\theta}{2\pi}\right) \right]^2}$$ $$=r\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}$$

Therefore the volume of the cone will be

$$V=\frac 13\pi R^2h=\frac 13\pi\left[r\left(1-\frac{\theta}{2\pi}\right)\right]^2\left[r\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}\right]$$ $$=\frac 13\pi r^3\left(1-\frac{\theta}{2\pi}\right)^2\left(\sqrt{\frac{\theta}{\pi}-\frac{\theta^2}{4\pi^2}}\right)$$

You could simplify that further, as you like.

Note that the answer would have been significantly easier if you defined $\theta$ to be the central angle of the sector left after cutting rather than of the sector that was cut. If we define $\Theta$ as that central angle that was left, we end up with

$$V=\frac 13\pi r^3\left(\frac{\Theta}{2\pi}\right)^2 \sqrt{1-\left(\frac{\Theta}{2\pi}\right)^2}$$

[Math] volume of a truncated cone that is not a frustum

A plane that meets one nappe of a right circular cone in an ellipse defines an oblique cone with an elliptical base. If the plane lies at distance $d$ from the cone's vertex, and if the base has semi-axes $a$ and $b$, the volume is $\frac{1}{3} \pi abd$.

If the cone has vertex half-angle $\phi$, the cutting plane crosses the cone axis at distance $h$ from the vertex, and the cutting plane makes angle $\theta$ with the flat base, with $0 \leq \theta < \frac{\pi}{2} - \phi$, then the volume of the truncated cone is $$ \frac{\pi}{3}\, \frac{h^{3} \cot\phi}{(\cot^{2} \phi - \tan^{2} \theta)^{3/2}}. $$ (If $\theta \geq \frac{\pi}{2} - \phi$, the volume is infinite.)

For brevity, let $k = \cot\phi$ denote the slope of the cone generator and $m = \tan\theta$ denote the slope of the cutting plane.

In Cartesian coordinates with the vertex at the origin and the cone opening around the positive $z$-axis, the cone and cutting plane have respective equations \begin{align*} z^{2} &= k^{2}(x^{2} + y^{2}), \tag{1a} \\ z &= mx + h. \tag{1b} \end{align*} The plane and cone meet where $k^{2}(x^{2} + y^{2}) = z^{2} = (mx + h)^{2}$, or $$ (k^{2} - m^{2}) x^{2} - 2mhx - h^{2} + k^{2} y^{2} = 0. \tag{2} $$ (This is the equation of the elliptical "shadow" of the base, which does not directly give the shape of the base.)

In the longitudinal plane $y = 0$ (shown), the cone and cutting plane meet when $$ (k^{2} - m^{2}) x^{2} - 2mhx - h^{2} = 0. $$ The quadratic formula gives the roots $$ x_{\pm} = \frac{h(m \pm k)}{k^{2} - m^{2}}. \tag{3} $$ The semi-major axis is the distance between the corresponding points on the cone, $$ a = \tfrac{1}{2} \sqrt{1 + m^{2}}(x_{+} - x_{-}) = \sqrt{1 + m^{2}}\, \frac{hk}{k^{2} - m^{2}} = \sec\theta\, \frac{hk}{k^{2} - m^{2}}. \tag{4a} $$ The semi-minor axis of the slant base is the semi-minor axis of the ellipse (2), namely the positive value of $y$ in equation (2) when $$ x = \frac{x_{-} + x_{+}}{2} = \frac{hm}{k^{2} - m^{2}}. $$ For this $x$, we have $mx + h = \dfrac{hk^{2}}{k^{2} - m^{2}}$, so \begin{align*} y^{2} &= \frac{1}{k^{2}} \bigl[(mx + h)^{2} - k^{2} x^{2}\bigr] \\ &= \frac{1}{k^{2}} \left[\frac{h^{2} k^{4}}{(k^{2} - m^{2})^{2}} - \frac{k^{2} h^{2} m^{2}}{(k^{2} - m^{2})^{2}}\right] \\ &= \frac{h^{2}}{k^{2} - m^{2}}. \end{align*} The semi-minor axis is therefore $$ b = \frac{h}{\sqrt{k^{2} - m^{2}}}. \tag{4b} $$ Trigonometry shows the distance from the vertex to the cutting plane is $d = h\cos\theta$. Combining with equations (4a) and (4b), the volume of the slant cone is $$ \frac{\pi}{3} abd = \frac{\pi}{3} \left(\sec\theta\, \frac{hk}{k^{2} - m^{2}}\right) \frac{h}{\sqrt{k^{2} - m^{2}}}\, (h\cos\theta) = \frac{\pi}{3}\, \frac{h^{3} k}{(k^{2} - m^{2})^{3/2}}, $$ as claimed.

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Related Solutions

[Math] How to find the volume of a cone in terms of theta

[Math] volume of a truncated cone that is not a frustum

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