convex-conegeometrylinear algebra
A cone with base radius 12 cm is sliced parallel to its base, as shown,
to remove a smaller cone of height 15 cm. If the height of the smaller
cone is three-fourths that of the original cone, what is the volume of the remaining frustum?
I set the frustum's top radius as x and its height as y. Using the cone volume formula, I have $$\frac{144\pi\cdot(y+15)}{4}=\frac{15x^2\pi}{3}$$$$\implies 36 y+540=5x^2$$
I am stuck here. How should I continue?
A vertical cross section of the solid gives two similar triangles, one inside the other. The base of the larger triangle is $16$ and its height is $12$; the base of the smaller triangle is $12$ and its height is, say, $h$.
$h$/$6$ = $12$/$8$
which gives $h$ = $9$.
Now the volume can be found using the formula stated in the problem.
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.
Best Answer
For the height of the bigger cone: We have $h=15 = \frac{3}{4}H \implies H = 20\text {cm}$
$$\tan z = \frac{r}{h} = \frac{R}{H}$$
We have $R = 12, H = 20 , h = 15 \text{ (all in cm) } \implies r = \frac{R}{H}h = \cdots$
Now the height of the frustrum is $H-h = \cdots$