Cone-cut part.

[123]

Height of the cone h and base radius r lengths ratio is equal to 4:3. From the cone was cut part, as shown in figure (a=120⁰).
How many times the cut part of the total surface area is less than the total surface area of ​​a cone?
________
Hello,
I need help to solve this problem.

So, $\displaystyle \frac{h}{r}=\frac{4}{3}$


The total surfase area of the cone is: SA = πr(r + l).
How get formula of the cut part?

 

[wjm11]

View attachment 2752
Height of the cone h and base radius r lengths ratio is equal to 4:3. From the cone was cut part, as shown in figure (a=120⁰).
How many times the cut part of the total surface area is less than the total surface area of ​​a cone?
________
Hello,
I need help to solve this problem.

So, $\displaystyle \frac{h}{r}=\frac{4}{3}$


The total surfase area of the cone is: SA = πr(r + l).
How get formula of the cut part?

Your formula for the SA of a cone is correct. Notice that 120 degrees is 1/3 of a circle (120/360 = 1/3). This means that you have cut 1/3 of the cone's volume away. If you neglected the cut surfaces, you would also have removed 1/3 of the base (circle) area and 1/3 of the "side" surface area.

From the way you have stated the problem, it is not clear to me if you must also include the "cut" surfaces into the calculation for the SA of the cut cone. If you are supposed to include those surfaces, remember that triangle area is (1/2)(base)(height) and that you have two triangles.

 

[HallsofIvy]

The slant length is $\displaystyle \sqrt{r^2+ h^2}$. If you cut along one side and flatten it out (Which you can do without distortion. The surface of a cone, unlike the surface of a sphere, is a "developable surface". Any surface, at every point of which there is at least one straight through that point in the surface, is a "developable surface" and can be "flattened". For a cone, at any point, the line through the point and the vertex lies in the cone.) you have part of a disk of radius $\displaystyle \sqrt{r^2+ h^2}$ and so circumference $\displaystyle 2\pi\sqrt{r^2+ h^2}$ and area $\displaystyle \pi(r^2+ h^2)$. The circumference of the base of the cone is $\displaystyle 2\pi r$ and so only covers a fraction $\displaystyle \frac{2\pi r}{2\pi\sqrt{r^2+ h^2}}= \frac{r}{\sqrt{r^2+ h^2}}$ of the circumference of the disk. The area of the cone is that same fraction of the area of the disk. Turning that over gives the number of times the area of that disk is of the area of the cone: $\displaystyle \frac{\sqrt{r^2+ h^2}}{r}$.

In particular, if $\displaystyle h=(4/3)r$ that becomes $\displaystyle \frac{\sqrt{r^2+ 16r^2/9}}{r}= \sqrt{1+ 16/9}= \sqrt{25/9}= 5/3$.

 

[123]

$\displaystyle h:r = 4:3 \Rightarrow h = 4k;r = 3k$

$\displaystyle l = \sqrt {25{k^2}} = 5k$

$\displaystyle S = \pi Rl + \pi R = 24\pi {k^2}$

$\displaystyle S1$ - cut part.
$\displaystyle S2$ - other part.

$\displaystyle \frac{{S1}}{{S2}} = \frac{{120}}{{360}} = \frac{1}{3}$

$\displaystyle S1 = \frac{1}{3} \cdot 24\pi {k^2} = 8\pi {k^2}$
$\displaystyle S1 = \frac{2}{3} \cdot 24\pi {k^2} = 16\pi {k^2}$

$\displaystyle \frac{{S2}}{{S1}} = \frac{{16\pi {k^2}}}{{8\pi {k^2}}} = 2$ times smaller.

 

[lookagain]

$\displaystyle h:r = 4:3 \Rightarrow h = 4k;r = 3k$

$\displaystyle l = \sqrt {25{k^2}} = 5k$

$\displaystyle S = \pi Rl + \pi R = 24\pi {k^2}$

$\displaystyle S1$ - cut part.
$\displaystyle S2$ - other part.

$\displaystyle \frac{{S1}}{{S2}} = \frac{{120}}{{360}} = \frac{1}{3}$

$\displaystyle S1 = \frac{1}{3} \cdot 24\pi {k^2} = 8\pi {k^2}$
$\displaystyle S2 = \frac{2}{3} \cdot 24\pi {k^2} = 16\pi {k^2} \ \ \$ I edited this show "S2" instead of "S1" that was there. (I don't know if your work is correct, though.)

$\displaystyle \frac{{S2}}{{S1}} = \frac{{16\pi {k^2}}}{{8\pi {k^2}}} = 2$ times smaller.

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