Hello, viet!
Your angles are incorrect . . .
You have a cone shaped bag.
At the bottom of the bag is an orange with a radius of 2 inches.
On top of the orange is a melon with a radius of 6 inches.
It touches the orange and fits snugly in the bag, touching it in a ring around the orange.
Its top is at the same level as the top of the bag.
Find height of the cone and its radius.
Code:
B D C
o - - - - - - - - - - *o* - - - - - - - - - - o
\ * | * /
\ | /
\ * 6| * /
\ * | * /
\ * | * /
\ * Ho * /
\* | \ 6 */
\* | \ */
\* 6| o/J
\ * | * /
\ * |G * /
\ *o* /
\ * | * /
\ * 2| * /
\ * | * /
\ * Fo * /
\* |\2 */
\* 2| \ */
\* | o/I
\ *o* /
\y|E/
\|/
o
A
\(\displaystyle ABC\) is the cone.
\(\displaystyle F\) is the center of the two-inch orange.
. . \(\displaystyle FE\,=\,FG\,=\,FI\,=\,2\)
\(\displaystyle H\) is the center of the six-inch melon.
. . \(\displaystyle HD\,=\,HG\,=\,HJ\,=\,6\)
Let \(\displaystyle EA\,=\,y.\)
In right triangle \(\displaystyle FIA:\;AF\,=\,y+2,\;FI\,=\,2\)
In right triangle \(\displaystyle HJA:\;AH\,=\,y+10,\;HJ\,=\,6\)
Since \(\displaystyle \Delta FIA\,\sim\,\Delta HJA:\;\frac{y+2}{2}\:=\:\frac{y+10}{6}\;\;\Rightarrow\;\;y\,=\,2\)
. . Therefore:
. \(\displaystyle h\,=\,DA\,=\,18\)
In right triangle \(\displaystyle FIA:\;AI^2\,+\,FI^2\:=\:AF^2\;\;\Rightarrow\;\;AI^2\,+\,2^2\:=\:4^2\;\;\Rightarrow\;\;AI\,=\,2\sqrt{3}\)
Let \(\displaystyle DC\,=\,r\) (radius of cone).
Since \(\displaystyle \Delta CDA\,\sim\,\Delta FIA:\;\frac{CD}{DA}\,=\,\frac{FI}{AI}\;\;\Rightarrow\;\;\frac{r}{18}\,=\,\frac{2}{2\sqrt{3}}\)
. . Therefore:
.\(\displaystyle r\,=\,6\sqrt{3}\)
