# Math Help - a few math team problems need help with

1. ## a few math team problems need help with

1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.

2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)

3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.

4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow

You don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!

-Amer

2. Originally Posted by amerlaw1
4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow
Best thing to do is draw a picture to truly understand this one.

Using similar triangles and equating ratios of side lengths you should get

$\displaystyle \frac{x}{6} = \frac{x+3}{9}$

Can you find $x$ ?

3. Originally Posted by amerlaw1
1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.

2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)

3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.

4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow

You don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!

-Amer
What is "mathteam"? Are these questions part of a competition?

4. thank you but can u show me how the are similar like what theorem. E.G sas or aa or sss????

yes math team is a competition but these questions came from a competition that happened 14 years ago so you need not to worry about cheating

5. 1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.

Let the side of the triangle be $a$. Then you can find the areas of AOB, BOC and AOC, which together give the area of ABC. On the other hand, the area of ABC is $\sqrt{3}a^2/4$. This gives you an equation that allows finding $a$.

6. Hello, amerlaw1!

2. What is the distance between the incenter and circumcenter
of a triangle with sides of the length 12, 16 and 20?
Code:
  _ A *
:   | *
:   |   *
:   |     *
:   |       *
:   |         *
:   |           *   20
:   |       * * * *
:   |   *           *   E
12   | *               o
:   |*          r   *  *
:   |             *       *
:   *           *       *   *
: F o - - - - o         *     *
:   *    r    |O        *       *
:   |         |                   *
:  r|*        |r       *            *
:   | *       |       *               *
:   |   *     |     *                   *
- C * - - - * o * - - - - - - - - - - - - * B
r     D
: - - - - - - - - 16  - - - - - - - - :

We have right triangle $ABC:\:AC = 12,\;BC = 16,\;AB = 20.$
Place the triangle on a coordinate system with $\,C$ at the Origin.

The incenter is $O(r,r).$
. . $OD = OE = OF = CD = CF = r.$

The circumcenter is the midpoint of $AB:\;P(8,6).$
. .
(Not shown on the diagram.)

Note that: . $AF \,=\, 12-r$
Since $AE$ is also tangent to the circle: $AE \,=\, 12-r$

Note that: . $BD \,=\,16-r.$
Since $BE$ is also tangent to the circle: $BE \,=\,16-r$

Since $AB = 20$, we have: . $(12-r) + (16-r) \:=\:20 \quad\Rightarrow\quad r \,=\,4$

We have: . $\begin{Bmatrix}\text{Incenter:} & O(4,4) \\ \text{Circumcenter:} & P(8,6) \end{Bmatrix}$

Therefore: . $\iverline{OP} \;=\;\sqrt{(8-4)^2 + (6-5)^2} \;=\;\sqrt{16+4} \;=\;\sqrt{20} \;=\;2\sqrt{5}$