# Math Help - 2 problems

1. ## 2 problems

Hi, here are two problems i am still having trouble with today.

1.1.ABCDEF is a regular hexagon and EDGH is a square. Find the measures of <EFH, <FHD and <DFH.

IDK if this is right but i found this:Angle EDH = 45 angle EHD=45 Ang. HDG = 45 angle DHG=4 all angles in that square are right angles as well

8.Two vertical poles have heights of 6 feet and 12 feet. A rope is stretched from the top of each pole to the bottom of the other. How far above the ground do the ropes cross? (hint: the lengths of y and z do not affect the answer.)

2. [quote=abc123;253849]Hi, here are two problems i am still having trouble with today.

1.1.ABCDEF is a regular hexagon and EDGH is a square. Find the measures of <EFH, <FHD and <DFH.

IDK if this is right but i found this:Angle EDH = 45 angle EHD=45 Ang. HDG = 45 angle DHG=4 all angles in that square are right angles as well

Find each interior angle of hexagon $= \frac{180(n-2)}{n} = \frac{180(6-2)}{6} = 120$

DEF = 120,

DEH = 90 (angles of a square.)

EHD = 45 = EDH

FE = EH ,so, EFH = EHF

FEH = 360 - 120 -90 = 150

In triangle EFH,

EFH = 15 = EHF

IN triangle FED,

EF = ED

EFD = EDF = 30

so,

FDH = 30 + 45 = 75

DFH = 30 + 15 = 45

3. Originally Posted by abc123
Hi, here are two problems i am still having trouble with today.

8.Two vertical poles have heights of 6 feet and 12 feet. A rope is stretched from the top of each pole to the bottom of the other. How far above the ground do the ropes cross? (hint: the lengths of y and z do not affect the answer.)
Use similar triangles properties,

$\frac{x}{6} = \frac{z}{y+z}$

and

$\frac{x}{12} = \frac{y}{y+z}$

$\frac{x}{6}+\frac{x}{12} = \frac{z}{y+z}+\frac{y}{y+z} = \frac{y+z}{y+z} = 1$

x = 4

4. Thanks a lot for the posts. they helped a lot. I'm still confused though on number 6. I don't really understand what you did at all. So, if you could explain it to me a bit I'd be very greatful

5. Hello again, abc123!

8. Two vertical poles have heights of 6 feet and 12 feet.
A rope is stretched from the top of each pole to the bottom of the other.
How far above the ground do the ropes cross?
Code:
                              * C
* |
*   |
*     |
*       |
*         |
A *           *           | 12
|   *   E *             |
|       *               |
6 |     * :   *           |
|   *   :x      *       |
| *     :           *   |
B *-------*---------------* D
: - y - F - - - z - - - :

$\Delta EFD \sim \Delta ABD \quad\Rightarrow\quad \frac{x}{z} \:=\:\frac{6}{y+z} \quad\Rightarrow\quad x(y+z) \:=\:6z\;\;{\color{blue}[1]}$

$\Delta EFB \sim \Delta CDB \quad\Rightarrow\quad \frac{x}{y} \:=\:\frac{12}{y+z} \quad\Rightarrow\quad x(y+z) \:=\:12y\;\;{\color{blue}[2]}$

Equate [1] and [2]: . $6z \:=\:12y \quad\Rightarrow\quad z \:=\:2y$

Substitute into [2]: . $x(y + 2y) \:=\:12y \quad\Rightarrow\quad x(3y) \:=\:12y \quad\Rightarrow\quad \boxed{x \:=\:4}$

6. Hello, abc123!

1) $ABCD{E}F$ is a regular hexagon and $EDGH$ is a square.
Find the measures of: $\angle EFH,\;\angle FHD,\;\angle DFH$
Code:
        A * - - - * F
/        :\*
/         : \ *
/          :  \  *
B * 120°      : E *   *
\          :  /90° * *
\         : /         * H
\        :/         /
C * - - - *         /
D   *    /
*
G

All interior angles of the hexagon are 120°.

We have: . $\angle FED = 120^o,\;\angle DEH = 90^o$
. . Hence: . $\angle FEH = 150^o$

$\Delta EFH$ is isosceles, $EF = EH$
. . Therefore: . $\angle EFH \:=\:\angle EHF \:=\:15^o$

In square $EDGH,\;\angle EHD = 45^o$
. . Therefore: . $\angle FHD \:=\:15^o + 45^o \:=\:60^o$

$\angle AFE = 120^o,\;\angle AFD = 90^o\quad\Rightarrow\quad \angle DFE = 30^o$
. . Therefore: . $\angle DFH \:=\:30^o + 15^o \:=\:45^o$