OMG

• Sep 23rd 2013, 09:49 PM
OMG
Iduj wears a very funny hat in the shape of a triangle OMG. The altitude from point M is drawn to point A on side OG. If OM is 13 inches, MG is 20 inches, and GA is 16 inches, find the area, in inches squared, of Iduj’s triangular hat.

*Hints or steps on how to get the answer is appreciated.*
• Sep 23rd 2013, 10:11 PM
votan
Re: OMG
You could use Pythagorean theorem to calculate the height of the triangle OMG. You will need to calculate AO, Can yo do it now. the are of the triangle is height*base/2. What is the base of the the triangle OMG?
• Sep 24th 2013, 07:31 PM
Re: OMG
Quote:

Originally Posted by votan
You could use Pythagorean theorem to calculate the height of the triangle OMG. You will need to calculate AO, Can yo do it now. the are of the triangle is height*base/2. What is the base of the the triangle OMG?

Is the base 16 inches? Height*16/2=A?
• Sep 24th 2013, 08:19 PM
Soroban
Re: OMG

Did you make a sketch?

Quote:

Iduj wears a very funny hat in the shape of a triangle OMG.
The altitude from point M is drawn to point A on side OG.
If OM is 13 inches, MG is 20 inches, and GA is 16 inches,
find the area, in inches squared, of Iduj’s triangular hat.

Code:

M
*
*| *
* |  *
13 *  |    *  20
*  |      *
*    |        *
*    |          *
*      |            *
O * * * * * * * * * * * * * G
A      16

In $\Delta M\!AG: \;M\!A^2 + 16^2 \:=\:20^2 \quad\Ruightarrow\quad M\!A^2 + 256 \:=\:400 \quad\Rightarrow\quad M\!A^2 \,=\,144$

. . Hence: . $M\!A \,=\,12$

In $\Delta M\!AO: \:OA^2 + 12^2 \:=\:13^2 \quad\Rightarrow\quad OA^2 + 144 \:=\:169 \quad\Rightarrow\quad OA^2 \,=\,25$

. . Hence: . $OA \,=\,5$

Therefore, triangle $OMG$ has:
. . base $OG \,=\,5+16\,=\,21$
. . height $M\!A \,=\,12$

Got it?
• Sep 25th 2013, 07:59 PM
Re: OMG
Quote:

Originally Posted by Soroban

Did you make a sketch?

Code:

M
*
*| *
* |  *
13 *  |    *  20
*  |      *
*    |        *
*    |          *
*      |            *
O * * * * * * * * * * * * * G
A      16

In $\Delta M\!AG: \;M\!A^2 + 16^2 \:=\:20^2 \quad\Ruightarrow\quad M\!A^2 + 256 \:=\:400 \quad\Rightarrow\quad M\!A^2 \,=\,144$

. . Hence: . $M\!A \,=\,12$

In $\Delta M\!AO: \:OA^2 + 12^2 \:=\:13^2 \quad\Rightarrow\quad OA^2 + 144 \:=\:169 \quad\Rightarrow\quad OA^2 \,=\,25$

. . Hence: . $OA \,=\,5$

Therefore, triangle $OMG$ has:
. . base $OG \,=\,5+16\,=\,21$
. . height $M\!A \,=\,12$

Got it?

Yes. I got it! I made a big mistake. I forgot to add the A!!! I' m sorry for causing sooo much trouble and thanks for the help!
• Sep 25th 2013, 11:50 PM
topsquark
Re: OMG
You could also "cheap out" and use Heron's formula.

-Dan
• Sep 26th 2013, 07:29 AM
bjhopper
Re: OMG
Quote:

Originally Posted by topsquark
You could also "cheap out" and use Heron's formula.

-Dan

Hello Dan,
Heron's formula gives area with 3 sides of triangle.

A quick answer using a scientific calculator

sin (arc cosine 16/20)=0.6
MA = 0.6*20=12
sin (arc cosine 12/13=0.3846
OA =0.3846*13=5
area =1/2*12*21