1. ## Geometry problem help!?!

http://i78.photobucket.com/albums/j95/ashree101/lmk.jpg

it's a picture of a problem i need help solving.

2. ## Solution

Please specify what exactly is 74 degree. Then only first three parts can be solved.

Now for next two parts...

a) To find the area of tr. ABC
-> As ABC is isosceles, therefore, its median can be dropped from A. This median would be the perpendicular bisector of BC. Let this median be AD.
So, BD = 1/2 x BC = 10 cm. So AD is 12.5 cm (approx) (Pythagoras theorem)
So area of tr. ABD = 1/2 x BD x AD = 1/2 x 10 x 12.5= 62.5 cm sq. and, so area of tr. ABC = 2 x ABD = 125 cm sq.

b) to find the area of circle
-> This circle is the circumcircle of ABC. So, O lies on AD as 2:1. Now, OA = 2/3 x AD = 8.3 cm (approx.)
area of circle = 22/7 x OA x OA = 216.3 sq. cm. (approx.)

3. Hello, freebrd01!

Code:
              o o o    C
o          ♠
o     16  *   * o
o       *       * o 74°
*           *
o  * 37°           *o
B ♠   *   *   *   *   ♠ A
o        20         o

o                 o
o               o
o           o
o o o

Find:
. . $(a)\;\angle C\quad (b)\;\angle A \quad (c)\;AC\quad (d)\:\text{area of }\Delta ABC \quad (e)\;\text{area of circle}$

Since $\text{arc}(CA) \,=\, 74^o,\;\angle B \,=\, 37^o$

Law of Cosines: . $AC^2 \:=\:16^2 + 20^2 - 2(16)(20)\cos37^o \;=\;144.8732736$

Hence: . $AC \;=\;\sqrt{144.8732736} \;=\;12.0363314 \quad\Rightarrow\quad AC\;\approx\;12$ .(c)

We have $\Delta ABC$ with: . $AC = 12,\;BC = 16,\;AB = 20$

Check it out! . $\Delta ABC$ is a right triangle!

(I assume that is the intent of the problem.)

Hence: . $\angle C \,=\,90^o$ (a)

. .And: . $\angle A \,=\,53^o$ (b)

$\text{Area }\Delta ABC \:=\:\tfrac{1}{2}(BC)(AC) \:=\:\tfrac{1}{2}(16)(12) \:=\:96$ (d)

Since the diameter is $AB = 20$, the radius is 10.

$\text{Area of circle} \:=\:\pi(10^2) \:=\:100\pi$ (e)