Thread: Geometry problem help!?!

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

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

Please see here ...

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.)

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:
. .

Since


Law of Cosines: .

Hence: . .(c)


We have with: .

Check it out! . is a right triangle!
(I assume that is the intent of the problem.)

Hence: . (a)

. .And: . (b)


(d)



Since the diameter is , the radius is 10.

(e)