http://i78.photobucket.com/albums/j95/ashree101/lmk.jpg
it's a picture of a problem i need help solving.
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:
. . $\displaystyle (a)\;\angle C\quad (b)\;\angle A \quad (c)\;AC\quad (d)\:\text{area of }\Delta ABC \quad (e)\;\text{area of circle}$
Since $\displaystyle \text{arc}(CA) \,=\, 74^o,\;\angle B \,=\, 37^o$
Law of Cosines: .$\displaystyle AC^2 \:=\:16^2 + 20^2 - 2(16)(20)\cos37^o \;=\;144.8732736$
Hence: .$\displaystyle AC \;=\;\sqrt{144.8732736} \;=\;12.0363314 \quad\Rightarrow\quad AC\;\approx\;12$ .(c)
We have $\displaystyle \Delta ABC$ with: .$\displaystyle AC = 12,\;BC = 16,\;AB = 20$
Check it out! .$\displaystyle \Delta ABC$ is a right triangle!
(I assume that is the intent of the problem.)
Hence: .$\displaystyle \angle C \,=\,90^o$ (a)
. .And: .$\displaystyle \angle A \,=\,53^o$ (b)
$\displaystyle \text{Area }\Delta ABC \:=\:\tfrac{1}{2}(BC)(AC) \:=\:\tfrac{1}{2}(16)(12) \:=\:96$ (d)
Since the diameter is $\displaystyle AB = 20$, the radius is 10.
$\displaystyle \text{Area of circle} \:=\:\pi(10^2) \:=\:100\pi$ (e)