Question 523965
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You are given a lot more than just the measure of two sides, though I have to simply assume that point D is on the segment *[tex \Large \overline{AB}] (you didn't specifically say so and you didn't provide a diagram).  With that presumption, knowing that *[tex \Large \overline{CD}] bisects *[tex \Large \overline{AB}] and that *[tex \Large \bigtriangleup{CDB}] is right, then CD is a perpendicular bisector of AB.  That means that *[tex \Large \bigtriangleup{CDA}] is congruent to *[tex \Large \bigtriangleup{CDB}] (by SAS) and ABC is an isosceles triangle with C as the apex and AB as the base.  Further, you know that *[tex \Large AD\ =\ DB\ =\ \frac{AB}{2}\ =\ 12].  From that you can see that *[tex \Large \bigtriangleup{CDB}] is a right triangle with a leg that measures 12 and a hypotenuse that measures 15.  You can either use Pythagoras or recognize a 3-4-5 proportion right triangle to determine that the short leg, specifically *[tex \Large \overline{CD}], measures 9.  From that you calculate the area by base times height divided by 2.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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