In the figure, circle O has radius 6 units. Chord CD has length 8 units and is parallel to KB segment . If KA= 12 units and points K,A,O and B are collinear, what is the area of triangle KDC ? Express your answer in simplest radical form.

please and thanks!

SydSu22 Apr 9, 2019

#3**+2 **

Not as difficult as it first seems......

From O, bisect chord CD and call the point of bisection, E

When a chord is bisected by segment drawn from a circle's center....the angle formed is = 90 °

So...ED = 4 and OD = 6

So OED is a right triangle with angle OED = 90 and ED a leg and OD the hypotenuse ....so.... OE = sqrt[ OD^2 - ED^2 ] = sqrt [ 6^2 - 4^2] = sqrt [ 20] = 2sqrt(5) = the height of obtuse triangle KCD

And the base of KCD = CD = 8

So....the area of KCD =

(1/2)* base * height =

(1/2 ) * 8 * 2 sqrt (5) =

[ 8 sqrt (5) ] units^2

CPhill Apr 9, 2019