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!
Where is the 'figure' ?
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