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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!

 Apr 9, 2019
edited by SydSu22  Apr 9, 2019
 #1
avatar+25918 
0

Where is the 'figure' ?

 Apr 9, 2019
 #2
avatar+256 
+1

Sorry... here it is 

 Apr 9, 2019
 #3
avatar+111437 
+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

 

 

cool cool  cool

 Apr 9, 2019
edited by CPhill  Apr 9, 2019

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