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Help.

NotSoSmart  Apr 7, 2017
 #2
avatar+89775 
+2

Tangents drawn to a circle from the same point outside the circle are equal.....therefore....

 

JA = JB, LA = LC, KC = KB  ..so

 

JA + JB + LA + LC + KC + KB  = perimeter of triangle JKL....so...by substitution

 

JA + JA + LA + LA + KC + KC   =  perimeter of triangle JKL

 

2 ( JA + LA + KC)  =  perimeter of triangle JKL

 

2 ( 13 + 19 + 7)  =  perimeter of triangle JKL

 

2 ( 39)  =  perimeter of triangle JKL

 

78  =  perimeter of triangle JKL

 

 

cool cool cool

CPhill  Apr 7, 2017

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