Triangle ABC has an area of 144. Point D is on BC such that BD : DC =1 : 3 and point F is on AB such that AF : FB= 1 : 4. Point E is on AC such that AD, BE, and CF intersect at point G. What is area of triangle GDC?
+600 This is a typical AMC 10/12 problem. We apply Mass points. Let C have mass 4. B will then have mass 4*3=12. A will then have mass 12*4=48. So F will have mass 12+48=60. So FG:GC=1:15. The height of GDC is 15/16 the large triangle, and the base is 4/5 the large triangle. Our answer is 144*15/16*4/5=108
+1639 Let Δ ABC be the right triangle with AB = 18 and BC = 16. 18 * 16 / 2 = 144 u2
∠ADB = arctan(AB / BD) 77.471º
∠GDC = 180 - 77.471 = 102.529º
∠BCF = arctan(BF / BC) = 41.987º
3 more steps:
1/ Use the law of sines to calculate the length of a side GC
2/ Use trigonometry to calculate the height from D to GC
3/ Multiply side GC by the height and then divide by 2
Have fun! (btw, the answer is not 108)
