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A Question related to triangle and centroid .

  1. Aug 10, 2013 #1
    1. The problem statement, all variables and given/known data

    The following is a geometry question I can't seem to get. "Consider an acute angle △ABC. Points D, E, F are mid points of sides BC, CA and AB respectively. G is the centroid of △ABC. Area of △AFG = 14, EC = 15/2. Perpendicular distance of F from BC = 6. Find BC2−AB2 "



    2. Relevant equations

    -

    3. The attempt at a solution

    Here is the figure that I have drawn -
    vXUkm.png

    From EC =15/2 I get AE=15/2 So AC =15 . But what now? How can I utilize the fact that FK=6 and Area(AFG)=14 in finding BC2−AB2 ?

    Is Area of AFG = Area of FGB = Area of GBD ... ? If so then area of ABC 14*6=84 . Is that right ? Then altitude from A = 6*2=12 Thus BC = 84/6 = 14 . But how to figure out AB ?
     
  2. jcsd
  3. Aug 10, 2013 #2
    Hint 1: ΔAGB, ΔBGC, and ΔCGA all have the same area. (This is a property of centroids)
    Hint 2: ΔAGF and ΔBGF have the same base (AF=BF) and height.

    Junaid Mansuri
     
  4. Aug 10, 2013 #3
    Thanks !
    Let me answer my own question :
    Consider triangles AFG and BFG . They have equal bases ( AF=BF , since F is midpoint. ) They also have equal corresponding heights ( Since they have a common vertex at G ) . So their areas are equal. Similarly areas of all the sub-triangles ( BGD , CGD ...) are equal. Thus they all have area of 14 . So the area of whole triangle ABC = 6 *14 ( since there are 6 sub-triangles ) = 84.

    Now draw AL perpendicular to BC . Triangle BFK ~ ( similar ) Triangle BAL . But AB/BF=2 thus AL/FK=2 and AL=2*6=12 . Now Area of triangle ABC = 1/2 * base BC * height AL . But we already know that area is 84 , thus 84=1/2*BC*12 and BC = 14 .

    In triangle ALC , by Pythagoras theorem CL2=225−144 ( Given EC=15/2 thus AC =15 ) . Thus CL=9. Then BL=BC-CL = 14-9=5 . Now in triangle ABL , by Pythagoras theorem AB2=AL2+BL2=122+52 Thus AB=13. Now BC2−AB2=196−169=27.

    Is this correct? Also is there any simpler way to get the answer ?
     
    Last edited: Aug 10, 2013
  5. Aug 10, 2013 #4
    Wow. I'm speechless. I think the way you solved it is excellent. There is a small typo. I think you meant to say: CL^2 = 225 − 144 but you still got CL = 9 which is correct. One way to check your answer is to plug in the sides: 13, 14, 15 into Heron's Formula and see if you get 84. Not sure if there is an easier way.
     
  6. Aug 10, 2013 #5
    Thanks for correcting my mistake ! Yes I meant 225-144 . I checked it using Heron's formula and it checks out ! ( Excellent method to check the answer by the way ) . I noticed that I could use AB2=BC2 + AC2-2*BC*CL ( Law of cosines ) since it is an acute angled triangle . But it will take nearly same amount of efforts .
     
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