A Question related to triangle and centroid .

[agoogler]

Homework Statement

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 "

Homework Equations

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The Attempt at a Solution

Here is the figure that I have drawn -

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 ?

 

[junaid314159]

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

 

[agoogler]

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

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 ?

 

[junaid314159]

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.

 

[agoogler]

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.

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 .