Geometry Problem | How To Find Area of the Triangle When Distance of Centroid From Vertices is Given

Geometry Problem | How To Find Area of the Triangle When Distance of Centroid From Vertices is Given
area of the triangle when distance of centroid from vertices is given
geometry
centroid
coordinate geometry
centroid of a triangle
centroid of triangle
median of a triangle
Join this channel to get access to perks:
/ @mathbooster
MEMBERS OF THIS CHANNEL
••••••••••••••••••••••••••••••••••••••••
Mr. Gnome
ஆத்தங்கரை சண்முகம்
Εκπαιδευτήρια Καντά
AR Knowledge
堤修一
Sidnei Medeiros Vicente
Mark Ludington
Arichai Vattanapat
Sambasivam Sathyamoorthy
Abebe Melesse
#geometry #matholympiad #areaoftriangle #mathbooster #premath #sybermath

[@jeanmarcbonici9525]

Nice thank you. For those who like calculations: We call I, J and K the respective midpoints of sides BC, AC and AB of triangle ABC. We know that GI=GA/2, GJ=GB/2 and GK=GC/2. We then apply the following formulas to calculate the area of triangle ABC: c=Sqrt[2*(GA^2+GB^2-2*GK^2)]; a=Sqrt[2*(GC^2+GB^2-2*GI^2)] ; b=Sqrt[2*(GA^2+GC^2-2*GJ^2)]; p=(a+b+c)/2 then AreaABC = Sqrt[p*(p-a)*(p-b)*(p-c)];

[@MorgKev]

I used sine rule and some other trig stuff but kept everything in radical form so ended up with the same exact answer. Got there but perhaps not as elegant as your method.

[@hanswust6972]

I love the logic you apply to solve this problem, it's incredible how much you can do with the sole meaning of *Centroid* or Gravity Center.

[@rdesouza25]

Your solution is much cleaner than mine, because I used the formula of the median for each side of the triangle ABC. Meadian for side a = (1/2)*sqrt{2(b^2+c^2)-a^2}. I used the same approach for each side and ended up with an easy to solve system for 3 variables (a, b and c). Using the same alternative as @jeanmarcbonici9525 below, I also extended the the segments in order to divide the triangle ABC into 6 equal areas. With that in mind I found out that triangle CGJ is right triangle. with sides sqrt6, 2 and sqrt2. As a result the area is (1/2)*2*sqrt2. This gives me sqrt2, which corresponds to (1/6) of total area of triangle ABC. Therefore Area of triangle ABC = 6sqrt2

[@padraiggluck2980]

There’s no need for an additional construction. BD is sqrt(3) -> the base is 2*sqrt(5) and area = (1/2)*(2*sqrt(5))*3*sqrt(3)) = 3*sqrt(15).

[@ROCCOANDROXY]

Letting A(0,0), B(x2,y), C(x1,0) implies G((x1 + x2)/3,y/3) implies (2 * x2 - x1)^2 + 4y1^2 = 72 (2 * x1 - x2)^2 + y^2 = 36 (x1 + x2)^2 + y^2 108 Solving the equations above implies x1 = x2 = 2 * sqrt(6) and y = 2 * sqrt(3) implies AB = 6, BC = 2 * sqrt(3) and AC = 2 * sqrt(6) in Right ABC) implies Area(Triangle(ABC)) = 6 * sqrt(2). Also centroid G(4 * sqrt(6)/3, 2 * sqrt(3)/3).

[@manasishastry345]

very nice explanation.I also tried by this method and got the ans.

[@manasishastry345]

enjoed the problem.Tried by co ordinate geometry also.

[@じーちゃんねる-v4n]

parallelogram rule for GAC 12+4=2(2+xx) x=√6 AC=2√6 by Herons rule GAC=2√2 ABC=6√2

[@giuseppemalaguti435]

Con solo le distanze il triangolo non è univocamente determinato