Question

In triangle ABC, BC=5, CA=8 and AB=7. If 'G'is the centroid of triangle ABC, then find GA2 + GB2 + GC2

Answer 1

Answer:

LetA(x1,y1),B(x2,y2)And C(x3,y3)be the vertices of △ABC.

Without the law of generality ,assume the centroid of the

△ABC to beat origin ,i.e.,G=(0,0).

Centroid of △ABC=

$△ABC=[ \frac{x1 + x2 + x3}3</p><p>, \frac{y1 + y2 + y3}{3} ]$

∴x1+x2+x3=0;y1+y2+ y3 =0

Squaring on both sides ,

x1² + x2² + x3² + 2x1.x2+2x2.x3+2x3.x1=0

y1² + y2² + y3² + 2y1.y2+2y2.y3+2y3.y1=0−(i)

AB² + BC² + CA²

=[(x2−x1) +(y2−y1)² ]+[(x3−x2) +(y3−y2)² ]+[(x1−x3)² +(y1−y3)² ]

=(x1² +x2² −2x1x2+y1² +y2² −2y1y2)+(x1² +x3² −2x1x3+y1² +y3² −2y2y3)+(x1² +x3²

−2x1x3+y1² +y3² −2y1y3)

=(2x1² +2x2² +2x3² −2x1x2−2x2x3−2x1x3)+

(2y1² +2y2² +2y3² −y1y2−2y2y3−2y1y3)

=(3x1² +3x2² +3x3² )+(3y1² +3y2² +3y3² )

=3(x1² +x2² +x3² )+3(y1² +y2² +y3² )−(ii)

3(GA² +GB² +GC² )

=3[(x1−0)² +(y1−0)² +(x2−0)² +(y2−0)² +(x3−0)² +(y3−0)² ]

=3[x1² +y1² +x2² +y2² +x3² +y3² ]

=3(x1² +x2² +x3² )+3(y1² +y2² +y3² )−(iii)

from(ii)&(iii)

AB² +BC² +CA²

=3(GA² +GB² +GC² )

Answer 2

Answer:

Step-by-step explanation:

GA²+GB²+GC²=1/3(AB²+BC²+CA²)

                        =1/3(7²+5²+8²)

                         =1/3(49+25+64)

                         =1/3(138)

                          =46

∴GA²+GB²+GC²=46