1.

If G be the centroid of a triangle ABC then prove that AB^2 + BC^2+ AC^2=3(AG^2+GB^2+GC^2)

Answer» Here,G is\xa0the centroid of a triangle ABC.Let A(b, c), B(0, 0) and C(a, 0) be the coordinates of {tex}\\Delta{/tex}ABC then coordinates of centroid are {tex}G\\left[ {\\frac{{a + b+0}}{3},\\frac{c+0+0}{3}} \\right]{/tex}To prove:-(AB)2 + (BC)2 + (CA)2 = 3(GA2 + GB2 + GC2)Consider : L.H.S.=(AB)2 +( BC)2 + (CA)2= b2 + c2 + a2 + (a - b)2 + c2= b2 + c2 + a2 + a2 + b2 - 2ab + c2= 2a2 + 2b2 + 2c2 - 2abConsider :\xa0R.H.S.=3(GA2 + GB2 + GC2){tex}=3\\left[ {{{\\left( {\\frac{{a + b}}{3} - b} \\right)}^2} + {{\\left( {c - \\frac{c}{3}} \\right)}^2} + {{\\left( {\\frac{{a + b}}{3}} \\right)}^2}} \\right.{/tex}{tex}\\left. { + {{\\left( {\\frac{c}{3}} \\right)}^2} + \\left( {\\frac{{a + b}}{3} - a} \\right) + {{\\left( {\\frac{c}{3}} \\right)}^2}} \\right]{/tex}{tex} = 3{\\left[ {{{\\left( {\\frac{{a - 2b}}{3}} \\right)}^2} + {{\\left( {\\frac{{2c}}{3}} \\right)}^2} + \\left( {\\frac{{a + b}}{3}} \\right)} \\right.^2}{/tex}{tex}\\left. { + {{\\left( {\\frac{c}{3}} \\right)}^2} + {{\\left( {\\frac{{b - 2a}}{3}} \\right)}^2} + {{\\left( {\\frac{c}{3}} \\right)}^2}} \\right]{/tex}{tex} = 3\\left[ {\\frac{{{a^2} + 4{b^2} - 4ab}}{9} + \\frac{{4{c^2}}}{9} + \\frac{{{a^2} + {b^2} + 2ab}}{9}} \\right.{/tex}{tex}\\left. { + {{\\frac{c^2}{9}}} + \\frac{{{b^2} + 4{a^2} - 4ab}}{9} + {{\\frac{c^2}{9}}}} \\right]{/tex}{tex}= 3 \\left[ {\\frac{{{a^2} + 4{b^2} - 4ab + 4{c^2} + {a^2} + {b^2} + 2ab + {c^2} + {b^2} + 4{a^2} - 4ab + {c^2}}}{9}} \\right]{/tex}{tex} = 3\\left[ {\\frac{{6{a^2} + 6{b^2} + 6{c^2} - 6ab}}{9}} \\right]{/tex}{tex} = 3 \\times 3\\left[ {\\frac{{2{a^2} + 2{b^2} + 2{c^2} - 2ab}}{9}} \\right]{/tex}= 2a2 + 2b2 + 2c2 - 2abL.H.S. = R.H.S.Therefore,\xa0(AB)2 + (BC)2 + (CA)2 = 3(GA2 + GB2 + GC2)


Discussion

No Comment Found

Related InterviewSolutions