Question 667067


A median in a triangle is a line segment drawn from an angle to the midpoint of the opposite side. The medians of a triangle all intersect in one point. This point divides each median in the ratio {{{1: 2}}}.

let {{{ABC}}} be a triangle 

if CM[1]=15...it is equal to the hight of triangle {{{h}}}, {{{AM[2]=12}}}, and {{{BM[3]=9}}}

if intersection point of all three medians is {{{T}}}, and {{{AM[2]}}} is median then 

{{{AT:TM[2]=2:1}}} .......=> {{{AT=(12/3)2=8}}} and {{{TM[2]=4}}}


and {{{CT:TM[1]=2:1}}} .......=>{{{CT:TM[1]=(15/3)2=10}}} and{{{TM[1]=(15/3)1=5}}}

now we can find {{{AM[1]}}} which is half of base {{{AB}}}

{{{AM[1]^2=(AT)^2-(TM[1])^2}}}

{{{AM[1]^2=8^2-5^2}}}

{{{AM[1]^2=64-25}}}

{{{AM[1]^2=39}}}

{{{AM[1]=sqrt(39)}}}

{{{AM[1]=6.25}}}...........so, the base {{{AB=6.25*2=12.5}}}


then the area will be:

{{{A=(1/2)12.5*15}}}

{{{A=6.2.5*15}}}

{{{A=93.75}}}