Question 1161458

A triangle has vertices {{{A}}}({{{3}}},{{{4}}}), {{{B}}}({{{-2}}},{{{0}}}), and {{{C}}}({{{5}}},{{{0}}}). 

Find the midpoint of each side, and label these midpoints {{{D}}},{{{ E}}}, and {{{F}}}.
the midpoint of {{{A}}}({{{3}}},{{{4}}}) and{{{ B}}}({{{-2}}},{{{0}}}) is {{{D}}}({{{(3-2)/2}}},{{{(4+0)/2}}})->{{{D}}}({{{1/2}}},{{{2}}})
the midpoint of  {{{A}}}({{{3}}},{{{4}}}) and {{{C}}}({{{5}}},{{{0}}}) is {{{E}}}({{{4}}},{{{2}}})
the midpoint of {{{B}}}({{{-2}}},{{{0}}}) and {{{C}}}({{{5}}},{{{0}}}) is {{{F}}}({{{3/2}}},{{{0}}})

 
{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(3,4,.12),circle(-2,0,.12),circle(5,0,.12),
circle(1/2,2,.12), circle(4,2,.12),circle(3/2,0,.12),
locate(3,4,A),locate(-2,-0.5,B),locate(5,-0.5,C),
locate(1/2,2,D),locate(4,2,E),locate(3/2,-0.5,F),
green(line(3,4,-2,0)),green(line(3,4,5,0)),green(line(5,0.1,-2,0.1)),
blue(line(1/2,2,4,2)),blue(line(1/2,2,3/2,0)),blue(line(4,2,3/2,0)),

 graph( 600, 600, -10, 10, -10, 10, 0)) }}}



that the area of triangle {{{ABC}}} is: 

{{{A=(b*h)/2}}}
{{{b}}}= distance between {{{A}}} and {{{C}}} which is {{{7}}} units
{{{h}}}= shortest distance between x-axis and {{{B }}}which is {{{4}}} units

{{{A=(7*4)/2}}}

{{{A=7*2}}}

{{{A=14 }}}square units

 the area of triangle {{{DEF}}} is:

 base is {{{DE}}} ->{{{b= 4-1/2=7/2}}}

{{{h=2}}}

{{{A=((7/2)*2)/2}}}

{{{A=7/2}}}

{{{A=3.5}}}

Verify that the area of triangle {{{ABC}}} is four times the area of triangle {{{DEF}}}:

{{{14=4*3.5}}}
{{{14=14}}}-> verified