Question 155021
Let known that {{{sideA=A}}} & {{{sideB=B}}} are congruent and the 3rd side={{{sideC=C}}}
So,
{{{A=(2/3)C}}}
{{{B=(2/3)C}}}
{{{C=C}}}
{{{Perimeter=A+B+C}}}, or since {{{A=B}}} -- congruent, then {{{P=2A+C}}}
{{{210m=(2/3)C+(2/3)C+C}}}
{{{210=(4/3)C+C}}}
{{{210=(4C+3C)/3}}}
{{{630=7C}}} ---> {{{cross(630)90/cross(7)=cross(7)C/cross(7)}}}
{{{C=90meters}}} ------------------------> side C
{{{A=B=(2/3)90=180/3=60meters}}} --------> side A & side B
{{{Area=(1/2)bh}}}
Remember: {{{base=sideC}}}={{{90}}}
Solve {{{h}}} by Pyth theorem:
{{{B^2=h^2+(C/2)^2}}} -----. we use only {{{1/2}}} of side C for a right triangle to use Pyth theorem:
{{{h=sqrt(B^2-(C/2)^2)}}}
{{{h=sqrt(60^2-(90/2)^2)}}}
{{{h=sqrt(3600-2025)}}}={{{sqrt(1575)}}}={{{39.7>=40m}}}
Then, {{{A=(1/2)90*40=1800m^2}}}
thank you,
Jojo