"area of triangle "=1/2bh
"where b is the base and h the height"
"expressing b in terms of h"
"using "color(blue)"Pythagoras' theorem"
b=sqrt(41^2-h^2)
"now area "=180
rArr1/2bh=180rArrbh=360
rArrhsqrt(41^2-h^2)=360
color(blue)"square both sides"
rArrh^2(41^2-h^2)=360^2
rArr1681h^2-h^4=129600
"multiply through by "-1" and equate to zero"
rArrh^4-1681h^2+129600=0
"use the substitution "u=h^2
rArru^2-1681u+129600=0
"solve using the "color(blue)"quadratic formula"
u=(1681+-sqrt2307361)/2=(1681+-1519)/2
u=(1681+1519)/2=1600" or "u=(1681-1519)/2=81
u=h^2rArrh^2=1600" or "h^2=81rArrh=40" or "h=9
h=9rArrb=360/9=40
"h=40rArrb=360/40=9
color(blue)"As a check"
h=40,b=9rArrsqrt(40^2+9^2)=41
"and area "=1/2xx40xx9=180" cm"^2
rArr"lengths of legs "=40" and "9 "cm"