#"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"#
