>>7613245Embed the triangle into a coordinate plane, with the short leg parallel to the y axis and the long leg parallel to x axis. The hypotenuse is a segment of the function f(x)=3x/4. To calculate the arc length we use the arc length integral. we need the arc length between x=0 and x=4 so the integral is set up as integrate sqrt(1+9/16) dx from 0 to 4 which is equal to 5.
Use herons formula:The area of a triangle is equal to sqrt(s*(s-a)(s-b)(s-c)) where a,b,c are lengths of the sides of the triangle and s is equal to (a+b+c)/2
We can calculate the area with base*height*.5=6
a=3, b=4, c=x
s=(3+4+x)/2=3.5+x/2
to simplify things x/2 will be replaced with y
the area by herons formula is
(y+3.5)(y+.5)(y-.5)(-y+3.5)
(3.5-y)(3.5+y)(y+.5)(y-.5)
(12.25-y^2)(y^2-.25)
sqrt(-y^4+12.5y^2-3.0625)
is the expression for the triangles area. previously we calculated that the area is 6 so
sqrt(-y^4+12.5y^2-3.0625)=6
-y^4+12.5y^2-3.0625=36
-y^4+12.5y^2-39.0625=0
to convert to reduced standard form we multiply by -1
y^4-12.5y^2+39.0625
and from there we can simply plug the coefficients of this equation into the quartic formula which is left as exercise for the anon.
