In the triangle ABC, a and B are its two sides, and S is the area of the triangle ABC. If s = 1 / 4 (a ^ 2 + B ^ 2), then the shape of the triangle ABC is?

isosceles right triangle
Analysis:
According to the triangle area formula s = 1 / 2 · ab · sinc ("·" is multiplied)
S = 1 / 4 (A & sup2; + B & sup2;)
So s = 1 / 2 · ab · sinc = 1 / 4 (A & sup2; + B & sup2;)
So the equation holds if and only if a = B and sinc = 1, that is, C = 90 degree
So the triangle line ABC is an isosceles right triangle

Let the two right angles of a triangle be a, B, and the hypotenuse be c. s be the area of the triangle and l be the perimeter of the triangle. A + B-C = M. prove s / L = 1 / 4m

A * a + b * b = C * C (right triangle)
(a+b)*(a+b)=c*c+2a*b
1/4(a+b)*(a+b)=1/4c*c+1/2a*b
1/4(a+b)*(a+b)-1/4c*c=1/2a*b
1/4[(a+b)*(a+b)-c*c]=1/2a*b
1/4(a+b-c)*(a+b+c)=1/2a*b
1/4(a+b-C)=1/2a*b/(a+b+c)
1/4M=S/L

In the image of function y = loga x (a > 1, X ≥ 1), there are three points a, B and C, whose abscissa are t, t + 2 and T + 4 respectively. If the area of triangle ABC is s, find s=

Make a straight line perpendicular to the x-axis through points a, B and C respectively, and the perpendicular feet are D, e and f respectively,
AD=logat,BE=loga(t+2),CF=loga(t+4),DE=EF=2,DF=4
S triangle ABC
=S-ladder abed + s-ladder bcfe-s-ladder ADFC
=1/2（AD+BE）DE+1/2（CF+BE）EF-1/2（AD+CF）DF
=logat+loga(t+2)+loga(t+4)+loga(t+2)-2[logat+loga(t+4)]
=loga[(t+2)^2/t(t+4)]

In the triangle ABC, a and B are its two sides, and S is the area of the triangle ABC. If s = 1 / 4 (a ^ 2 + B ^ 2), then the shape of the triangle ABC is?