Question

已知：在△ABC中,AB=30,BC=28,AC=26,求△ABC的面积.

Answer 1

过A做AH垂直BC交BC于H;

设BH=X;CH=Y;AH;=Z;

X^2+Z^2=AB^2=30^2;---(1)

Y^2+Z^2=AC^2=26^2;----(2)

(1)-(2)

X^2-Y^2=30^2-26^2;

(X-Y)(X+Y)=4*56;X+Y=BC=28;

所以X-Y=8;

X=18,Y=10;

Z^2=AB^2-X^2=30^2-18^2=24^2;

Z=24;

△ABC的面积=1/2BC*Z=1/2*28*24=12*28=336