Question

知道三边求三角形面积

一 a=32 b=21 c=21.3 ,求面积
二 a=10.3 b=33.2 c=38.9 求面积

Answer 1

1.p=1/2(a+b+c)=0.5*(32+21+21.3)=37.15
S=√[p(p-a)(p-b)(p-c)]=√[37.15(37.15-32)(37.15-21)(37.15-21.3)]= √[37.15*5.15*16.15*15.85]=√48974.25532=221.3012772670

2.p=1/2(a+b+c)=0.5*(10.3+33.2+38.9)=41.2
S=√[p(p-a)(p-b)(p-c)]=√[41.2(41.2-10.3)(41.2-33.2)(41.2-38.9)]= √[41.2*30.9*8*2.3]=√23424.672=153.0512071171

Answer 2

cos∠C=(a^2+b^2-c^2)/(2ab)=(32^2+21^2-21.3^2)/(2*32*21)=0.752
所以，三角形面积=abcosC/2=32*21*0.752/2=252.672

cos∠B=(a^2+c^2-b^2)/(2ab)=(10.3^2+38.9^2-33.2^2)/(2*10.3*38.9)=0.645
所以，三角形面积=accosB/2=10.3*38.9*0.645/2=129.216

Answer 3

到网上去查一下什么叫海伦公式