Question

在平面直角坐标系内描出点,A(-2,0),B(0,4),C(6,0),D(5,-5),并求出四边形ABCD的面积

Answer 1

SABCD=S△ABC+S△ACD

S△ABC=1/2x8x4=16

S△ACD=1/2x8x5=20

SABCD=16+20=36

周长=AB+BC+CD+AD

AB^2=2^2+4^2

AB=√20

BC^2=6^2+4^2

BC=√52

CD^2=1^2+5^2

CD=√26

AD^2=3^2+5^2

AD=√34

周长=√20+√52+√26+√34