⑴在直角△ACD中,由勾股定理得:AC=10.△ACD面积=½×6×8=24,
⑵再求△ABC的面积:过C点作AB的垂线,垂足为E点,
设AE=x,则BE=13-x,
由勾股定理得:AC²-AE²=CE²=CB²-BE²,
∴10²-x²=12²-﹙13-x﹚²,
解得:x=125/26,
∴CE²=10²-﹙125/26﹚²,
∴CE=15√231/26,
∴△ABC面积=½AB×CE
=½×13×15√231/26
=15√231/4,
⑶阴影面积=△ABC面积-△ACD面积
=15√231/4-24