a, SABC = \(\dfrac{1}{2}\).AB.AC=\(\dfrac{6.8}{2}\)=24(\(^{cm^2}\))
b, 2SABC =AB.AC =AH.BC
=>AH=\(\dfrac{AB.AC}{BC}\)=\(\dfrac{6.8}{10}=4,8\)(cm)
c,kẻ AM là tiếp tuyến ABC suy ra AM=BM=CM=\(\dfrac{BC}{2}\)=5
xét tam giác AHM vuông có AM=5;AH=4,8
AD Py ta go ta có MH=\(\sqrt{AM^2-AH^2}\)=\(\sqrt{5^2-4,8^2}\)=1,4(cm)
SAMH= \(\dfrac{AH.HM}{2}\)=\(\dfrac{4,8.1,4}{2}\)=3,36 \(\left(cm^2\right)\)