a) Xét \(\Delta ABC\) có
Góc A =900
=> BC2=AB2+AC2
BC2= 62+82
BC2=100
BC=\(\sqrt{100}\)=10
Xét\(\Delta ABC\) và \(\Delta DEC\)
Góc A= Góc E (=90o)
Góc C chung
=>\(\Delta ABC\)~\(\Delta DEC\) (g.g)
=>\(\dfrac{AH}{AC}=\dfrac{BC}{DC}\)
=>\(\dfrac{AH}{8}=\dfrac{6}{10}\)
=> AH=\(\dfrac{8.6}{10}\)=4,8
S\(\Delta ADC\)=\(\dfrac{1}{2}AH.BC\)
=\(\dfrac{1}{2}4,8.10\)
=24(cm2)
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