Question

Answer 1

Xét ΔHAB vuông tại H và ΔHCA vuông tại H có

\(\hat{HAB}=\hat{HCA}\left(=90^0-\hat{HBA}\right)\)

Do đó: ΔHAB~ΔHCA

=>\(\frac{HA}{HC}=\frac{HB}{HA}\)

=>\(HA^2=HB\cdot HC\)

ΔHAB~ΔHCA

=>\(\frac{S_{HAB}}{S_{HCA}}=\left(\frac{HA}{HC}\right)^2=\frac{HA^2}{HC^2}=\frac{HB\cdot HC}{HC^2}=\frac{HB}{HC}\)

=>\(\frac{HB}{HC}=\frac{54}{96}=\frac{9}{16}\)

=>\(\frac{HB}{9}=\frac{HC}{16}=k\)

=>HB=9k; HC=16k

\(HA^2=HB\cdot HC=9k\cdot16k=144k^2\)

=>HA=12k

ΔHAB vuông tại H

=>\(S_{HAB}=\frac12\cdot HA\cdot HB=\frac12\cdot9k\cdot12k=9k\cdot6k=54k^2\)

=>\(54k^2=54\)

=>\(k^2=1\)

=>k=1

=>BH=9cm; CH=16cm

BC=BH+CH=9+16=25(cm)