a) Theo bài ra, ta có:
\(\frac{AB}{AC}=\frac{8}{15}\Rightarrow\frac{AB}{8}=\frac{AC}{15}=k\Rightarrow\left\{{}\begin{matrix}AB=8k\\AC=15k\end{matrix}\right.\)
Áp dụng định lý Pytago vào △ABC vuông tại A, ta có:
\(BC^2=AB^2+AC^2\Rightarrow51^2=\left(8k\right)^2+\left(15k\right)^2=64k^2+225k^2=289k^2\Rightarrow2601=289k^2\Rightarrow k^2=9\Rightarrow k=3\left(k>0\right)\)\(\Rightarrow\left\{{}\begin{matrix}AB=8.k=8.3=24\left(cm\right)\\AC=15.k=15.3=45\left(cm\right)\end{matrix}\right.\)
b)Ta có:
S△ABC=\(\frac{AB.AC}{2}=\frac{24.45}{2}=540\left(cm^2\right)\)
