Cho tam giac ABC vuong tai A phan giac AD biet BD=3/14/17cm CD=9/3/17cm tinh do dai cac canh goc vuong cua tam giac Tinh do dai BC Su dung tinh chat duong phan giac BD/AB=CD/AC BC/AB+AC
2 câu trả lời
Đáp án:
Giải thích các bước giải: \(\begin{array}{l} BD = 3\frac{{14}}{{17}}cm = \frac{{65}}{{17}}cm\\ CD = 9\frac{3}{{17}}cm = \frac{{156}}{{17}}cm\\ \Rightarrow BC = CD + BD = 13cm \end{array}\) \(AD\) là phân giác \(\widehat {CAB}\) nên \(\frac{{DB}}{{AB}} = \frac{{CD}}{{AC}} \Leftrightarrow \frac{{65}}{{17AB}} = \frac{{156}}{{17AC}} \Leftrightarrow AC = \frac{{12}}{5}AB\) mà \(A{B^2} + A{C^2} = B{C^2}\) (định lí Pi-ta-go) \(\begin{array}{l} \Leftrightarrow A{B^2} + \frac{{144}}{{25}}A{B^2} = {13^2}\\ \Leftrightarrow A{B^2} = 25\\ \Leftrightarrow AB = 5 \Rightarrow AC = 12cm. \end{array}\)
\[\begin{array}{l}
BD = 3\frac{{14}}{{17}}cm = \frac{{65}}{{17}}cm\\
CD = 9\frac{3}{{17}}cm = \frac{{156}}{{17}}cm\\
\Rightarrow BC = CD + BD = 13cm\\
\Rightarrow \frac{{DB}}{{AB}} = \frac{{CD}}{{AC}}\\
\Rightarrow \frac{{65}}{{17AB}} = \frac{{156}}{{17AC}}\\
\Rightarrow AC = \frac{{12}}{5}AB\\
m\`a \,A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow A{B^2} + \frac{{144}}{{25}}\\
\Rightarrow \frac{{12}}{5}A{B^2} = {13^2}\\
\Leftrightarrow A{B^2} = 25\\
\Leftrightarrow AB = 5\\
\Rightarrow AC = 12cm
\end{array}\]