Question

Answer 1

Câu 4 :

a) Do M là trung điểm của MN (gt) \(\Rightarrow MQ=QN\)

Ta có: \(\Delta ABC\) cân tại A ( gt )\(\Rightarrow\left\{{}\begin{matrix}AB=AC\\\widehat{ABC}=\widehat{ACB}\end{matrix}\right.\)

Xét \(\Delta DMQ\) và \(\Delta DNQ\) có:\(\left\{{}\begin{matrix}MQ=MN\left(cmt\right)\\AB=AC\left(cmt\right)\\QDchung\end{matrix}\right.\)

\(\Rightarrow\Delta DMQ=\Delta DNQ\left(c.c.c\right)\)

b) Do \(QH\perp MD\left(gt\right)\Rightarrow\widehat{DHQ}=90^o\)

         \(QE\perp DN\left(gt\right)\Rightarrow\widehat{DEQ}=90^o\)

Do \(\Delta DMQ=\Delta DNQ\left(cmt\right)\Rightarrow\widehat{MDQ}=\widehat{EDQ}hay\widehat{HDQ}=\widehat{EDQ}\)

Xét \(\Delta DQH\) và \(\Delta DQE\) có : \(\left\{{}\begin{matrix}DQchung\\\widehat{DHQ}\\\widehat{HDQ}=\widehat{EDQ}\left(cmt\right)\end{matrix}\right.=\widehat{DEQ}\left(=90^o\right)\)

\(\Rightarrow\Delta DQH=\Delta DQE\left(ch-gn\right)\)

 

 

Answer 2

Câu 1 :

Trong \(\Delta ABC\) vuông tại A ( gt )

\(\Rightarrow BC^2=AB^2+AC^2\) ( Py-ta-go)

\(\Leftrightarrow BC^2=3^2+4^2\)

\(\Leftrightarrow BC^2=9+16\)

\(\Leftrightarrow BC^2=25\)

\(\Rightarrow BC=5\left(cm;BC>0\right)\)

Vậy BC = 5 cm