Giúp minhg bài 5. CẢM ƠN

a)

Xét $\Delta MNP$ và $\Delta KQP$, ta có:

+   $\widehat{NMP}=\widehat{QKP}=90{}^\circ $

+   $\widehat{MPN}=\widehat{KPQ}$ (hai góc đối đỉnh)

$\to \Delta MNP\backsim\Delta KQP\left( g.g \right)$

$\to \dfrac{MN}{QK}=\dfrac{NP}{QP}$

$\to MN.QP=NP.QK$

 

b)

Vì $\Delta MNP\backsim\Delta KQP\left( cmt \right)$

$\to \widehat{MNP}=\widehat{KQP}$

Mà $\widehat{MNP}=\widehat{KNQ}$ (vì $NP$ là phân giác)

$\to \widehat{KNQ}=\widehat{KQP}$

 

c)

Xét $\Delta KNQ$ và $\Delta MNP$, ta có:

+   $\widehat{KNQ}=\widehat{MNP}$ (vì $NP$ là phân giác)

+   $\widehat{NKQ}=\widehat{NMP}=90{}^\circ $

$\to \Delta KNQ\backsim\Delta MNP\left( g.g \right)$

$\to \dfrac{NQ}{NP}=\dfrac{QK}{MP}$

$\to NP.QK=MP.NQ$

d)

Có ${{S}_{\Delta KQP}}=\dfrac{1}{2}QK.PK=\dfrac{1}{2}\cdot 3\cdot 4=6c{{m}^{2}}$

Có $\Delta KNQ\backsim\Delta MNP\left( cmt \right)$

Có $\Delta MNP\backsim\Delta KQP\left( cmt \right)$

$\to \Delta KNQ\backsim\Delta KQP$

$\to \dfrac{{{S}_{\Delta KNQ}}}{{{S}_{\Delta KQP}}}={{\left( \dfrac{QK}{PK} \right)}^{2}}={{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{9}{16}$

$\to {{S}_{\Delta KNQ}}=\dfrac{9}{16}{{S}_{\Delta KQP}}=\dfrac{9}{16}\cdot 6=\dfrac{27}{8}c{{m}^{2}}$