1)
a,
$\Delta MPQ$ vuông tại $Q$, $QI\bot MP$ có:
$MQ^2=MI.MP$
$\Delta MNQ$ vuông tại $M$, $MI\bot QN$ có:
$MQ^2=QI.QN$
Vậy $MI.MP=QN.QI$
b,
$\Delta MNQ$ và $\Delta QMP$ có:
$\widehat{NMQ}=\widehat{MQP}=90^o$
$\widehat{MNQ}=\widehat{QMP}=90^o-\widehat{MQN}$
$\to \Delta MNQ\backsim\Delta QMP$ (g.g)
$\to \dfrac{MQ}{MN}=\dfrac{PQ}{MQ}$
$\to MQ^2=MN.PQ$
2)
a,
$MQ^2=MN.PQ\to MQ=\sqrt{4,5.8}=6(cm)$
$\to S_{MNPQ}=\dfrac{1}{2}(4,5+8).6=37,5(cm^2)$
b,
Kẻ $NH\bot PQ$
Tứ giác $MNHQ$ có $\widehat{MQH}=\widehat{QHN}=\widehat{NMQ}=90^o$
$\to MNHQ$ là hình chữ nhật
$\to MN=HQ=4,5(cm); MQ=HN=6(cm)$
$\to HP=8-4,5=3,5(cm)$
$\Delta NHP$ vuông tại $H$ có:
$\tan\widehat{NPQ}=\dfrac{NH}{HP}=\dfrac{12}{7}$
$\to \widehat{NPQ}\approx 60^o$
$\to \widehat{MNP}\approx 180^o-60^o=120^o$
$\widehat{NMQ}=\widehat{MQH}=90^o$
c,
Theo định lí Pytago:
$MP=\sqrt{6^2+8^2}=10(cm)$
$NP=\sqrt{6^2+3,5^2}=\dfrac{\sqrt{193}}{2}(cm)$
Chu vi $\Delta MNP$:
$MN+NP+MP=14,5+\dfrac{\sqrt{193}}{2}=\dfrac{29+\sqrt{193}}{2}(cm)$
