Question

Fie ABCD un dreptunghi cu AB=10,BC=4.Stiind ca M si N se gasesc pe latura DC astfel incat DM=CN=3 cm.
a.Aratati ca AMNB este trapez isoscel si determinati perimetrul
Am nevoie urgent! Multumesc

Answer 1

$m( \angle MAB)\ \textless \ m( \angle DAB)=90 si~m( \angle MBA)\ \textless \ m( \angle ABC)= 90 \textdegree . \\ \\ Deci~m( \angle MAB)+m( \angle MBA)\ \textless \ 180 \textdegree \Rightarrow MA \not ||~MB~(1). \\ \\ MN || AB ~(2). \\ \\ (1)~si~(2) \Rightarrow ABNM-trapez~(3). \\ \\ AD=BC~si~DM=NC \Rightarrow \Delta ADM \equiv \Delta BCN ~(C.C) \Rightarrow AM=BN~ \\ (4).$

$Din~(3)~si~(4)~rezulta~ABNM-trapez~isoscel.$

$T.Pitagora~in~\Delta ADM (dreptunghic~in~D): \\ \\ AM^2=AD^2+DM^2 \Rightarrow AM= \sqrt{AD^2+DM^2}= \sqrt{16+9}= \sqrt{25}=5~ \\ (cm). \\ \\ AM=BN=5~cm. \\ \\ MN=CD-DM-NC=10-3-3=4~(cm). \\ \\ P_{ABNM }=AM+MN+BN+AB=5+4+5+10=24~(cm).$