Question

1. Im praralelogramul ABCD m(,B)=120 grade,AB=8 cm si AD=4 cm.Calculati lungimea diagonalelor
2. se da dreptunghiul ABCD cu AB=20 cm si BC= 15 cm. Perpendiculara din D pe Ac intersecteaza AC,AB,BC respectiv in E,F.G. Se cer lungimile segmentelor
a) [DE] b) [DF] c) [BG]

Answer 1

[tex]BD^2=AB^2+AD^2-2AD\cdot AB\cos120=64+16-64(-\frac{1}{2})=\\ 80+32=112 \Rightarrow BD=\sqrt{112}=4\sqrt7\\ AC^2=BA^2+BC^2-2BA\cdot BD\cos 60=64+16-64\frac{1}{2}=48\\ AC=\sqrt{48}=4\sqrt3\\ \text{Problema 2} \\ \text{Aflam lungimea diagonalei $AC$ cu teorema lui Pitagora si obtinem}\\ AC=25\\ \text{$DE$ este inaltime in triunghiul dreptunghic $DAC$}\\ DE=\frac{DA\cdot DC}{AC}=\frac{15\cdot20}{25}=12\\ AE=\frac{AD^2}{AC}=\frac{15^2}{25}=9 \\ \Delta AEF\sim\Delta CDA\Rightarrow \\ [/tex]
[tex]\frac{AE}{DC}=\frac{EF}{AD}\Rightarrow EF=\frac{15\cdot 9}{20}=\frac{27}{4}\\ DF=DE+EF=12+\frac{27}{4}=\frac{75}{4}[/tex]
[tex]BG=GC-BC\\ \text{Pe $GC$ il scoti din asemanrea }\Delta ADC\sim\Delta DCG [/tex]