Question

Fie ABCD un trapez cu AB||CD, AB=13 cm, DC=5 cm, AD=BC=2√13.
a) Calculați aria lui ABCD.
b) Arătați că AC _|_ CB.

Fie ABCD un paralelogram cu AB=6 cm, AD=4 cm și m(A)=60°.
a) Determinați aria paralelogramului.
b) Calculați lungimea diagonalei BD.

Answer 1

   
Desenele sunt in fisierele atasate.

[tex]\displaystyle\\ 1)\\ AD=BC=2\sqrt{13}~\Longrightarrow~\text{Trapezul este isoscel}\\ \text{Din D coboram inaltimea}~CE\perp AB\\ \text{In triunghiul dreptunghic}~\Delta BCE~avem:\\ BC=2\sqrt{13}~cm=ipotenuza\\\\ BE=\frac{AB-CD}{2}=\frac{13 -5}{2}=\frac{8}{2}=4~cm=cateta\\\\ CE=cateta=?\\\\ CE=\sqrt{BC^2-BE^2}=\sqrt{(2\sqrt{13})^2-4^2}=\sqrt{52-16}=\sqrt{36}= 6~cm\\\\ a)\\ A=\frac{(B+b)\cdot h}{2}=\frac{(AB+CD)\cdot CE}{2}=\frac{(13+5)\cdot6}{2}= \frac{18\cdot6}{2}=54~cm^2 [/tex]

[tex]b)\\ AE = AB - BE = 13-4 = 9~cm \\ \text{Din triunghiul dreptunghic}~\Delta ACE~\text{calculam latura AC.}\\\\ AC = \sqrt{AE^2 + CE^2}= \sqrt{9^2 + 6^2}=\sqrt{81 + 36} =\sqrt{117}~cm \\\\ \text{In triunghiul }~\Delta ABC ~\text{verificam daca }~AC \perp BC.\\\\ AC^2 + BC^2 = (\sqrt{117})^2 +(2 \sqrt{13} )^2 = 117 + 52 = 169\\\\ AB^2 = 13^2 = 169\\\\ \text{Observam ca suma patratelor a 2 latur = patratul celei de-a 3-a.}\\\\ \Longrightarrow~~~AC \perp BC[/tex]



[tex]\displaystyle\\ 2)\\ \text{Din D coboram inaltimea } ~DE \perp AB, E \in AB\\\\ \text{In triunghiul dreptunghic }~ADE~\text{avem:}\\ AD = 4~ cm = \text{ipotenuza}\\ m(\ \textless \ A) = 60^o\\ AE = AD\cos 60^o = 4\times\frac{1}{2} = \frac{4}{2} = 2~cm\\\\ DE = AD\sin 60^o = 4\times\frac{ \sqrt{3} }{2} =\frac{4\times \sqrt{3} }{2} =2\sqrt{3}~cm\\\\ BE = AB - AE = 6-2 = 4~cm\\\\ a)\\ A = B \times h = AB \times DE = 6 \times 2\sqrt{3} = 12\sqrt{3} ~cm [/tex]


[tex]\displaystyle\\ b)\\ \text{Din triunghiul dreptunghic }\Delta BDE ~\text{calculam latura }BD.\\\\ BD = \sqrt{BE^2 +DE^2} = \sqrt{4^2 +(2 \sqrt{3})^2} =\\\\ =\sqrt{16 +12} =\sqrt{28} = 2 \sqrt{7} ~cm[/tex]