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Problemă geometrică.Cum se rezolv&#259

 
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pviorel
Arhimede


Joined: 08 Dec 2012
Posts: 7

PostPosted: Sat Dec 08, 2012 2:09 pm    Post subject: Problemă geometrică.Cum se rezolv&#259 Reply with quote

Un trapez isoscel ABCD(AB||CD,AB<CD)are [AD]≡[AB]≡[BC] şi măsura unghiului C=60.Linia mijlocie a trapezului este egală cu 15 cm.Calculaţi Perimetrul ABCD.
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Radu Lecoiu



Joined: 07 Sep 2017
Posts: 3

PostPosted: Fri Sep 08, 2017 4:08 pm    Post subject: Reply with quote

\angle C =60 \Rightarrow \angle B=120.
\bigtriangleup ABC isoscel deci \angle BAC = \angle BCA = 30
Cum \angle BAD =120 \Rightarrow CAD =90
In fine, \bigtriangleup CDA = 30-60-90.
Notam AD=a atunci DC=2a , iar cum l.m. a trapezului e 15 avem \frac{a+2a}{2} = 15 . a=10 deci perimetrul este 3*10+20=50
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Radu Lecoiu
Clasa a VII-a
Colegiul National "Ienachita Vacarescu"
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