Question

3. Fie ABCD un trapez (AB || CD), AB= 21 cm, CD=6 cm, BC = 12 cm, AD = 9 cm.

c) calculati oc/oa unde {0}= AC intersectat BD. d) Calculaţi perimetrul triunghiului DEC unde AD intersectat cu BC =E.
e) Dacă MN || AB, M aparține AD, N aparține BC, D aparține MN .arătaţi că MO=ON​

Answer 1

Răspuns:

Sper ca te-am ajutat! Mult succes în continuare!

La punctul c ma mai gandesc.

Answer 2

Explicație pas cu pas:

c)

$DC || AB \implies\Delta DOC \sim \Delta BOA$

$\dfrac{OC}{OA} = \dfrac{OD}{OB} = \dfrac{CD}{AB} \iff \dfrac{OD}{OB} = \dfrac{OC}{OA} = \dfrac{6}{21}$

$\implies \dfrac{OD}{OB} = \bf \dfrac{OC}{OA} = \dfrac{2}{7}$

d)

$DC || AB \implies\Delta DEC \sim \Delta AEB$

$\dfrac{DE}{AE} = \dfrac{CE}{BE} = \dfrac{CD}{AB}$

$\dfrac{DE}{AE - DE} = \dfrac{CE}{BE - CE} = \dfrac{CD}{AB - CD}$

$\dfrac{DE}{AD} = \dfrac{CE}{BC} = \dfrac{6}{21 - 6} \iff \dfrac{DE}{9} = \dfrac{CE}{12} = \dfrac{2}{5}$

$\dfrac{DE}{9} = \dfrac{2}{5} \implies DE = \dfrac{18}{5} = 3.6 \ cm \\ \dfrac{CE}{12} = \dfrac{2}{5} \implies CE = \dfrac{24}{5} = 4.8 \ cm$

$\mathcal{P}_{\triangle DEC} = DE+CE+CD= \\ = 3,6 + 4,8 + 6 = \bf 14,4 \ cm$

e)

din c)

$\dfrac{OD}{OB} = \dfrac{OC}{OA} \iff \dfrac{OB + OD}{OB} = \dfrac{OA + OC}{OA} \\ \dfrac{BD}{OB} = \dfrac{AC}{OA} \iff \dfrac{OA}{AC} = \dfrac{OB}{BD} \ \ (1)$

$OM || CD \implies\Delta AMO \sim \Delta ADC \\\implies \dfrac{OM}{CD} = \dfrac{OA}{AC} \ \ (2)$

$ON || CD \implies\Delta BNO \sim \Delta BCD \\\implies \dfrac{ON}{CD} = \dfrac{OB}{BD} \ \ (3)$

$(1);(2);(3) \implies \dfrac{OM}{CD} = \dfrac{ON}{CD} \\ \implies \bf OM = ON$

$q.e.d.$