Question

Ex 5 va rog mult
Dau coroana

Answer 1

a) <FDC=45°(bisectoare)

<ADF=45°(Bisectoare)

tg C=AD/DC

tg C=√3/3

=> <C=30°

=> <DFC=180°-(45°+30°)

<DFC=180°-75°

<DFC=105°

=> <CFE=180°-<DFC

=> <CFE=180°-105°

=> <CFE=75°

b) <AFE=<DFC=105°

<AED=45°

=> <FAE=30°

înalțimea (h) în ΔAFE= AE× sin(<45)

h=3√3×√2/2

h=3√6/2

$AF=\frac{h}{ sin (180°- <DFC)} \\\\ AF=\frac{ \frac{3\sqrt{6}}{2} }{ \frac{\sqrt{2}+\sqrt{6}}{2} } \\\\ AF=\frac{6\sqrt{6}}{\sqrt{2}+\sqrt{6}}\\\\ AF=\frac{3\sqrt{12}-18}{-2}\\\\ AF=-3\sqrt{3}+9\\ AF=3(3-\sqrt{3}) cm$

Answer 2

Răspuns:

Explicație pas cu pas:

AD ≡ BC; AB  ≡ DC

T.Pitagora in ΔADC:

$AC^{2} = AD^{2} +DC^{2} = 6^{2} +(6\sqrt{3}) ^{2} = 144 \implies \bf AC = 12 \ cm$

a)

$AD = \dfrac{AC}{2} \implies \measuredangle ACD = 30^{ \circ} \implies \measuredangle CAD = 60^{ \circ}$

DE este bisectoare => ∠ADE = 45°

$\measuredangle AFD = 180^{ \circ} - \measuredangle CAD - \measuredangle ADE = 180^{ \circ} - 60^{ \circ} - 45^{ \circ} = 75^{ \circ}$

$\measuredangle CFE \equiv \measuredangle AFD \ (op.vf.) \implies \measuredangle CFE = 75^{ \circ}$

b)

∠ADE = 45° => ΔADE este dreptunghic isoscel => AE ≡ AD => AE = 6 cm

AB || DC => ΔAFE ≈ ΔCFD

$\dfrac{AF}{CF} = \dfrac{AE}{CD} \iff \dfrac{AF}{CF} = \dfrac{6}{6\sqrt{3} } \iff \dfrac{AF}{CF} = \dfrac{1}{\sqrt{3} }$

$\dfrac{AF}{AF+CF} = \dfrac{1}{1+\sqrt{3}} \iff \dfrac{AF}{AC} = \dfrac{1}{1+\sqrt{3}}$

$\dfrac{AF}{12} = \dfrac{1}{1+\sqrt{3}} \iff AF = \dfrac{12}{1+\sqrt{3}} = \dfrac{12(\sqrt{3} - 1)}{(1+\sqrt{3})(\sqrt{3} - 1)} = \dfrac{12(\sqrt{3} - 1)}{3-1}$

$\implies \bf AF = 6(\sqrt{3} - 1) \ cm$