Question

Problema 4, va roggg muuuultttt!!!

Answer 1

Răspuns:

a) 12 cm²

Explicație pas cu pas:

∢DAM = 30° => DM = ½×AM

=> AM = 2DM

T.P. în ΔDAM: AM² = AD²+DM²

4DM² = 6²+DM² => 3DM² = 36

=> DM² = 12 => DM = 2√3 cm

CM = DC - DM = 6 - 2√3 cm

idem, pentru:

BN = 2√3 cm și CN = 6 - 2√3 cm

$\mathcal{A}_{\triangle ADM} = \mathcal{A}_{\triangle ABN} = \dfrac {6 \cdot 2 \sqrt{3} }{2} = 6 \sqrt{3} \ {cm}^{2}$

$\mathcal{A}_{\triangle MCN} = \dfrac {(6 - 2 \sqrt{3} )^{2}}{2} = \dfrac {36 - 24 \sqrt{3} + 12}{2} = 24 - 12 \sqrt{3} \ {cm}^{2}$

$\mathcal{A}_{\triangle MAN} = \mathcal{A}_{ABCD} - (\mathcal{A}_{\triangle ADM} + \mathcal{A}_{\triangle ABN} + \mathcal{A}_{\triangle MCN}) =$

$= {6}^{2} - (2 \cdot 6 \sqrt{3} + 24 - 12 \sqrt{3})$

$= 36 - (12 \sqrt{3} + 24 - 12 \sqrt{3})$

$= 36 - 24 = \bf 12 \ {cm}^{2}$

sau:

AM = 2DM = 4√3 cm

∢MAN = 90° - 2×30° = 30°

$\mathcal{A}_{\triangle MAN} = \dfrac {AM \cdot AN \cdot \sin \angle MAN}{2} = \\ = \dfrac {(4 \sqrt{3})^{2} \cdot \sin 30 \degree}{2} = \dfrac {48 \cdot \frac{1}{2}}{2} = 12 \ {cm}^{2}$

Answer 2

Răspuns:

Ai răspuns atașat pe foaie.