Question

Dau coroană!!!!!Vă rog​

Answer 1

Explicație pas cu pas:

a)

AM = 3 cm => BM = 1 cm

BN = CN = 2 cm

$\mathcal{A}_{ABCD} = AB^{2} = 4^{2} = 16 \ cm^{2}$

$\mathcal{A}_{\triangle ADM} = \dfrac{AD \cdot AM}{2} = \dfrac{4 \cdot 3}{2} = 6 \ cm^{2} \\ \mathcal{A}_{\triangle DCN} = \dfrac{DC \cdot CN}{2} = \dfrac{4 \cdot 2}{2} = 4 \ cm^{2} \\ \mathcal{A}_{\triangle BMN} = \dfrac{BM \cdot BN}{2} = \dfrac{1 \cdot 2}{2} = 1 \ cm^{2}$

$\mathcal{A}_{\triangle DMN} = \mathcal{A}_{ABCD} - \Big(\mathcal{A}_{\triangle ADM} + \mathcal{A}_{\triangle DCN} + \mathcal{A}_{\triangle BMN}\Big) =$

$= 16 - (6 + 4 + 1) = \bf 5 \ cm^{2}$

b)

DM² = AD²+AM² = 4²+3² = 16+9 = 25 = 5²

=> DM = 5 cm

DN² = DC²+CN² = 4²+2² = 16+4 = 20

=> DN = 2√5 cm

$\mathcal{A}_{\triangle DMN} = \dfrac{DM \cdot DN \cdot \sin \angle MDN}{2} = \\ = \dfrac{5 \cdot 2\sqrt{5} \cdot \sin \angle MDN}{2} = 5\sqrt{5} \sin \angle MDN$

$5\sqrt{5} \sin \angle MDN = 5 \implies \sin \angle MDN = \dfrac{1}{ \sqrt{5} }$

q.e.d.