Question

Va roog ex 4 trebuie sa trimit într-o oră va rog din suflet dau 80 puncte​

Answer 1

Răspunsul se află în poză.

Answer 2

Explicație pas cu pas:

a)

$BM = BC - MC = \ell - \dfrac{1}{3}\ell = \dfrac{2\ell}{3} \\ DN = DC - NC = \ell - \dfrac{1}{3}\ell = \dfrac{2\ell}{3}$

$\mathcal{A}_{\triangle ABM} = \dfrac {AB \cdot BM}{2} = \dfrac {\ell \cdot \dfrac{2\ell}{3}}{2} = \dfrac{ {\ell}^{2} }{3}$

$\mathcal{A}_{\triangle ADN} = \dfrac {AD \cdot DN}{2} = \dfrac {\ell \cdot \dfrac{2\ell}{3}}{2} = \dfrac{ {\ell}^{2} }{3}$

$\mathcal{A}_{\triangle MCN} = \dfrac {MC \cdot NC}{2} = \dfrac {\dfrac{\ell}{3} \cdot \dfrac{\ell}{3}}{2} = \dfrac{ {\ell}^{2} }{18}$

$\mathcal{A}_{AMCN} = \mathcal{A}_{ABCD} - (\mathcal{A}_{\triangle ABM} + \mathcal{A}_{\triangle ADN}) =$

$= {\ell}^{2} - \Big(\dfrac{ {\ell}^{2} }{3} + \dfrac{ {\ell}^{2} }{3}\Big) = {\ell}^{2} - \dfrac{ {2\ell}^{2} }{3} = \dfrac{{\ell}^{2} }{3}$

b)

$\mathcal{A}_{\triangle AMN} = \mathcal{A}_{AMCN} - \mathcal{A}_{\triangle MCN} =$

$= \dfrac{{\ell}^{2} }{3} - \dfrac{ {\ell}^{2} }{18} = \dfrac{5{\ell}^{2} }{18}$

T.P. în ΔADN:

AN² = AD² + DN²

${AN}^{2} = {\ell}^{2} + \dfrac{4}{9}{\ell}^{2} = \dfrac{13{\ell}^{2}}{9}$

$\implies AN = \dfrac{\ell \sqrt{13} }{3}$

notăm MP = d(M; AN)

$\dfrac{MP \cdot AN}{2} = \dfrac{5{\ell}^{2} }{18} \iff MP \cdot \dfrac{\ell \sqrt{13} }{3} = \dfrac{5{\ell}^{2} }{9}$

$MP = \dfrac{3 \cdot 5{\ell}^{2} }{9 \cdot \ell \sqrt{13}} \implies MP = \dfrac{5\ell \sqrt{13} }{39}$