Question

În figura este reprezentat un dreptunghi ABCD cu AB=15 cm și perimetrul egal cu 42 cm. Fie M un punct pe segmentul CD astfel încât DM=3cm. a) Demonstrați ca AM perpendicular pe BM b) Calculați lungimea segmentului CP, unde P este punctul de intersecție a dreptelor AC și MB. Dau coroana!!!

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it \mathcal{P} =2\cdot AB+2\cdot BC \Rightarrow 42=2\cdot15+2\cdot BC \Rightarrow 42=30+2\cdot BC|_{-30} \Rightarrow\\ \\ \Rightarrow 12=2\cdot BC \Rightarrow BC =6 cm \Rightarrow AD = 6 cm\\ \\ CD=AB=15cm;\ \ MC=CD-DM=15-3=12cm$

$\it Vom\ compara\ \Delta CMB\ cu\ \Delta DAM:\\ \\ \left.\begin{aligned}\dfrac{CM}{DA}= \dfrac{12}{6}=2\\ \\ \widehat{BCM}\equiv \widehat{MDA}\ (=90^o)\\ \\ \dfrac{BC}{MD}= \dfrac{6}{3}=2\end{aligned}\right\}\ \stackrel{(LUL)}{\Longrightarrow}\ \Delta CMB\sim \Delta DAM \Rightarrow \begin{cases}\it \widehat{CMB}\equiv\widehat{DAM}\ \ \ \ (1)\\ \\ \it \widehat{MBC}\equiv\widehat{AMD}\ \ \ \ (2)\end{cases}$

$\it \Delta DAM\ \Rightarrow\ m(\widehat{DAM})+m(\widehat{AMD})=90^o\stackrel{(1)}{\Longrightarrow} m(\widehat{CMD})+m(\widehat{AMD})=90^o \ \ \ \ (3)\\ \\ m(\widehat{BMA}) =180^o-\Big(m(\widehat{CMD})+m(\widehat{AMD})\Big) \stackrel{(3)}{=}\ 180^o-90^o=90^o\Rightarrow \\ \\ \Rightarrow AM\perp BM$

$\it b)\\ \\ \left.\begin{aligned}\widehat{CMB}\equiv \widehat{ABM}\ (alterne\ intrene)\\ \\ \widehat{MPC}\equiv \widehat{BPA}\ (opuse\ la\ v\hat arf)\\ \\ \widehat{PCM}\equiv \widehat{PAB}\ (alterne\ intrene)\end{aligned} \right\} \stackrel{(UUU)}{\Longrightarrow}\Delta PCM\sim \Delta PAB \Rightarrow\\ \\ \\ \Rightarrow \dfrac{PC}{PA}=\dfrac{CM}{AB} \Rightarrow \dfrac{PC}{PA}=\dfrac{12}{15} \Rightarrow \dfrac{PC}{PA+PC} =\dfrac{12}{15+12} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{PC}{AC} =\dfrac{12}{27}$

$2\it ABC-dreptunghic,\ m(\hat B)=90^o\stackrel{T.Pitagora}{\Longrightarrow}\ AC^2=AB^2+BC^2 \Rightarrow \\ \\ \Rightarrow AC^2=15^2+6^2=225+36=261\ \ \ \ \ (4)\\ \\ \dfrac{CP}{AC}= \dfrac{12^{(3}}{27} \Rightarrow \dfrac{CP}{AC}= \dfrac{4}{9} \Rightarrow \dfrac{CP^2}{AC^2} =\dfrac{4^2}{9^2} \stackrel{(4)}{\Longrightarrow} \dfrac{CP^2}{261}=\dfrac{16}{81} \Rightarrow\\ \\ \Rightarrow CP^2=\dfrac{261\cdot16}{81} =\dfrac{29\cdot16}{9}=\dfrac{4^2\cdot(\sqrt{29})^2}{3^2} \Rightarrow$

$\it \Rightarrow CP=\dfrac{4\sqrt{29}}{3}$