Question

VA ROG AJUTATI-MA!!! DAU COROANA!

Answer 1

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Answer 2

$\it a)\ Ducem\ \ AF\perp BC,\ F\in BC.\\ \\ Din\ \Delta FAB \Rightarrow sinB=\dfrac{AF}{AB} \Rightarrow AF = AB\cdot sinB=12\sqrt3\cdot sin60^o=\\ \\ \\ =12\sqrt3\cdot\dfrac{\sqrt3}{2} =\dfrac{12\cdot3}{2} =6\cdot3=18\ cm$

$\it AF\ este\ \c{s}i\ median\u{a} \Rightarrow BF=\dfrac{BC}{2}=\dfrac{12\sqrt3}{2}=6\sqrt3\ cm\\ \\ Fie\ AF\cap QP = \{D\},\ DF=NP=x \Rightarrow AD=18-x\\ \\ QP||BC \Rightarrow QD||BF \Rightarrow \Delta AQD \sim \Delta ABF \Rightarrow \dfrac{AD}{AF}=\dfrac{QD}{BF} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{18-x}{18}=\dfrac{QD}{6\sqrt3}\Rightarrow QD=\dfrac{6\sqrt3(18-x)}{18}=\dfrac{\sqrt3(18-x)}{3}=\dfrac{18\sqrt3-\sqrt3x}{3}\\ \\ \\ MN=QP=2\cdot QD=2\cdot\dfrac{18\sqrt3-\sqrt3x}{3}=\dfrac{36\sqrt3-2\sqrt3x}{3}$

$\it b)\ MNPQ-p\u{a}trat \Rightarrow MN=NP \Rightarrow\dfrac{36\sqrt3-2\sqrt3x}{3}=x \Rightarrow\\ \\ \\ \Rightarrow 36\sqrt3-2\sqrt3x=3x \Rightarrow 36\sqrt3=2\sqrt3x+3x|_{:{\sqrt3}} \Rightarrow\\ \\ \\ \Rightarrow36=2x+\sqrt3x \Rightarrow 36=x(2+\sqrt3) \Rightarrow x=\dfrac{^{{2-\sqrt3})}36}{\ \ 2+\sqrt3} =\\ \\ \\ =\dfrac{36(2-\sqrt3)}{1}=36(2-\sqrt3)\ cm$

$\it c)\ \mathcal{A}_{ABC} =\dfrac{BC\cdot AF}{2}=\dfrac{12\sqrt3\cdot18}{2} =6\sqrt3\cdot18 =108\sqrt3\ cm^2\\ \\ MN=MQ\sqrt3\ \ \ (1)\\ \\ \mathcal{A}_{MNPQ} =MN\cdot MQ\stackrel{(1)}{=}\ MQ\sqrt3\cdot MQ=MQ^2\sqrt3\ \ \ \ (2)$

$\it (1) \Rightarrow \dfrac{36\sqrt3-2\sqrt3x}{3}=x\sqrt3|_{:\sqrt3} \Rightarrow \dfrac{36-2x}{3}=x \Rightarrow\\ \\ \\ 36-2x=3x \Rightarrow36=5x \Rightarrow x = \dfrac{36}{5}\ \ \ \ (3)$

$\it (2),\ (3) \Rightarrow \mathcal{A} =\dfrac{36^2}{5^2}\sqrt3 =\dfrac{1296\sqrt3}{25}\ cm^2$

$\it p\%\ din\ \mathcal{A}_{ABC} =\mathcal{A}_{MNPQ} \Rightarrow \dfrac{p}{100}\cdot108\sqrt3 =\dfrac{1296\sqrt3}{25} \Rightarrow\\ \\ \\ \Rightarrowp=\dfrac{1296\sqrt3}{25}\cdot\dfrac{100}{108\sqrt3}= \dfrac{1296\cdot100}{2700} =\dfrac{1296}{27} =48\\ \\ \\ Deci,\ \mathcal{A}_{MNPQ}= 48\%\ din\ \mathcal{A}_{ABC}$