Question

Problema 52 , va rooog :)

Answer 1

$a)~ \Delta DBC~este~dreptunghic~in~B \Rightarrow m( \angle BDC) + m( \angle BCD)=90 \textdegree. \\ \\ Dar~ \frac{1}{2}m(\angle BCD)=m( \angle BDC) \Leftrightarrow m( \angle BCD)=2 m( \angle BDC). \\ \\ Deci~m( \angle BDC)+2m( \angle BDC)=90 \textdegree \Leftrightarrow 3m( \angle BDC) =90 \textdegree \Rightarrow \\ \\ \Rightarrow m( \angle BDC)=30 \textdegree \Rightarrow m( \angle BCD)=60 \textdegree.$

$Fie~S \in (CD~astfel~incat~AS \perp CD. \\ \\ AD~||~BC \Rightarrow m( \angle ADS)=m(\angle BCD)=60 \textdegree. \\ \\ m( \angle SAD)=90 \textdegree - m( \angle ADS)=30 \textdegree. \\ \\ Notez~SD=k. \\ \\ In~triunghiul~ADS~(dreptunghic~in~S),~cateta~[SD]~se~opune~un-~ \\ \\ ghiului~cu~masura~de~30 \textdegree.~Deci~SD= \frac{AD}{2}=k \Rightarrow AD=2k. \\ \\ Din~T.Pitagora~in~ \Delta ASD,~avem:~AS^2+SD^2 =AD^2 \Rightarrow$

$\Rightarrow AS= \sqrt{AD^2-SD^2}= \sqrt{4k^2-k^2}= \sqrt{3k^2}=k \sqrt{3}. \\ \\ In \Delta BCD~(dreptunghic~in~B),~cateta~[BC]~se~opune~unghiului \\ \\ cu~masura~de~30 \textdegree \Rightarrow CD=2 BC=2AD=2 \cdot 2k=4k. \\ \\ SC=SD+CD=k+4k=5k. \\ \\ Aplic~T.Pitagora~in ~ \Delta ASC~(dreptunghic~in~S): \\ \\ AS^2+SC^2=AC^2 \Leftrightarrow (k \sqrt{3})^2+(5k)^2=(18 \sqrt{7})^2 \Leftrightarrow \\ \\ \Leftrightarrow 3k^2+25k^2=2268 \Leftrightarrow 28k^2=2268 \Rightarrow k^2=81 \Rightarrow k=9.$

$Deci~AD=2k=2 \cdot 9 =18~(cm)~si~CD=4k=4 \cdot 9=36~(cm). \\ \\ P_{ABCD}=2(AD+CD)=2(18+36)=108~(cm).$

$b)~A_{ABCD}= d(A,CD) \cdot CD=AS \cdot CD=9 \sqrt{3} \cdot 36=324 \sqrt{3}~(cm^2). \\ \\ c)~Se~pare~ca~aici~s-a~omis~un~lucru:~nu~s-a~precizat~unde~se \\ \\ afla~punctul~O,~insa~sunt~sigur~ca~este~;a~intersectia~diagonalelor. \\ \\ O~este~mijlocul~segmentelor ~[AC]~si~[BD].~Deci~AO~este~mediana \\ \\ in \Delta ABD \Rightarrow A_{AOD}=A_{AOB}~(1). \\ \\ BO-mediana~in~ \Delta ABC \Rightarrow A_{AOB}=A_{BOC} ~(2).\\ \\ CO-mediana~in~ \Delta BCD \Rightarrow A_{BOC}=A_{COD}~(3).$

$Din~(1),~(2)~si~(3)~rezulta:~A_{AOB}=A_{BOC}=A_{COD}=A_{AOD}= \\ \\ = \frac{A_{ABCD}}{4} \Rightarrow A_{AOB}= 25 \%~din~A_{ABCD}.$