Question

Va rog foarte mult dau multe puncte coroană va rog din sufletul meu am test luni ​
Ex 9 si desen va rog

Answer 1

Răspuns:

ve zi atasament

pr lui A pe BC este C deci AC⊥BC deciΔ ACB este dreptunghic in C

Explicație pas cu pas:

Answer 2

$\it a)\ pr_{BC}A=C \Rightarrow AC\perp BC \Rightarrow \Delta CAB-\ dreptunghic,\ m(\hat C)=90^o\\ \\ T.\ catetei \Rightarrow BC^2=AB\cdot EB \Rightarrow EB=\dfrac{BC^2}{AB}=\dfrac{15\cdot15}{25}=9\ cm\\ \\ \\ b)\ AECD-dreptunghi \Rightarrow DC=AE=AB-EB=25-9=16\ cm\\ \\ \\ c)\ T.\ Pitagora\ \^in\ \Delta CAB \Rightarrow AC^2=AB^2-BC^2 \Rightarrow AC^2=25^2-15^2=$

$\it =(25-15)(25+15)=10\cdot40=400 =20^2\Rightarrow AC=20\ cm\\ \\ CE-\^in\breve{a}l\c{\it t}ime\ corespunz\breve{a}toare\ ipotenuzei\ \^{i}n\ \Delta CAB \Rightarrow CE=\dfrac{CA\cdot CB}{AB}\Rightarrow$

$\it\Rightarrow CE=\dfrac{20\cdot15}{25}=12\ cm\\ \\ \\ d) AECD-dreptunghi \Rightarrow AD=CE=12\ cm\\ \\ \mathcal{P}_{ABCD}=AB+BC+CD+DA=25+15+16+12=68\ cm$

$\it \mathcal{A}_{ACD}=\dfrac{c_1\cdot c_2}{2}=\dfrac{CD\cdot AD}{2}=\dfrac{16\cdot12}{2}=96\ cm^2\\ \\ \\ f)\ \mathcal{A}_{ABCD}=\dfrac{\mathcal{B}+b}{2}\cdoth=\dfrac{AB+CD}{2}\cdot AD =\dfrac{25+16}{2}\cdot12=41\cdot6=246\ cm^2$