Question

dau 50 puncte va rog ajutor​

Answer 1

 

Se da:

ΔABC in care AB = BC = 13 cm si AC = 10 cm

NP = linie mijlocie.

M si Q sunt proiectiile lui N respectiv P pe AB

Rezulta dreptunghiul MNPQ

Se cere:

a) Aria ΔABC = ?

b) Aria  dreptunghiului MNPQ

Rezolvare:

.

$\displaystyle\bf\\a)\\\\\textbf{Calculam aria triunghiului cu formula lui Heron.}\\\\P=AB+BC+AC=13+13+10=36~cm~~(perimetrul)\\\\p=\frac{P}{2}=\frac{36}{2}==18~cm~~(semiperimetrul)\\\\A_{\Delta ABC}=\sqrt{p(p-AB)(p-BC)(p-AC)}\\\\A_{\Delta ABC}=\sqrt{18(18-13)(18-13)(18-10)}\\\\A_{\Delta ABC}=\sqrt{18\times5\times5\times8}\\\\A_{\Delta ABC}=\sqrt{5^2\times9\times2\times8}\\\\A_{\Delta ABC}=\sqrt{5^2\times3^2\times2^4}\\\\A_{\Delta ABC}=5\times3\times2^2\\\\\boxed{\bf A_{\Delta ABC}=60~cm^2}$

.

$\displaystyle\bf\\b)\\Calculam~inaltimea~triunghiului~dusa~din~punctul~C.\\\\A_{\Delta ABC}=\frac{AB\times h}{2}=60~cm^2\\\\ \frac{13\times h}{2}=60~cm^2\\\\13h=2\times60\\13h=120\\\\h=\frac{120}{13}~cm\\\\\textbf{Linia mijlocie imparte ln 2 parti egale laturile AC,~BC~si~inaltimea.}\\\\\implies~MN=\frac{h}{2}=\frac{60}{13}\\\\NP=\frac{AB}{2}=\frac{13}{2}\\\\Aria~dreptunghiului~= MN\times NP=\frac{60}{13}\times \frac{13}{2}=\frac{60}{2}=30~cm^2$

 

 

 

Answer 2

a) Ducem perpendiculara BT din B pe AC.

Pentru ca AB=BC ⇒ BT este si mediana ⇒ AT=TC=1/2AC=5.

Folosind teorema lui pitagora intr-unul dintre triunghiurile BTC sau BTA obtinem ca BT=√(13²-5²)=12.

Cum A(ABC)=BTxAC/2 ⇒ A(ABC)=12x10/2=12x5=60u².

b) Daca NP este linie mijlocie rezulta ca NP=1/2AB=13/2.

AP este mediana in triunghiul ABC de unde A(APB)=1/2A(ABC)=30, dar

A(APB)=PQxAB/2 ⇔ 13PQ=60 ⇒ PQ=60/13.

Deci A(ABCD)=PNxPQ=30u².