Question

8. în triunghiul dreptunghic isoscel ABC, <A=90°, ipotenuza BC are lungimea de 8 cm. Calculati perimetrul triunghiului ABC. ​

Answer 1

ABC; A = 90⁰

BC=8 cm

==>Teorema lui pitagora

BC²=AB²+AC²

8²=AB²+AC²

64=AB²+AC²

AB=AC

==>2x²=8²

x²=64/2=32

x=√32=4√2

Perimetru=2*4√2+8=8√2+8=8(√2+1)

Answer 2

$\it \Delta ABC-dreptunghic\ isoscel,\ \widehat A=90^0 \Rightarrow AB=AC\ \stackrel{not}{=}\ x\\ \\ Th.\ Pitagora\ \Rightarrow x^2+x^2=8^2 \Rightarrow 2x^2=64|_{:2} \Rightarrow x^2=32 \Rightarrow \\ \\ \Rightarrow x=\sqrt{32}=\sqrt{16\cdot2}=4\sqrt2\ cm\\ \\ \mathcal{P}=AB+AC+BC=4\sqrt2+4\sqrt2+8=8\sqrt2+8\ cm$