Chizmada ABC toʻgʻri burchakli uchburchak berilgan. Agar AD ED, ∠ B=90^°, ED=6, S_BEDC=20 va S_AED=16…

$\triangle ADE$ va $\triangle ABC$ o‘xshash ($\angle A$ umumiy, $\angle D=\angle B=90^\circ$), bunda $ED$ tomon $BC$ ga mos. $S_{ABC}=16+20=36$. O‘xshashlik koeffitsiyenti: $\left(\frac{ED}{BC}\right)^2=\frac{16}{36}$, $\frac{6}{x}=\frac{2}{3}$, $x=9$. Tekshirish: $AD=\frac{2\cdot16}{6}=\frac{16}{3}$, $AB=\frac{3}{2}AD=8$, $S_{ABC}=\frac{8\cdot9}{2}=36$.

Manba: UZBMB namunaviy test topshiriqlari (2022, mutaxassislik)