Question

Urgent imi trebuie, dau 50 de puncte pentru raspunsul dat

Answer 1

Răspuns:

$AD^{2} = BD * DC => (teorema~intaltimi)\\\\AD^{2} = 4 * 9\\AD^{2} = 36~|:\sqrt{~} \\AD = 6 \\\\ADC~dr.~m(<D)=90\\AC^{2} = CD^{2}-AD^{2}\\AC^{2} = 9^{2}-6^{2}\\AC^{2} = 81 - 36\\AC^{2} = 45\\AC = 3\sqrt{5}\\\\ ADB~dr.~m(<D)=90\\AB^{2} = AD^{2} + BD^{2}\\AB^{2} = 6^{2} + 4^{2}\\AB^{2} = 36 + 16\\AB = 2\sqrt{13}$