a) Xét ΔABC , `\hat{BAC}` = $90^{o}$ có :
BC² = AB² + AC² ( pytago )
=> BC² = 3² + 3² = 9 + 9 = 18
=> BC = 3√2 ( cm )
b) Xét ΔABC , `\hat{BAC}` = $90^{o}$ có :
BC² = AB² + AC² ( pytago )
=> BC² = 4² + 6² = 16 + 36 = 52
=> BC = 2√13 ( cm )
c) Xét ΔABC , `\hat{BAC}` = $90^{o}$ có :
BC² = AB² + AC² ( pytago )
=> BC² = (2,3)² + (3,9)² = 5,29 + 15,21 = 20,5
=> BC = $\frac{√82}{2}$ ≈ 4,52 ( cm )
d) Xét ΔABC , `hat{BAC}` = $90^{o}$ có :
BC² = AB² + AC² ( pytago )
=> BC² = (√5)² + 3² = 5 + 9 = 14
=> BC = 2 ( cm )
