Answer: B
In the right-angled triangle \(\triangle ABM\): BM = \(\frac{1}{2}\) × BC = 6 cm AM = 8 cm \[ \therefore AB^2 = BM^2 + AM^2 = 6^2 + 8^2 = 36 + 64 = 100 \] \[ \therefore AB = \sqrt{100} = 10 \text{ cm} \] \[ \therefore AB = AC = 10 \text{ cm} \]
In the right-angled triangle \(\triangle ABM\): BM = \(\frac{1}{2}\) × BC = 6 cm AM = 8 cm \[ \therefore AB^2 = BM^2 + AM^2 = 6^2 + 8^2 = 36 + 64 = 100 \] \[ \therefore AB = \sqrt{100} = 10 \text{ cm} \] \[ \therefore AB = AC = 10 \text{ cm} \]