(a) ΔABC is right-angled at B.

Therefore, by applying Pythagoras theorem, we obtain

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (6 cm)^{2} + (8 cm)2

AC^{2} = (36 + 64) cm^{2} =100 cm2

AC = (√100) cm = (√10x10) cm

AC = 10 cm

(b) ΔABC is right-angled at B.

Therefore, by applying Pythagoras theorem, we obtain

AC^{2} = AB^{2} + BC^{2}

(13 cm)^{2} = (AB)^{2} + (5 cm)^{2}

AB^{2} = (13 cm)^{2} − (5 cm)^{2} = (169 − 25) cm^{2} = 144 cm2

AB = (√144) cm = (√12x12) cm

AB = 12 cm