Question

in adjoining figure find the distance AC

Answer 1

by Pythagoras theorem

AE^2 + AB^2 = EB^2

AB^2 = EB^2 - AE^2

         = 100 - 36

         = 64

AB = 8

BC^2 = BD^2 - DC^2

         = 169 - 144

         = 25

BC = 5

AC = AB + BC

     = 8 + 5

     = 13

Answer 2

let AB=x, BC=y

x²+6²=10² (pythagoras thm)

x²+36=100

x²=64

x=√64=8m

AB=8m

y²+12²=13²

y²=144=169

y²=25

y=√25=5m

BC=5m

AC=AB+BC

AC=8+5

AC=13m