Answer:
see explanation
Step-by-step explanation:
Using Pythagoras' identity in the right triangles.
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
DF² = DE² + EF² , then
(a)
DF² = 16² + 12² = 256 + 144 = 400 ( take square root of both sides )
DF = [tex]\sqrt{400}[/tex] = 20
(b)
DF² = 7² + 2² = 49 + 4 = 53 ( take square root of both sides )
DF = [tex]\sqrt{53}[/tex] ≈ 7.28 ( to 2 dec. places )
