Question

HD, JD, and KD are perpendicular bisectors of EFG find each length

Answer 1

Part 1)
`HD=√(105)\ units`

Part 2)
`JD=2√(34)\ units`

Part 3)
`KD=6√(2)\ units`

Explanation:

step 1

Find the length HD

In the right triangle HDF

Applying the Pythagoras Theorem

`FD^(2) =HF^(2) +HD^(2)`

substitute the values and solve for HD

`19^(2) =16^(2) +HD^(2)`

`HD^(2)=19^(2)-16^(2)`

`HD^(2)=105`

`HD=√(105)\ units`

step 2

Find the length JD

In the right triangle JDF

Applying the Pythagoras Theorem

`FD^(2) =JD^(2) +FJ^(2)`

we have

`FD=19\ units`

`FJ=JG=15\ units` ----> because JD is a perpendicular bisector

substitute the values and solve for JD

`19^(2) =JD^(2) +15^(2)`

`JD^(2)=19^(2)-15^(2)`

`JD^(2)=136`

`JD=√(136)\ units`

`JD=2√(34)\ units`

step 3

Find the length KD

In the right triangle KDG

Applying the Pythagoras Theorem

`GD^(2) =KD^(2) +GK^(2)`

we have

`GD=FD=19\ units`

`GK=17\ units`

substitute the values and solve for KD

`19^(2) =KD^(2) +17^(2)`

`KD^(2)=19^(2)-17^(2)`

`KD^(2)=72`

`KD=√(72)\ units`

`KD=6√(2)\ units`