Answer:
Part 1) [tex]HD=\sqrt{105}\ units[/tex]
Part 2) [tex]JD=2\sqrt{34}\ units[/tex]
Part 3) [tex]KD=6\sqrt{2}\ units[/tex]
Step-by-step explanation:
step 1
Find the length HD
In the right triangle HDF
Applying the Pythagoras Theorem
[tex]FD^{2} =HF^{2} +HD^{2}[/tex]
substitute the values and solve for HD
[tex]19^{2} =16^{2} +HD^{2}[/tex]
[tex]HD^{2}=19^{2}-16^{2}[/tex]
[tex]HD^{2}=105[/tex]
[tex]HD=\sqrt{105}\ units[/tex]
step 2
Find the length JD
In the right triangle JDF
Applying the Pythagoras Theorem
[tex]FD^{2} =JD^{2} +FJ^{2}[/tex]
we have
[tex]FD=19\ units[/tex]
[tex]FJ=JG=15\ units[/tex] ----> because JD is a perpendicular bisector
substitute the values and solve for JD
[tex]19^{2} =JD^{2} +15^{2}[/tex]
[tex]JD^{2}=19^{2}-15^{2}[/tex]
[tex]JD^{2}=136[/tex]
[tex]JD=\sqrt{136}\ units[/tex]
[tex]JD=2\sqrt{34}\ units[/tex]
step 3
Find the length KD
In the right triangle KDG
Applying the Pythagoras Theorem
[tex]GD^{2} =KD^{2} +GK^{2}[/tex]
we have
[tex]GD=FD=19\ units[/tex]
[tex]GK=17\ units[/tex]
substitute the values and solve for KD
[tex]19^{2} =KD^{2} +17^{2}[/tex]
[tex]KD^{2}=19^{2}-17^{2}[/tex]
[tex]KD^{2}=72[/tex]
[tex]KD=\sqrt{72}\ units[/tex]
[tex]KD=6\sqrt{2}\ units[/tex]
