Answer:
Part 4) [tex]ER=3\ units[/tex]
Part 5) [tex]DF=9\sqrt{10}\ units[/tex]
Part 6) [tex]DE=30\ units[/tex]
Step-by-step explanation:
Part 4) Find ER
we know that
In the right triangle ERF
Applying the Pythagorean Theorem
[tex]EF^2=ER^2+RF^2[/tex]
substitute the given values
[tex](3\sqrt{10})^2=ER^2+9^2[/tex]
solve for ER
[tex]ER^2=(3\sqrt{10})^2-9^2[/tex]
[tex]ER^2=90-81\\ER^2=9\\ER=3\ units[/tex]
Part 5) Find DF
we know that
In the right triangle DRF
Applying the Pythagorean Theorem
[tex]DF^2=DR^2+RF^2[/tex]
substitute the given values
[tex]DF^2=27^2+9^2[/tex]
[tex]DF^2=810\\DF=\sqrt{810}\ units[/tex]
simplify
[tex]DF=9\sqrt{10}\ units[/tex]
Part 6) Find DE
we know that
[tex]DE=DR+RE[/tex] ----> by segment addition postulate
we have
[tex]DR=27\ units\\RE=ER=3\ units[/tex]
substitute
[tex]DE=27+3=30\ units[/tex]
