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wolverine [178]
3 years ago
11

How do I solve for the missing lengths?

Mathematics
1 answer:
Advocard [28]3 years ago
3 0

Answer:

Part 4) ER=3\ units

Part 5) DF=9\sqrt{10}\ units

Part 6) DE=30\ units

Step-by-step explanation:

Part 4) Find ER

we know that

In the right triangle ERF

Applying the Pythagorean Theorem

EF^2=ER^2+RF^2

substitute the given values

(3\sqrt{10})^2=ER^2+9^2

solve for ER

ER^2=(3\sqrt{10})^2-9^2

ER^2=90-81\\ER^2=9\\ER=3\ units

Part 5) Find DF

we know that

In the right triangle DRF

Applying the Pythagorean Theorem

DF^2=DR^2+RF^2

substitute the given values

DF^2=27^2+9^2

DF^2=810\\DF=\sqrt{810}\ units

simplify

DF=9\sqrt{10}\ units

Part 6) Find DE

we know that

DE=DR+RE ----> by segment addition postulate

we have

DR=27\ units\\RE=ER=3\ units

substitute

DE=27+3=30\ units

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<em></em>

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