Question

Need Geometry Help!

Answer 1

@jim_thompson5910 has you covered.

Answer 2

Use the pythagorean theorem

\[\Large a^2 + b^2 = c^2\]

Answer 3

In this case,

a = unknown (we're solving for it)
b = GH = 18
c = FH = 21

\[\Large a^2 + b^2 = c^2\]

\[\Large a^2 + 18^2 = 21^2\]

\[\Large a = ???\]

Answer 4

Sorry just got these! one second!

Answer 5

117?

Answer 6

how are you getting that

Answer 7

21x21=441-18x18=324=117

Answer 8

\[\Large a^2 + b^2 = c^2\]

\[\Large a^2 + 18^2 = 21^2\]

\[\Large a^2 + 324 = 441\]

\[\Large a^2 = 441 - 324\]

\[\Large a^2 = 117\]

\[\Large a = ???\]

Answer 9

13689

Answer 10

you have to undo squaring to isolate 'a'

Answer 11

what operation undoes squaring?

Answer 12

I've never heard of it

Answer 13

square roots undo squaring

Answer 14

example:

6 squared = 6^2 = 6*6 = 36

square root of 36 = sqrt(36) = 6

Answer 15

10.81

Answer 16

\[\Large a^2 + b^2 = c^2\]

\[\Large a^2 + 18^2 = 21^2\]

\[\Large a^2 + 324 = 441\]

\[\Large a^2 = 441 - 324\]

\[\Large a^2 = 117\]

\[\Large a = \sqrt{117}\]

Now simplify

Answer 17

I recommend you factor 117 into two numbers where one number is a perfect square.

Answer 18

\[3\sqrt{13}\]

Answer 19

hopefully you see how to get there?

Answer 20

Yes I got it. Thank you so much!

Answer 21

you're welcome