Intermediate Algebra: Find third side, given a = 48, b = 20

[Helen]

I think that I am coming up with the wrong answer.
This is a right triangle. I can't draw it so you will need to opicture it.
a=48, b=20
c^(2) = a^(2) + b^(2)
c^(2) = 48^(2) + 20^(2)
c^(2) = 2304 + 400
c^(2) = 2704
c = radical 2704
c = radical 16 * 169
c = radicals 16 * 169
c = radical 4 radicand 169
I hope this can be understood. Helen

 

[Deleted member 4993]

Re: Intermediate Algebra

I think that I am coming up with the wrong answer.
This is a right triangle. I can't draw it so you will need to opicture it.
a=48, b=20
c^(2) = a^(2) + b^(2)
c^(2) = 48^(2) + 20^(2)
c^(2) = 2304 + 400
c^(2) = 2704
c = radical 2704
c = radical 16 * 169
c = radicals 16 * 169
c = radical 4 radicand 169
I hope this can be understood. Helen

\(\displaystyle \sqrt{169} = 13\)

so

\(\displaystyle \sqrt{16\cdot\ 169} = 4\cdot\ 13 = 52\)

 

[soroban]

Re: Intermediate Algebra

Hello, Helen!

You quit too soon . . .

This is a right triangle: .\(\displaystyle a = 48,\;b=20\)

\(\displaystyle c^2 \:=\: a^2 + b^2 \quad\Rightarrow\quad c^2 \;=\;48^2 + 20^2 \;=\;2304 + 400 \;=\;2704\)

\(\displaystyle c \;= \;\sqrt{2704} \;=\;\sqrt{16\cdot169} \;=\;\sqrt{16}\cdot\sqrt{169} \;\Rightarrow\;\boxed{4\cdot13}\;\Rightarrow\;\boxed{52}}\)

 

[Helen]

Re: Intermediate Algebra

subhotosh khan,
soroban,
Thank you for your help. Very much appreciated. Helen