How do you use the Pythagorean Theorem to find the missing side of the right triangle with the given measures given c is the hypotenuse and we have b=3x,c=7x?

The Pythagorean Theorem states that

#a^2+b^2=c^2#

in a triangle with legs #a,b# and hypotenuse #c#, as you've already described in the problem.

With #b=3x# and #c=7x#, we have the relation:

#a^2+(3x)^2=(7x)^2#

Now, recall that when we have something like #(3x)^2#, we have to square both the #3# and the #x#:

#(3x)^2=3^2*x^2=9x^2#

Similarly, for #(7x)^2#:

#(7x)^2=7^2*x^2=49x^2#

Substituting these back in to the Pythagorean Theorem equation, we see that

#a^2+9x^2=49x^2#

Subtract #9x^2# from both sides of the equation.

#a^2=40x^2#

Take the square root of both sides.

#a=sqrt(40x^2)#

We can rewrite #sqrt(40x^2)# as a product of mostly squared terms in order to simplify. For example, it's important to note that #40=4xx10#.

#a=sqrt4*sqrt(x^2)*sqrt10#

#a=2xsqrt10#

Using the principle of proportionality disregard the #x's# for now.
Think if it as working on a triangle that has been reduced in scale but is of the same ratio.

By Pythagoras #A^2+B^2 = C^2#

So #A->sqrt(7^2-3^2)" "=" "sqrt(49-9)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Technically we could write #A=xsqrt(7^2-3^3)#
It is simpler just to leave it out for now but incorporate it at the end.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#A->sqrt(2^2xx10)#

#A->2sqrt(10)#

Scaling back up we have

#A=2xsqrt(10)#