If a TV has a diagonal length of 42in. (For the viewing area), How wide and how high is the viewing area of this TV?

You seem to have asked this again. I've already given my answer in response to that question.

this is your answer...Modern Tv s have a ratio of 16:9 for the width:height.

Draw the TV and the diagonal. Consider one of the (right) triangles it forms.

Width & Base & hypotenuse (the diagonal).

You have W² + B² = 42² [why ?]

And you are also told that W/B = 16/9.

The above is a system of two equations & two unknowns ( the W & B ). So just solve the system!

Here is a starter: From the second eqs, you get W = (16/9)B. Replacing this into the first eqs you

get: ( (16/9)B )² + B² = 42² thus (16/9)²B² + B² = 42². Now solve for B. Then W.

Once found, the area of the TV is W*B.

Done!

There are other ways to go about this problem where you dont even need to set up the equations, nor to solve for W & B. I suggest you try to discover that technique too.

42 (16:9) 36.61"wide 20.59" height

42 (21:9) 38.6" wide 16.54" height