Modern televisions have a ratio of 16:9 for the width:height. If a television has a diagonal lenght of 42 in.?

Question

(for the viewing area), how wide and how high is the viewing area of this television?
I'm stuck on this problem.

Answer 1

Let the width be 16x and the height 9x. Then the diagonal length is √((16x)^2 + (9x)^2) = 42 in.
so √(256x^2 + 81x^2) = 42
so √(337x^2) = 42
so 337x^2 = 42^2 = 1764
so x^2 = 1764 / 337
so x = √(1764/337) = 2.29 in.
So the width is 16×2.29 = 36.6 in and the height is 9×2.29 = 20.6 in.

Answer 2

The TV is approximately 36.61 inches wide and 20.59 inches high.

this requires the pythagorus theory -- the square of the hypotenuse is equal to the sum of the squares of the other two sides

The tv is 42 inches diagonally, so the hypotenuse is 42 inches. the square of this is 42x42=1764

then the sum of (16x) x (16x) + (9x) x (9x) has to equal 1764
where 'x' is the multiplier for the ratio

thus (256x-squared) + (81x- squared)=1764,
or 337x-squared =1764

x-squared is thus 1764/337 =5.2344203
and x = square root of 5.2344203 = 2.2878857

Therefore the width is 16 x 2.2878857 =36.606 inches
and the height is 9 x 2.2878857=20.590971 inches

There you have it!