Question

What are the actual dimensions of a television that has a 15:9 aspect ratio and a screen that is 42 inches?

Answer 1

Answer:

Let x be the number.

It is given that a 15 : 9 aspect ratio and a screen that is 42 inches.

Aspect ratio states that  the ratio of the display image’s width to its height.

Then;

Width of the television of(w) = 15x

and height of the television(h) = 9x.

Since, television is in the form of Rectangle.

And it is also, given that the screen is 42 inches .

⇒Diagonal of the television(D) = 42 inches

Using formula of diagonal of the rectangle :

$D^2 =w^2+h^2$

Substitute the given values to solve for x;

$42^2 = (15x)^2+(9x)^2$

$1764 = 225x^2 + 81x^2 = 306x^2$

or

$x^2 = \frac{1764}{306} = 5.76$

$x = \sqrt{5.76} = 2.4$

Then:

width of the television (15x) = 15(2.4) = 36 inches and

Height of the television (9x) = 9(2.4) = 21.6 inches.

Therefore, the actual dimensions of the television are;

width = 36 inches and height = 21.6 inches.