Answer:
[tex]Width = 36in[/tex]
[tex]Size = 42.02[/tex]
Step-by-step explanation:
Given
[tex]x \to width[/tex]
[tex]y \to height[/tex]
[tex]x : y = 16 : 9[/tex]
Required
The size of [tex]screen[/tex] of height [tex]20,6in[/tex]
First, we calculate the width using the following equivalent ratios
[tex]16: 9 = x : 20.6[/tex]
Express as fraction
[tex]\frac{16}{ 9} = \frac{x }{ 20.6}[/tex]
Solve for x
[tex]x = 20.6 * \frac{16}{ 9}[/tex]
[tex]x = 36.62[/tex]
Hence:
[tex]Width = 36in[/tex] --- approximated
So, we have:
[tex]x = 36.62[/tex]
[tex]y =20.6[/tex]
The size (diagonal) is then calculated using:
[tex]Size = \sqrt{x^2 + y^2[/tex]
[tex]Size = \sqrt{36.62^2 + 20.6^2[/tex]
[tex]Size = \sqrt{1765.3844[/tex]
[tex]Size = 42.02[/tex]
[tex]Size =42in[/tex] --- approximated