In movies and television, the ratio of the width of the screen to the height is called the aspect ratio. Television screens usually have an aspect ratio of 4 : 3, while movie screens usually have an aspect ratio of 1.85 : 1. However, if a movie is made for television in "Letterbox" format, it retains the 1.85 : 1 aspect ratio and fills in the top and bottom parts of the screen with black bars. What would be the height of a movie in "Letterbox" format on a television screen that measures 25 inches along its diagonal? (Hint: First find the width and height of the television screen.)
Right triangle: width and height as your two legs, and the diagonal as your hypotenuse You don't know the width and height but you know that they are in a 4:3 ratio. meaning the width is 4x and the height is 3x. a^2 + b^2 = c^2 (4x)^2 + (3x)^2 = 25^2 16x^2 + 9x^2 = 625 25x^2 = 625 x^2 = 25 x = 5 so your TV width is 4x = 4*5 = 20 inches your TV height is 3x = 3 * 5 = 15 inches set up a proportion width on movie. . . .......................width on TV ----------------------- = --------------------- height on movie. . ......................height on TV 1.85 . . . ...........20 ------- = --------- . 1 . . . . ........h 1.85h = 20 h = 10.81
The choices are 13.51 in, 10.81 in, 15 in, and 8.12 inches
Screen's diagonal 25 inches, we need to find its height and width \[\frac{width}{height}=\frac{4}{3}\] \[width = 4\times \frac{height }{3}\] We know tv is a rectangle in shape so \[width^2+height^2= diagonal^2\] \[{(\frac{4}{3} \times height)}^2+ height ^2=625\] \[\frac{25}{9} height^2=625\] \[height^2= {225}\] \[height= 15\] \[width=\frac{4}{3} \times 15=20\] width= 20 inches height= 15 inches so movie with 1.85:1 aspect ratio will occupy width of 20 inches, so its height \[\frac{width}{height} =\frac{1.85}{1}\] \[height= width/1.85\] \[height= 20/1.85=10.25 inches\] so movie will occupy a height 0f 10.256 inches on tv
Sorry made a mistake it'd be 10.81 inches
@penguinup did you understand the solution?
ok, thanks guys, I get it
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