Question 1207448
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Comparing TVs The screen size of a television is determined by the length of the diagonal 
of the rectangular screen. Traditional TVs come in a 4 :3 format, meaning the ratio of the length 
to the width of the rectangular screen is 4 to 3.
(a) What is the area of a 50-inch traditional TV screen? 
(b) What is the area of a 50-inch LCD TV whose screen is in a 16 :9 format? Which screen is larger?
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(a)  Consider 4:3 format.

     It means that if W inches is the width, then the length is  L = {{{(4/3)W}}} inches.


     Then the square of the diagonal is

          {{{W^2}}} + {{{L^2}}} = {{{W^2}}} + {{{((4/3)W)^2}}} = {{{W^2 + (16/9)W^2}}} = {{{(1 + 16/9)W^2}}} = {{{(25/9)W^2}}}  square inches.


     Thus  

          {{{(25/9)W^2}}} = {{{50^2}}}  square inches.


     It implies

          {{{W^2}}} = {{{(9/25)*50^2}}} = {{{(9/25)*50*50}}} = {{{9*(50/25)*50}}} = 9*2*50 = 9*100 = 900.


     Hence,  W = {{{sqrt(900)}}} = 30  inches.


     Thus the width is 30 inches, the length is  ((((4/3)*30}}} = 40 inches.


     So, this TV screen has dimensions 30 inches by 40 inches. 
     Its area is 30*40 = 1200 square inches.


     At this point, part (a) is solved, in full.



(b)  Solve (b) by the same (= the similar) way.
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