Lets lenght of widescreen = w
so height will be : 2w

so area = 2w^2

lets length of standard = s
so height = 4s/3

area = 4s^2/3



we need relation between W and S or relation between W^2 and S^2

use diagonal information:
w^2 + 2w^2 = 30^2
5w^2 = 30^2 ...............(1)


s^2+ (4s/3)^2 = 3-^2

25s^2/9 = 30^2...........(2)

use 1 and 2

5W^2 = 25S^2/9

9W^2 = 5S^2

W^2/S^2 = 5/9

W^2 = 5/9 * S^2

2*W^2 = 2*5/9 * S^2

2*W^2 = 2*5*3/9*4 * 4S^2/3

area of wide screen = 5/6 * area of standard

= 83% ..I got D as answer.

Got D correct

1) Find sides of two TV's: standart Tv - sides 24 and 18 diag 30 (special triangle 3:4:5)
widescreen TV - sides 2sqr45 and 4sqr45 (4x^2 +2x^2 = 900)

Compare areas: (2sqr45*4sqr45)/24*18 = 2*45*4/24*18 = 45/54 = 0.83, or 83%

Unfortunately, I was unable to follow your explanation sunnyjohn.

Ershovici, where did you get "2x^2 and 4x^2" for the lengths of the sides of the widescreen?

I was able to recognize that the standard was a 3-4-5 triangle, but I could not figure out how to find the sides of the widescreen TV.

Please provide more detail with your explanation, as I am not understanding right now ... sorry.

a/b = 2/3 is same as a = 2k and b = 3k. You can transform two unknowns into one unknown.



TV1:

d^2 = l^2 + b^2
30^2 = k^2 + 4k^2
k^2 = 30^2/5



TV2:

30^2 = 9m^2 +16m^2

m^2 = 30^2/25



Find Area of TV2/Area of TV1

= (3m)(4m)/[k(2k)] = 6m^2/k^2 = 6(5)/25 = 6/5


Answer = 5/6 = 83/100