stephny wrote:The size of a flat-screen television is given as the length of the screen's diagonal. how many square inches greater is the screen of a square 34-inch flat screen television than a square 27- inch flat screen television?

If d is the diagonal of a square then its area is ½ d^2.

A square 34-inch flat screen television would have ½ (34)^2 square inches of area, whereas a square 27-inch flat screen television would have ½ (27)^2 square inches of area.

The excess is


½ (34)^2 - ½ (27)^2

= ½ (34 - 27) (34 + 27)

= ½ (7) (61)

= ½ (427)

= 213 ½ square inches.

Diagnol = Side * sqrt(2)

To find: Area of 34 - Area of 27

(34/sqrt(2))^2 - (27/sqrt(2))^2

1/2 (34-27)(34+27)

1/2 * (7) * (61) = 427/2 = 213.5

Some consistent and accurate answers given, however they both assume the TV to be square.
Have you ever seen a square TV? I know I haven't.

Two typical TV screen proportions are 16:9 and 4:3 (one of which would have been helpful in the question).

Below are two different solutions, one for each case:

Case 1 (16:9 aspect ratio)

Pythagoras' Theorem gives

(16c)^2 + (9c)^2 = D^2, where c = constant and D = diagonal length

Simplified this gives:

c^2 = D^2/337

Also, Area = 16c x 9c = 144c^2

Therefore Area = 144D^2/337

Substituting D1 = 34 gives Area1 = 493.96
Substituting D2 = 27 gives Area2 = 311.50

Therefore D2 - D1 = 182.5 (to 1 decimal place)


Case 2 (4:3 aspect ratio)

Pythagoras' Theorem gives

(3c)^2 + (4c)^2 = D^2, where c = constant and D = diagonal length

Simplified this gives:

c^2 = D^2/25

Also, Area = 3c x 4c = 12c^2

Therefore Area = 12D^2/25

Substituting D1 = 34 gives Area1 = 554.88
Substituting D2 = 27 gives Area2 = 349.92

Therefore D2 - D1 = 204.96 (to 1 decimal place)

OK, to be fair the question does stipulate that the screen is square, but I thought this made the question a bit more exciting!