The size of a television screen is given as the length of

the diagonal of a square is always \(side*\sqrt{2}\)
and the side of a square is vice versa always \(\frac{diagonal}{\sqrt{2}}\)

Therefore:
area of the bigger one is \((\frac{21}{\sqrt{2}})^2\)
area of the smaller one is \((\frac{19}{\sqrt{2}})^2\)
and the difference is = 40

The area of a square is \(side^2\), but since here the length of the diagonal is given, then it's better to use another formula \(area_{square}=\frac{diagonal^2}{2}\).

The size of a television screen is given as the length of the screen's diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

A. 2
B. 4
C. 16
D. 38
E. 40

We're dealing with squares here. When you take the diagonal of a square it creates two isosceles triangles with a ratio of x: x: x√2. Because the hypotenuse is 21 inches it is equal to x√2. To find the length of the legs, we set 21 = x√2. x = 21/√2
To find the area we square 21/√2 which equals 441/2. Similarly, for the 19 inch screen we follow the same steps. We find that the legs of the triangle (or the sides of the square screen) are 19/√2 inches. To find the area we square this to get 361/2. Finally, subtract 361/2 from 441/2 to get 80/2 = 40.

E

Just to refresh my understanding, how is the \(area_{square}=\frac{d^2}{2}\)

The side x side is simple enough. Given that a square is essentially 2 isoceles triangles, I was going with 21=\(x\sqrt{2}\)

That's where I got stuck. But I still don't get your formula...

Let’s determine the side of the square 21-inch screen (i.e., the diagonal of the screen is 21 inches). Recall that the diagonal of a square is equal to side√2.

21 = side√2

21/√2 = side

Since area is side^2, the area of the 21-inch screen is (21/√2)^2 = 441/2.

Let’s determine the side of the square 19-inch screen:

19 = side√2

19/√2 = side

The area of the 19-inch screen is (19/√2)^2 = 361/2.

Thus, the difference is 441/2 - 361/2 = 80/2 = 40.

Alternate solution:

We are given two square TV screens with diagonals 21 and 19, respectively. We have to determine the difference between the areas of the screens. Recall that the area of a square, given its diagonal d, is A = d^2/2. Thus, the area of the 21-inch screen is 21^2/2 = 441/2 and the area of the 19-inch screen is 19^2/2 = 361/2. Therefore, the difference in areas is 441/2 - 361/2 = 80/2 = 40.

Answer: E

area of a square 21-inch screen = (21/\sqrt{2})^2 = 441/2
area of a square 19-inch screen = (19/\sqrt{2})^2 = 361/2

Diff = 80/2 = 40
E

If the size of a square television screen is given by its diagonal, we need side lengths to calculate area.

The relationship between the a square's side and its diagonal, d, is given by

\(s\sqrt{2} = d\)
\(s = \frac{d}{\sqrt{2}}\)

The side of the 21-inch size television (d = 21), therefore, is

\(\frac{21}{\sqrt{2}}\). Square that to find area:

\((\frac{21}{\sqrt{2}}\) * \(\frac{21}{\sqrt{2}})\) = \(\frac{21*21}{2}\) = \(\frac{441}{2}\)

The side of the 19-inch size television (d = 19) is

\(\frac{19}{\sqrt{2}}\). Square that to find area:

\((\frac{19}{\sqrt{2}}\) * \(\frac{19}{\sqrt{2}})\) = \(\frac{19*19}{2}\) = \(\frac{361}{2}\)

Difference in area between larger and smaller, in square inches:

\((\frac{441}{2} - \frac{361}{2}) =\frac{80}{2} = 40\)

Answer E

Kinda late to the party, but how do we know that the screen is a square?

I don't see many perfect square screens around to be honest.

Is this question ancient OG, like pre 2005 when TVs were huge square bricks?

That would explain it

I love this question because it's quite literally the definition of the difference of two square (i.e. X^2 - Y^2 = (X+Y)(X-Y).

So start off remember that if you take the diagonal of a square then you get two triangles each with side ratios of X, X and Xsqrt(2). So this means that the sides of each of the TVs can be determined by the equation of Side of Square X Sqrt(2) = Diagonal.

Side of first TV = 21/Sqrt(2)
Side of second TV = 21/Sqrt(2)

Thus we want to take the difference of the AREA of the first and second TV. So (21/sqrt(2))^2 - (19/sqrt(2))^2, which can be solved further as (21^2 - 19^2)/2, so (21+19)(21-19)/2 = 40(2)/2 so 40 which is answer choice E is the answer.

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