Which has a larger area, a 4:3 aspect ratio 32 inch TV or a 16:9 aspect ratio 32 inch TV? Find the side lengths of each of the TV's and the area of each TV to compare.
Explain your reasoning and show all mathematical calculations.

Answers:

Explanation:

Let x be some positive real number

The 4:3 aspect ratio means the width (horizontal) portion of the tv is 4x inches while the height (vertical) portion is 3x inches

The ratio 4x:3x reduces to 4:3 after dividing both parts by x.

The 4x by 3x rectangle has the diagonal 32 inches as the instructions state. The tv size is always measured along the diagonal.

So effectively, we have two identical right triangles with legs 4x and 3x, and hypotenuse 32.

Apply the pythagorean theorem to find x

a^2+b^2 = c^2

(4x)^2+(3x)^2 = 32^2

16x^2+9x^2 = 1024

25x^2 = 1024

x^2 = 1024/25

x = sqrt(1024/25)

x = 32/5

x = 6.4

Recall that x is positive, so we ignore the negative square root here.

This 4x by 3x tv then has dimensions of

These values are exact.

The area is therefore base*height = 25.6*19.2 = 491.52 square inches

This of course only applies to the 4:3 tv that's 32 inches in diagonal.

Now onto the 16:9 tv.

We'll follow the same steps as the last section. We'll use y this time

The 16:9 ratio becomes 16y:9y

a^2+b^2 = c^2

(16y)^2 + (9y)^2 = 32^2

256y^2 + 81y^2 = 1024

337y^2 = 1024

y^2 = 1024/337

y = sqrt(1024/337)

y = 1.7431510742491 which is approximate

y = 1.74315

So,

the area is approximate since the width and height are approximate. It rounds to about 437.55 square inches

To recap, we found the following:

lol

step-by-step explanation: