Question

8. The size of TV screens is usually given by the length of its diagonal. Suppose that

the ratio of the width to height of a TV is 16:9. [3 points]
a) What size is the TV if it has a width of 32 inches?
b) What are the width and height dimensions of a TV that is said to be 60 inches?

Answer 1

To find the size of the TV, use the ratio given for the width and height. For a) with a width of 32 inches, the height is 18 inches. For b) with a TV size of 60 inches, the width is approximately 106.67 inches and the height is 60 inches.

Step-by-step explanation:

To find the size of the TV, we need to use the ratio given for the width and height. The ratio is 16:9, which means that for every 16 units of width, there are 9 units of height.

a) If the width of the TV is 32 inches, we can set up a proportion using the ratio. 16/9 = 32/x. Cross multiplying, we get 16x = 9 * 32. Dividing both sides by 16, we find that x = 9 * 32 / 16 = 18. Therefore, the height of the TV is 18 inches.

b) If the TV is said to be 60 inches, we can set up a proportion again. 16/9 = x/60. Cross multiplying, we get 16 * 60 = 9x. Dividing both sides by 9, we find that x = (16 * 60) / 9 = 106.67. Therefore, the width of the TV is approximately 106.67 inches and the height is 60 inches.

Answer 2

a) Width to height is 16:9

Width = 32

32 / 16 = 2

2 * 9 = 18

Size of TV = Length of diagonal (use pythagorean theorem)

Length of diagonal =
`\sqrt{Width^(2) + Height^(2) }` =
`\sqrt{32^(2) + 18^(2) }` = 36.72 (round to the nearest hundredth)

b) Diagonal = 60 and width to height is 16:9

`\sqrt{16^(2) + 9^(2) }` =
`√(337)`

16 /
`√(337)` = 0.87 (round to the nearest hundredth)

Width = 60 * 0.87 = 52.2

52.2 / 16 = 3.26 (round to the nearest hundredth)

Height = 3.26 * 9 = 29.34

Checking:
`\sqrt{52.2^(2) + 29.34^(2) }` = 59.88 (round to the nearest hundredth)

It is close to 60, because of some rounding to the nearest hundredth