A business recently upgraded a conference TV they had in their conference room. Originally the TV was 32 inches (across the diagonal) and had a perimeter of 87.16 inches. The new TV is 70 inches across the diagonal. Since the new TV will be recessed in the wall, it will need trim put on the border. What is the perimeter of the new TV? Round your answer to two decimal places.

Answer:

The perimeter of the new TV is approximately  95.32 inches

Step-by-step explanation:

The diagonal dimension of the original TV = 32 inches

The perimeter of the original TV = 87.16 inches

The diagonal dimension of the new TV = 70 inches

The perimeter of the new TV = 87.16 inches

For the original TV, we have;

Let w be the width of the original TV and l be the length of the original TV, we have;

2·w + 2·l = 87.16

√(w² + l²) = 32

Therefore, we have;

32² = w² + l²

w = (87.16 - 2·l)/2

w = 43.58 - l

32² = l² + (43.58 - l)²

l² + (43.58 - l)² - 32² = 0

l² + 43.58² - 2 × 43.85 × l + l² - 32² = 0

2·l² - 87.16·l + 875.2164 = 0

By the quadratic formula, we have;

l = (87.16 ± √((-87.16)² - 4×2 × 875.2164))/(2 × 2)

Therefore, l = 27.89 inches or l = 15.69

Therefore, w = (87.16 - 2·l)/2 = (87.16 - 2× 27.89)/2 = 15.69

or w = (87.16 - 2× 15.69)/2 = 27.89

Whereby the ratio of the width to the length of the original TV and the new TV are the same, we have;

15.69/27.89

w₂/l₂ for the new TV = 15.69/27.89

∴ w₂ = l₂ × 15.69/27.89

w₂² + l₂² = 70²

∴ (l₂ × 15.69/27.89)² + l₂² = 70²

l₂² × (15.69/27.89)² + l₂² = 70²

l₂²((15.69/27.89)² + 1) = 70²

l₂² = 70²/((15.69/27.89)² + 1) = 3722.04

l₂ = √(3722.04) ≈ 61.0

w₂ = l₂ × 15.69/27.89 = 61 × 15.69/27.89 ≈ 34.32

Therefore;

The perimeter of the new TV = 34.32 + 61 = 95.32 inches.

In this exercise we have to use the knowledge and information given to calculate the perimeter that will correspond to:

The perimeter of the new TV is approximately  95.32 inches

So the data for this exercise is from:

Let w be the width of the original TV and l be the length of the original TV, we have;

[tex]2*(w) + 2*(l) = 87.16\\\sqrt{(w^2 + l^2)} = 32[/tex]

Therefore, we have:

[tex]32^2 = w^2 + l^2\\w = (87.16 - 2*l)/2\\w = 43.58 - l\\32^2 = l^2 + (43.58 - l)^2\\l^2 + (43.58 - l)^2 - 32^2 = 0\\l^2 + 43.58^2 - (2 * 43.85 * l)+ l^2 - 32^2 = 0\\(2)*(l)^2 - 87.16*l + 875.2164 = 0[/tex]

By the quadratic formula, we have;

[tex]l = (87.16 +/- \sqrt{((-87.16)^2 - 4*2 * 875.2164))/(2 * 2)} )[/tex]

Therefore:

[tex]w = (87.16 - (2)*(l))/2 = (87.16 - 2* 27.89)/2 = 15.69\\w = (87.16 - 2* 15.69)/2 = 27.89[/tex]

Whereby the ratio of the width to the length of the original TV and the new TV are the same, we have;

w₂/l₂ for the new TV = [tex]15.69/27.89[/tex]

[tex]w_2 = (l_2)* (15.69/27.89)\\w_2^2 + l_2^2 = 70^2\\(l_2 *15.69/27.89)^2 + l^2_2 = 70^2\\l^2_2 * (15.69/27.89)^2 + l^2_2 = 70^2\\l^2_2((15.69/27.89)^2 + 1) = 70^2\\l^2_2 = 70^2/((15.69/27.89)^2 + 1) = 3722.04\\l_2 = \sqrt{(3722.04)} = 61.0\\w_2= l_2 * 15.69/27.89 = 61 * 15.69/27.89 = 34.32\\[/tex]  

The perimeter of the new TV :

[tex]P = 34.32 + 61 = 95.32[/tex]

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