Q:

Pls explain Pythagorean theorem Determine the missing measurements for each TV.32” TV height: 16” width: _____• _____ TV height: 34” width: 61”• 60” TV height: 30” width: _____• _____ TV height: 20” width: 35”• 52” TV height: _____ width: 45”If a TV has the following dimensions 48” wide, 27” height, and 55”diagonal. Respond to the following questions.If the TV is 5” wider and 3” higher, what is the new diagonal measurement? (Show work) If the TV is 3” wider and 5” higher, what is the new diagonal measurement? (Show work)

Accepted Solution

A:
Answer:All answers are to 2 decimal places for accuracy.1. Width is 27.71"2. TV Size is 69.84"3. Width is 51.96"4. TV Size is 40.13"5. Height is 26.06"6.Part 1: Diagonal is 61.91"Part 2: Diagonal is 60.21"Step-by-step explanation:THe pythagorean theorem tells us that the sum of two legs of a triangle squared equal the hypotenuse squared.Note: This applies for right triangles and the side opposite of the right angle is the hypotenuseAlso, to solve the first 5 problems, we can say:[tex]Height^2 + Width^2 = TV \ Size^2[/tex]#1  32” TV height: 16” width: _____Let width be w[tex]16^2 + w^2 = 32^2\\w =\sqrt{32^2 - 16^2} \\w=27.71[/tex]Width is 27.71"#2  _____ TV height: 34” width: 61”Let tv size be t[tex]34^2 + 61^2 = t^2\\t= \sqrt{34^2 + 61^2} \\t=69.84[/tex]TV Size is 69.84"#3  60” TV height: 30” width: _____Let width be w[tex]30^2 + w^2 = 60^2\\w =\sqrt{60^2 - 30^2} \\w=51.96[/tex]Width is 51.96"#4  _____ TV height: 20” width: 35”Let tv size be t[tex]20^2 + 35^2 = t^2\\t= \sqrt{20^2 + 35^2} \\t=40.31[/tex]TV Size is 40.13"#5  52” TV height: _____ width: 45”Let height be h[tex]h^2 + 45^2 = 52^2\\h =\sqrt{52^2 - 45^2} \\h=26.06[/tex]Height is 26.06"#6Part1:New dimensions would be48+5 = 53" Width27+3 = 30" HeightLet diagonal be d, so[tex]d=\sqrt{53^2+32^2} \\d=61.91[/tex]Diagonal is 61.91"Part 2:New dimensions are:48+3 = 51" Width27 + 5 = 32" HeightLet diagonal be d, so[tex]d=\sqrt{51^2+32^2} \\d=60.21[/tex]Diagonal is 60.21"