Question 1208776
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Part A


If you were to use that hint, then you'll have this diagram.
{{{
drawing(400,400,-4,4,-3,3,
line(-3,-2,3,-2),line(3,-2,3,2),line(3,2,-3,2),line(-3,2,-3,-2),line(-3,-2,3,2),
locate(-0.48,0.34,"37"),locate(-0.62,-2.22+0.15,"x"),locate(3.26-0.2,0.26,"(3/4)x"),
locate(-3.5-0.3,-2.6,matrix(1,4,"Diagram","not","to","scale"))
)
}}}
The diagonal splits the rectangle into two congruent right triangles.
Use the Pythagorean Theorem {{{a^2+b^2=c^2}}} to form this equation {{{x^2+( (3/4)x )^2 = 37^2}}} where a sub-goal is to solve for x.
I'll leave the scratch work for the student to do. 
You should find the two solutions to that equation are x = 148/5 and x = -148/5
Ignore the negative x value since we cannot have a negative width.


x = 148/5 = 29.6
(3/4)x = (3/4)*(148/5) = 111/5 = 22.2
For a 37 inch TV to have a 4:3 aspect ratio, the screen needs to be 29.6 inches across horizontally and 22.2 inches tall.
Note that (29.6)^2+(22.2)^2 = 37^2 due to the Pythagorean Theorem.
The area of this TV is therefore 29.6*22.2 = 657.12 square inches.


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Another approach for part A


The use of a fraction in (3/4)x is a bit clunky in my opinion. 
We can avoid that fraction by letting 4x be the horizontal portion of the screen and 3x be the vertical height, as shown here,
{{{
drawing(400,400,-4,4,-3,3,
line(-3,-2,3,-2),line(3,-2,3,2),line(3,2,-3,2),line(-3,2,-3,-2),line(-3,-2,3,2),
locate(-0.48,0.34,"37"),locate(-0.62,-2.22+0.15,4x),locate(3.26-0.2,0.26,3x),
locate(-3.5-0.3,-2.6,matrix(1,4,"Diagram","not","to","scale"))
)
}}}
Note the ratio 4x:3x reduces to 4:3


If you were to solve {{{(3x)^2+(4x)^2 = 37^2}}} you would get x = -37/5, x = 37/5 as the two solutions. We ignore the negative solution.


horizontal = 4x = 4*37/5 = 148/5 = 29.6 inches
vertical = 3x = 3*37/5 = 111/5 = 22.2 inches
area = 29.6*22.2 = 657.12 square inches.

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Part B


This is one possible diagram to draw, when having a 16:9 screen
{{{
drawing(400,400,-4,4,-3,3,
line(-3,-2,3,-2),line(3,-2,3,2),line(3,2,-3,2),line(-3,2,-3,-2),line(-3,-2,3,2),
locate(-0.48,0.34,"37"),locate(-0.62,-2.22+0.15,16x),locate(3.26-0.2,0.26,9x),
locate(-3.5-0.3,-2.6,matrix(1,4,"Diagram","not","to","scale"))
)
}}}
The length of the diagonal is the same as before.
The equation to solve would be {{{(16x)^2+(9x)^2 = 37^2}}}
The positive solution from this equation is x = 37*sqrt(337)/337
That would approximate to roughly x = 2.01551843
area = 16x*9x = 16*2.01551843*9*2.01551843 = 584.973294000432
This is roughly a 584.97 square inch screen depending how you round that decimal value.
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