SOLUTION: I'm having some difficulty with the following problem. I'm stuck in the middle of this problem. Two television monitors sitting beside each other on a shelf in an appliance stor

Question 196580: I'm having some difficulty with the following problem. I'm stuck in the middle of this problem.
Two television monitors sitting beside each other on a shelf in an appliance store have the same screen height. One has a conventional screen, which is 6 inches wider than it is high. The other has a wider, high-definition screen, which is 1.8 times as wide as it is high. The diagonal measure of the wider screen is 10 inches more than the diagonal measure of the smaller. What is the height of the screens, correct to the nearest 0.1 inch?
thanks!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Two television monitors sitting beside each other on a shelf in an appliance store have the same screen height. One has a conventional screen, which is 6 inches wider than it is high. The other has a wider, high-definition screen, which is 1.8 times as wide as it is high. The diagonal measure of the wider screen is 10 inches more than the diagonal measure of the smaller. What is the height of the screens, correct to the nearest 0.1 inch?
thanks!
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Draw a picture of each of the screens.
The conventional screen has height "x"
The conventional screen has width = "x+6"
----
The conventional screen diagonal = sqrt[x^2 + (x+6)^2]
= sqrt[2x^2 + 12x + 36]
-------------------------------
The HD screen has height "x"
The HD screen has width = "1.8x"
---
The HD screen has diagonal = sqrt[x^2 + (1.8x)^2]
= sqrt[4.24x^2]
============================
Equation to solve for "x":
sqrt[4.24x^2] - sqrt[2x^2 + 12x + 36] = 10
----
Comment:
That is a mess to solve, so I graphed it and found
x = 22.4707
Rounding to 0.1 you get height = 22.5 inches
--
Cheers,
Stan H.

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