# help with measurements?

The diagonal front face of the box is 31.11"
Tom Baker
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One last try...
Start with a square 34" on each side. Choose one corner and measure along the side 16" and make a mark. From the same corner measure 16" along the other side and mark it. Connect the two marks forming a 'diagonal' line. Its length is 22 5/8". This should satisfy all the conditions set down by the OP. Does this solve her problem?
TW
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TinWoodsmn wrote...

(g)
Well, the problem was that the 22" she gave was wrong. She actually had counted 22 diagonals, but they are 1.414" each, so as Tom noted,

Cheers!
Jim
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It isn't the diagonal of 31.11" which is the governing number, it's the TV. To enclose it we need a trapezoid with 25" (front of TV) and 19" (rear of TV) parallel sides with a height of at least 19" (length of TV). In addition, the overall container can have one side 34" long maximum (door jamb). Again start with a 34" square. Proceed as in my last post, except change the 16" dimension to 17 3/4". When completed, the box has a 25" wide face for the TV screen, two rear adjacent 34" sides and will contain a TV with a 25" wide face, 19" wide rear and about 26" long.
TW
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TinWoodsmn wrote...

But she wasn't trying to design the cabinet anymore by then. She already had the shape she liked. She was wondering how the TV was able to fit in the scale mock-up when the front dimension of the cabinet seemed smaller than the TV.
Jim