Greatest area of a Quadrilateral, given constant diagonal length 10 inches

[Leead Moyal]

Hello, I am wondering, given a constant diagonal, let's say 10 inches, will the area of a Quadrilateral be greater shaped as a square or a rectangle? I believe it will always be a square, when the angle is tan45°.
This question came across my mind as a result of smartphone screen sizes as they are calculated by the size of the diagonal, which isn't the max size, instead of square inch area. I know the screens are, until the current generation of smartphone s, aspect ratio is 16 by 9.
Thxs alot

 

[Deleted member 4993]

Hello, I am wondering, given a constant diagonal, let's say 10 inches, will the area of a Quadrilateral be greater shaped as a square or a rectangle? I believe it will always be a square, when the angle is tan45°.
This question came across my mind as a result of smartphone screen sizes as they are calculated by the size of the diagonal, which isn't the max size, instead of square inch area. I know the screens are, until the current generation of smartphone s, aspect ratio is 16 by 9.
Thxs alot

Are you talking about a general quadrilateral or a rectangle (a special type of quadrilateral)?

 

[stapel]

Hello, I am wondering, given a constant diagonal, let's say 10 inches, will the area of a Quadrilateral be greater shaped as a square or a rectangle? I believe it will always be a square, when the angle is tan45°.

This question came across my mind as a result of smartphone screen sizes as they are calculated by the size of the diagonal, which isn't the max size, instead of square inch area. I know the screens are, until the current generation of smartphone s, aspect ratio is 16 by 9.

I'll assume, from the smartphone discussion, that you're talking only about squares and rectangles, rather than any of the other various sorts of quadrilaterals. So your question is:

Find the rectangle (of which the square is a special case) having the largest area, given that the diagonal is ten units.

So set up the geometry and algebra, and solve. You've got the unknown length L and width w, and the Pythagorean Theorem which tells you that L^2 + w^2 = (10)^2 = 100. Solve this for either one of the variables, in terms of the other. Then plug into the formula for the area A in terms of the variable and the expression, and find the maximizing input values. (You can do this with either of calculus or algebra.)

If you get stuck, please reply showing all of your steps so far. Thank you!

 

[Leead Moyal]

Thanks for your response

I'll assume, from the smartphone discussion, that you're talking only about squares and rectangles, rather than any of the other various sorts of quadrilaterals. So your question is:

Find the rectangle (of which the square is a special case) having the largest area, given that the diagonal is ten units.

So set up the geometry and algebra, and solve. You've got the unknown length L and width w, and the Pythagorean Theorem which tells you that L^2 + w^2 = (10)^2 = 100. Solve this for either one of the variables, in terms of the other. Then plug into the formula for the area A in terms of the variable and the expression, and find the maximizing input values. (You can do this with either of calculus or algebra.)

If you get stuck, please reply showing all of your steps so far. Thank you!

You were the closest to understand my unclear question .
Let me retort, for any square or rectangle, with the same diagonal length, will the square always have a larger area? If so, is that correct for all quadrilaterals (The square will always have the largest area given the identical diagonal length)?

 

[stapel]

You were the closest to understand my unclear question .
Let me ask in reply: For any square or rectangle, with the same diagonal length, will the square always have a larger area?

Form the generic "area" function, differentiate, and find the maximum. You'll find that the answer is "yes" (by also remembering a little trig, maybe).

If so, is that correct for all quadrilaterals (The square will always have the largest area given the identical diagonal length)?

My guess is "yes", but the proof would likely be much more complex, since we wouldn't be able to assume right angles (goodbye, Pythagorus) or convexity (goodbye, shapes without divots).