What is the greatest possible area for a rhombus constructed on an 8 1/2 -by-11- inch sheet of paper?

Answer 1

It depends how you draw out the rhombus, but what ever area it covers, It will be less then the area of the page.

Answer 2

The largest rhombus has the two opposite vertices of the 8.5 x 11 rectangle as two of its vertices. We'll call these vertices A and C.

The other two vertices, which we'll call B and D, are on each of the 11 inch sides. To find where B and D are (note the length of AB must be the length of BC for it to be a rhombus):

Let x be the length of a side, i.e. the length of AB and BC. Then, we have a right triangle outside the rhombus but inside the rectangle. The short sides of the right triangle are 8.5 and (11-x), while the hypotenus is x. So, using the Pythagorean Theorem:

(11-x)^2 + 8.5^2 = x^2

Solve this, and you get x = 773/88.

The area of the rhombus, then, is 773/88 * 8.5 = 74.66. Note that this rhombus has a larger area than the square, since its base is larger than 8.5.

Answer 3

Let corners of your sheet be A,B,C,D. AB=8.5, BC=11;
Assuming that longer diagonal of rhombus in question is AC, let’s divide AC by 2 with point E. Assume EF is normal to AC, F being on line BC. Thus EA & EF are half-diagonals of our rhombus. If F’ is a mirror point of F (belonging to AD), then AFCF’ is the biggest rhombus.
Can anybody prove the contrary? I hope you calculate triangles ABF & CDF’ to be cut off, but you did not ask me to do it.

Answer 4

8.5 inches times 8.5 inches. Its a rhombus with angles of 90 degrees.

There is actually no other way to answer this question unless you define a square as a rhombus.

bah I suck, the other answers are right.

Answer 5

If one is to consider that a rhombus is a 'special' square, (look up the definitions of both) the answer is 72.25 sq ins - a square of 8.5 ins