KCC's Quizzes AQQ261 about a rectangle area

The area of the small rectangle is y*(y-8)

y*(y-8) = 9

y^2 -8y = 9

y^2 -8y -9 = 0

Factoring:

(y - 9)(y + 1) = 0

So y has to be 9 or -1, but in this case has to be the positive integer

y + 2 = 11 so the length of side AD, x = 12

The length of side AB = x - 4 = 8

So the area of the big rectangle ABCD = 8*12 = 96

The area of the small rectangle is y*(y-8)

y*(y-8) = 9

y^2 -8y = 9

y^2 -8y -9 = 0

Factoring:

(y - 9)(y + 1) = 0

So y has to be 9 or -1, but in this case has to be the positive integer

y + 2 = 11 so the length of side AD, x = 12

The length of side AB = x - 4 = 8

So the area of the big rectangle ABCD = 8*12 = 96

Fantastic Martin! You are, again, the first one, congratulations!

Hi Martin, I was too fast when granting your answer! 96 cannot be the area!

Look: you obtained y=9 while segment AB is 8: that's visualy impossible...

What a paradox, isn't it?

Yes, x (length of BC) = 2y - 6 (length of AD) if this is a rectangle.  This leads to x = 12.  However, length of AB = x - 4 = 8 but this cannot be the case if y = 9. 

One of the conditions must be wrong or ABCD is not a rectangle.

If the length of CD was y instead, then the rectangle would be 9 * 12 = 108 but the values of x would not make sense in this case.

Yeah! In fact the equations used to computed the area are in fact valid! By staying in a pure math world, that's fine, but not in our physical world negative length and area have no sense. 

What a amazing quiz, isn't it?

What is interesting is that the solution of the quadratic equation for y = -1, 9. Substituting it into x = 2*y-6, then x = -8, 12. Substituting either the negative or positive results into the area = x^2 - 4*x results in 96 for both cases.

Yeah, DennisFeucht ! Mathematics offer much more possibilities in both real and virtual worlds!