Find the area of the rectangle

Question

The image below shows a half circle, and a rectange DBFE. Your task is simply to calculate the area of the rectangle, based on the information given in the image.

Answer 1

If C = E then the area is 18, since both will be at the maximum point of the semicircle and, therefore, 6 is the hypotenuse of a square.

If C = A then the area is 18, since then 6 is the long side of the rectangle, and the diameter of the semicircle, and the smaller side of the rectangle will be the radius of the semicircle, that is, 3.

If C belongs to the arc between A and the maximum point of the semicircle, it will take values between those, so must be 18.

Another way, with more geometric, then:

The triangle D-C-Center of the circumference is a right triangle whose hypotenuse is the radius.
DB minus radius squared plus DC squared is, therefore, the radius squared. Since DB squared plus DC squared is 6 squared, then DB multiplied by radius is 18. And since the small side of the rectangle is the radius, then the area is 18.

Formulas:
$(DB-r)^2 + DC^2 = r^2$ (because Center to B and Center to C are radius)
$DB^2 + DC^2 = 6^2$
=> $DB^2 - 2DBr + r^2 + DC^2 = r^2$
=> $6^2 - 2DBr = 0$
=> $DBr = 18$

Answer 2

Let $DB = a$, $AD = b$. So, $CD = \sqrt{ab}$ $$$$ By Pythagoras theorem, $$$$ $6^2 = a^2 + ab = a(a+b)$ $$$$ Also, radius $r = \frac{a+b}{2}$ = breadth of the rectangle. $$$$ So, Area$$A= a(a+b)/2 = 36/2 = 18$$

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PROOF for $CD = \sqrt{ab}$ $$$$ In a semicircle the angle touching the circle at any point from the two ends of the diameter is $90^o$

$$AC^2 = b^2 +c ^2$$ $$6^2 + AC^2 = AB^2$$ $$6^2 + b^2 +c ^2 = (a+b)^2$$

Also, $6^2 = a^2 + c^2 $

So, $$a^2 + c^2 + b^2 + c^2 = (a+b)^2 = a^2 +b^2 +2ab$$ $$c^2 = ab$$ $$c = \sqrt{ab}$$

Answer 3

My answer:

As the diameter $d$ of the semicircle is not given,
the answer must be the same for all $d \geq 6$.

This includes $d = 6$, when the area of the rectangle will be

$ \frac{d^2}{2} = 18$

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More explanation as requested:

The question says "based on the information given in the image" where the only numerical information is $6$ the length of the line. Clearly the circle can have a larger diameter than the one shown and there will be a solution for a line length $6$.

Also for smaller ones, the smallest of which has diameter $6$, in which case the line is on the horizontal diameter. Here, the rectangle exactly encloses the semicircle, and therefore has dimensions $6 \times 3$.

As no information was given as to the diameter, the answer (if there is an answer) must be $18$ for all possible semicircles, including the special case.