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.
3 Answers
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)
$DB^2 + DC^2 = 6^2$
=> $DB^2 - 2DBr + r^2 + DC^2 = r^2$
=> $6^2 - 2DBr = 0$
=> $DBr = 18$
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$\begingroup$ Ok, I'm adding the formulas to make it more understandable. Sorry. $\endgroup$– HermesMay 15, 2019 at 8:17
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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$$
PROOF for $CD = \sqrt{ab}$ $$$$ In a semicircle the angle touching the circle at any point from the two ends of the diameter is $90^o$
$$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}$$
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$\begingroup$ @Ak19 You're welcome! It's a brilliant answer! $\endgroup$– gaborschMay 15, 2019 at 11:21
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$\begingroup$ The proof you provide at the end is known as rot13(Trbzrgevp zrna gurberz) (hiding its name is probably excessive, but it could still be a hint, so...) $\endgroup$ May 16, 2019 at 0:07
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$
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.
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2$\begingroup$ Can you please elaborate on this? $\endgroup$ May 15, 2019 at 9:12
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$\begingroup$ @PrameshBajracharya I have extended the answer. $\endgroup$ May 15, 2019 at 9:35
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1$\begingroup$ Could not the answer be such that it depended on the radio? It would still be a valid answer, and would continue to fit the special cases, but it would not always be exactly the value of any of the special cases. $\endgroup$– HermesMay 15, 2019 at 10:56
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$\begingroup$ @Hermes if that were true, then the question did not contain enough information to answer it. The algebraic solution has already been posted, where the radius $r$ cancels out. $\endgroup$ May 15, 2019 at 11:03
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2$\begingroup$ @Hermes after all this is a puzzle site, not a mathematics site! I rather like puzzles which apparently have insufficient information. $\endgroup$ May 15, 2019 at 11:12