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.

enter image description here

  • $\begingroup$ The solution is very interesting, but I'm afraid I must VTC - this is simply a mathematics exercise, not a puzzle. Interesting, but not a puzzle. $\endgroup$
    – Brandon_J
    Commented May 16, 2019 at 3:10

3 Answers 3


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.
enter image description here
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.
enter image description here
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.
enter image description here
$(DB-r)^2 + DC^2 = r^2$ (because Center to B and Center to C are radius)
$DB^2 + DC^2 = 6^2$
=> $DB^2 - 2DBr + r^2 + DC^2 = r^2$
=> $6^2 - 2DBr = 0$
=> $DBr = 18$

  • $\begingroup$ Ok, I'm adding the formulas to make it more understandable. Sorry. $\endgroup$
    – Hermes
    Commented May 15, 2019 at 8:17
  • $\begingroup$ Your answer is also short and nice!!!(+1) $\endgroup$
    – 19aksh
    Commented May 15, 2019 at 8:41

enter image description here

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 $90^o$

enter image description here

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

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

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

  • $\begingroup$ This answer is really cool. $\endgroup$
    – Hermes
    Commented May 15, 2019 at 8:33
  • $\begingroup$ @Ak19 You're welcome! It's a brilliant answer! $\endgroup$
    – gaborsch
    Commented May 15, 2019 at 11:21
  • $\begingroup$ 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...) $\endgroup$ Commented 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.

enter image description here

  • 2
    $\begingroup$ Can you please elaborate on this? $\endgroup$ Commented May 15, 2019 at 9:12
  • $\begingroup$ @PrameshBajracharya I have extended the answer. $\endgroup$ Commented May 15, 2019 at 9:35
  • 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. $\endgroup$
    – Hermes
    Commented May 15, 2019 at 10:56
  • $\begingroup$ @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. $\endgroup$ Commented May 15, 2019 at 11:03
  • 2
    $\begingroup$ @Hermes after all this is a puzzle site, not a mathematics site! I rather like puzzles which apparently have insufficient information. $\endgroup$ Commented May 15, 2019 at 11:12

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