# Find the area of the rectangle

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. • 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. – Brandon_J May 16 '19 at 3:10

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)
$$DB^2 + DC^2 = 6^2$$
=> $$DB^2 - 2DBr + r^2 + DC^2 = r^2$$
=> $$6^2 - 2DBr = 0$$
=> $$DBr = 18$$

• Ok, I'm adding the formulas to make it more understandable. Sorry. – Hermes May 15 '19 at 8:17
• Your answer is also short and nice!!!(+1) – Ak. May 15 '19 at 8:41 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, $$a^2 + c^2 + b^2 + c^2 = (a+b)^2 = a^2 +b^2 +2ab$$ $$c^2 = ab$$ $$c = \sqrt{ab}$$

• This answer is really cool. – Hermes May 15 '19 at 8:33
• @Ak19 You're welcome! It's a brilliant answer! – gaborsch May 15 '19 at 11:21
• 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...) – Fabio says Reinstate Monica May 16 '19 at 0:07

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. • Can you please elaborate on this? – Pramesh Bajracharya May 15 '19 at 9:12
• @PrameshBajracharya I have extended the answer. – Weather Vane May 15 '19 at 9:35
• 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. – Hermes May 15 '19 at 10:56
• @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. – Weather Vane May 15 '19 at 11:03
• @Hermes after all this is a puzzle site, not a mathematics site! I rather like puzzles which apparently have insufficient information. – Weather Vane May 15 '19 at 11:12