Rhombus Riddle

Let the minor diagonal be (2 m) then the major diagonal is (4 m). It follows that $\tan \angle DBF = 2$ and therefore, $\tan \angle BDF = \frac{1}{2}$ . From this, we deduce that $\cos^2 \angle BDF = \frac{4}{5}$ and that $\sin^2 \angle BDF = \frac{1}{5}$ . As mentioned above the minor diagonal $BD = 2 m$ . Therefore, $BF = \frac{2}{\sqrt{5}} m$ , and $DF = \dfrac{4}{\sqrt{5}} m$ . Thus, $24 = \dfrac{8}{5} m^2$ , from which, we have $m^2 = 15$ . Finally the area of the rhombus is $\frac{1}{2} (2 m)(4 m) = 4 m^2 = 60$ . Thus the yellow area is $60 - 24 = 36$ .

Suppose the length $BD$ = $x$ . Since $AC = 2\times BD$ , its area equals $\dfrac{1}{2}(2x)(x) = x^2$ . Then we can draw a square, depicted as red, of side length $x$ , which will have the same area as the rhombus, as shown below:

Now let $BE$ = $h$ . Since $BEDF$ is a rectangle, we can conclude that $24 = h\times ED$ . Thus, $ED = \dfrac{24}{h}$ .

Also, by Pythagorean theorem, regarding the quarter congruent triangle of the rhombus, the squared side length of the rhombus equals $x^2 + (\dfrac{x}{2})^2 = \dfrac{5}{4}(x^2)$ . The rhombus side length is $x\sqrt{\dfrac{5}{4}}$ .

Then considering the $\triangle ABD$ , there are two ways to calculate its area, depending on the either base $AD$ or $BD$ used. We can set equation of such area:

$\dfrac{1}{2}(h)(AD) = \dfrac{1}{2}(\dfrac{AC}{2})(BD) = \dfrac{1}{2}(x^2)$

Substituting $AD = x\sqrt{\dfrac{5}{4}}$ obtained earlier, we will get:

$hx\sqrt{\dfrac{5}{4}} = x^2$

$h = x\sqrt{\dfrac{4}{5}}$

Furthermore, using Pythagorean theorem again, we can set up the equation regarding the side lengths, regarding $\triangle BED$ :

$h^2 + (\dfrac{24}{h})^2 = x^2$

Substituting $h = x\sqrt{\dfrac{4}{5}}$ obtained before, we will get:

$\dfrac{4}{5} x^2 + \dfrac{24^2}{\dfrac{4x^2}{5}} = x^2$

$\dfrac{5\times 24^2}{4x^2} = \dfrac{x^2}{5}$

$24^2 = (\dfrac{2x^2}{5})^2$

$24 = \dfrac{2x^2}{5}$

$x^2 = 60$

Therefore, the rhombus has an area of $60$ . Thus, the yellow region has an area of $60-24 = \boxed{36}$ .

Alternatively, the rectangle has an area of $\dfrac{2}{5}$ of the rhombus, thus leaving the rest of the yellow region for $\dfrac{3}{5}$ . That is the area ratio between the blue to the yellow is $2:3$ .

Finally, the yellow region has an area of $\dfrac{3}{2}\times 24 = \boxed{36}$ .

Let the center of the rhombus be $O$ . Then we note that $\triangle BDF$ and $\triangle BOC$ are similar. Given that the major diagonal is twice the length of the minor diagonal, $\implies \dfrac {BO}{CO} = \dfrac {BF}{DF} = \dfrac 12$ . Let $BF=a$ . Then $DF=2a$ . Since the area of the blue rectangle $BEDF$ is $BF \times DF = 2a^2 = 24$ $\implies a = 2 \sqrt 3$ . By Pythagorean theorem we have $BD^2 = BF^2 + DF^2 = 5a^2$ $\implies BD = 2\sqrt{15}$ , $BO = \frac {BD}2 = \sqrt {15}$ , $AC = 4\sqrt{15}$ , the area of the rhombus $2 \times \frac 12 \times BO \times AC = 60$ , and the area of the yellow region $60-24 = \boxed {36}$ .