Rectangles in rectangle

Let heights of top row to the bottom row be $a$ , $b$ , $c$ and $d$ respectively and the unknown area of yellow rectangle be $x$ . Since rectangle in the same column have the same width, their areas are directly proportional to their heights. For example, in the first column, rectangle in the second and third rows $\frac bc = \frac {12}8$ . The same for other columns. Therefore, we have:

$\begin{cases} \dfrac bc =\dfrac {12}{8} = \dfrac 32 &…(1) \\ \dfrac ad =\dfrac {20}{25} = \dfrac 45 &…(2) \\ \dfrac ac =\dfrac {14}{15} &…(3) \\ \dfrac bd =\dfrac {x}{21} &…(4) \end{cases}$

From (4):

$\begin{aligned} \dfrac bd & =\dfrac {x}{21} \\ \implies x &= 21 \times \frac bd \\ &= 21 \times \frac bc \times \frac ad \times \frac ca \\ &= 21 \times \frac 32 \times \frac 45 \times \frac {15}{14} \\ &=\boxed{27} \end{aligned}$

Let $x_1,x_2,x_3,x_4$ and $y_1,y_2,y_3,y_4$ the respective horizontal and vertical lengths of the rectangles. As an example, the red square has area $x_2*y_1$ . The following holds :

$\begin{array}{lll} x_1*y_2 &=&12\\ x_1*y_3 &=&8\\ x_2*y_1 &=&20\\ x_2*y_4 &=& 25\\ x_3*y_1 &=& 14\\ x_3*y_3 &=& 15\\ x_4*y_4 &=& 21\\ x_4*y_2 &=& ? \end{array}$

The game is now to multiply and divide terms to get to $x_4*y_2$ . The only one with $x_4$ is the orange square so we start with $x_4*y_4$ and remove $y_4$ by dividing by the 4th row. Then, we repeat the process.

There is only one possibility (life is beautiful), leading to $x_4*y_2$ : $\begin{array}{lllllllllllllllll} (x_4*y_4) &*& (x_2 * y_4)^{-1} &*& (x_2 * y_1) &*& (x_3 * y_1) ^ {-1} &*& (x_3*y_3) &*& (x_1*y_3)^{-1} &*& (x_1*y_2) &=&x_4*y_2\\ 21 & / & 25 & * & 20 & / & 14 & * & 15 & / & 8 & * & 12 & = & 27 \end{array}$

Considering the two rectangles with known areas in the topmost row, they have areas in the ratio $\frac{20}{14}=\frac{10}{7}$ . In other words, the rectangle immediately to the right of the one with area 25 has area $\frac{25}{\frac{10}{7}}=\frac{5}{\frac{2}{7}}=5 \times \frac{7}{2}=\frac{5\times7}{2}=\frac{35}{2}$ .

Next considering the two rectangles with areas 8 and 15, the rectangle immediately below the one with area 8 has area $\frac{8}{15} \times \frac{35}{2}=\frac{4}{3} \times 7=\frac{4\times7}{3}=\frac{28}{3}$ .

Finally considering this rectangle with area $\frac{28}{3}$ and the one with area 21, they have areas in the ratio $\frac{\frac{28}{3}}{21}=\frac{\frac{4}{3}}{3}=\frac{4}{3\times 3}=\frac{4}{9}$ . Then considering the rectangle with area 12, the unknown area would be $\frac{12}{\frac{4}{9}}=12 \times \frac{9}{4} = \frac{12}{4} \times 9 =3 \times 9 =\boxed{27}$

@Chew-Seong Cheong provided a solution based on the height of rows. An entirely analogous solution can be given based on the width of columns. Let $p, q, r, s$ be the widths for the columns, from left to right, and $x$ the unknown area. Then $\begin{cases} \dfrac s q = \dfrac{21}{25} \\ \dfrac q r = \dfrac{20}{14} = \dfrac {10}7 \\ \dfrac r p = \dfrac{15}{8} \\ \dfrac s p = \dfrac{x}{12}. \end{cases}$ Since $\frac s p = \frac s q \cdot \frac q r \cdot \frac r p,$ we have $\frac{x}{12} = \frac{21}{25}\cdot\frac{10}{7}\cdot\frac{15}8 = \frac 9 4,$ so that $x = 12\cdot \frac 9 4 = 3\cdot 9 = \boxed{27}.$

Looking at column 2: C2 * R1 = 20 C2 * R4 = 25 So that's easy, the common factor is 5, so: C2 = 5 R1 = 4 R4 = 5 The rest is inevitable! C4 = 21 ÷ 5 = 4 1/5 (or 4.2) C3 = 14 ÷ 4 = 3 1/2 (or 3.5) R3 = 15 ÷ 3 1/2 = 4 2/7 (which is why I prefer to work with fractions!) C1 = 8 ÷ 4 2/7 = 28/15 = 1 13/15 (just less than two, sounds about right!) R2 = 12 ÷ 28/15 = 45/7 = 6 3/7 So the area of our mystery box will be: C4 * R2 = 4 1/5 * 6 3/7 = 21/5 * 45/7 - which reduces very nicely! 3/1 * 9/1 = 3 * 9 = 27 Q.E.D. (all of which was very nicely formatted until I hit the "Publish" button!)

This one is really challenging to do mentally. So, we can apply a little trick!

$\frac{a}{b}=\frac{A}{B}$

All the bold edge is doing is multiplying both a and b by the same constant.

This was the toughest part to get. Now, we will build up the calculation using that diagram:

$20\times\frac{15}{14}\times\frac{12}{8}\times\frac{21}{25}=x$

We can simplify a bit :

$20\times\frac{15}{14}\times\frac{3}{2}\times\frac{21}{25}=x$

Then further :

$20\times\frac{15}{2}\times\frac{3}{2}\times\frac{3}{25}=x$

And again

$20\times\frac{3}{2}\times\frac{3}{2}\times\frac{3}{5}=x$ $4\times\frac{3}{2}\times\frac{3}{2}\times 3=x$ $3\times3\times 3=x$

And finally: $27=x$

So, we have found our target: 27 .

Let X1,X2,X3,X4 and Y1,Y2,Y3,Y4 the respective horizontal and vertical lengths of the rectangles

Here is a shorter solution We have

21 * 20 * 12 * 15 = product of all small edges = X1 * X2 * X3 * X4 * Y1 * Y2 * Y3 * Y4

We also have 12 * 8 * 20 * 25 * 14 * 15 * 21 * orange area = (product of all small edges) ^2 Therefore Orange Area = (21 * 20 * 12 * 15)^2 / (12 * 8 * 20 * 25 * 14 * 15 * 21) = 27

Area of rectangle above yellow.
=(21/25)(20)=84/5,
Area of White rectangle below yellow.
=(15/14)(84/5)=18,
Area of yellow=(18/8)(12)=27