Area Of Colored Squares

Let the red square have a side length of $x$ and the purple square have a side length of $y$ , so that the blue square has a side length of $x + y$ , and label the diagram as follows:

Since $QR$ is a diameter, $\angle QPR$ is a right angle, and $\triangle QSP \sim \triangle PSR$ by AA similarity.

This means that $\cfrac{QS}{PS} = \cfrac{PS}{SR}$ , or $\cfrac{y - x}{y} = \cfrac{y}{y + 2x}$ , which simplifies to $y = 2x$ for $x \neq 0$ .

Also, $QR = QT + TR = y + (x + y) = x + 2y = 5$ .

The two equations $y = 2x$ and $x + 2y = 5$ solve to $x = 1$ and $y = 2$ .

Therefore, the three squares have side lengths of $1$ , $2$ , and $1 + 2 = 3$ , and the sum of their areas is $1^2 + 2^2 + 3^2 = \boxed{14}$ .

$h^2 = x(5 - x) \implies h = \sqrt{x(5 - x)}$

$m = 5 - h \implies m - h = 5 - 2h$

$A_{pink} = h^2$

$A_{lightblue} = m^2 = (5 - h)^2$

$A_{red} = (m - h)^2 = (5 - 2h)^2$

$A_{white} = x(m - h) = x(5 - 2h)$

$\implies 5m = 5(5 - h) = h^2 + 25 - 10h + h^2 + 25 - 20h + 4h^2 + 5x - 2xh \implies$

$6h^2 + 5x + 25 = (2x + 25)h$

Using $h = \sqrt{x(5 - x)} \implies$

$-6x^2 + 35x + 25 = (2x + 25)\sqrt{x(5 - x)}$

Squaring both sides and simplifying we obtain:

$8x^4 - 68x^3 + 210x^2 - 275x + 125 = 0$

By inspection $x = 1$ is a root to the above equation and dividing $8x^4 - 68x^3 + 210x^2 - 275x + 125$ by $x - 1$ we have:

$(x - 1)(8x^3 - 60x^2 + 150x - 125) = 0 \implies (x - 1)(2x - 5)^3 = 0$

$\implies x = 1, x = \dfrac{5}{2}$

$x = 1 \implies h = 2 \implies m = 3 \implies m - h = 1$

$x = \dfrac{5}{2} \implies h = \dfrac{5}{2} \implies m = \dfrac{5}{2} \implies m - h = 0$

$\therefore$ drop $x = \dfrac{5}{2}$ and pick $x = 1$

$\therefore$ the desired area $A = 2^2 + 3^2 + 1 = \boxed{14}$ .

Drop $PH$ perpendicular to the diameter $AB$ .
Denote by $x$ and $y$ the side lengths of the light blue and red squares respectively and by $O$ the center of the semicircle.
Then, $EL=PH=AL=5-x$ . Moreover, $DE+EL=DL\Rightarrow y+\left( 5-x \right)=x\Rightarrow y=2x-5$ Since, $y>0$ we also have $2x-5>0\Leftrightarrow x>\frac{5}{2}$ . By Pythagoras’ theorem on the right-angled $\triangle OPH$ , $\begin{aligned} P{{H}^{2}}+H{{O}^{2}}=P{{O}^{2}} & \Rightarrow {{\left( 5-x \right)}^{2}}+{{\left( y+x-\frac{5}{2} \right)}^{2}}={{\left( \frac{5}{2} \right)}^{2}} \\ & \overset{(1)}{\mathop{\Leftrightarrow }}\,{{\left( 5-x \right)}^{2}}+{{\left( 2x-5+x-\frac{5}{2} \right)}^{2}}=\frac{25}{4} \\ & \Leftrightarrow 2{{x}^{2}}-11x+15=0 \\ & \Leftrightarrow x=\frac{5}{2}\text{ or }x=3 \\ \end{aligned}$ Due to the restriction, the value $\frac{5}{2}$ is rejected. Hence $x=3$ and this gives $y=1$ .
Thus, the total area of the three squares is $A={{x}^{2}}+{{y}^{2}}+{{\left( 5-x \right)}^{2}}={{3}^{2}}+{{1}^{2}}+{{2}^{2}}=\boxed{14}.$

We have squares of three colors: pink, red, and cyan. Let's refer to the side lengths of each square as side lengths $p$ (pink square), $r$ (red square), and $c$ (cyan square).

The following statements are what we know to be true about the figure, where the $x$ and $y$ values refer to the horizontal and vertical side lengths, respectively.

$x_p + x_c = 5$

$x_r = \frac{1}{2} x_p$

$y_p + y_r = y_c$

We know, though, that the $x$ and $y$ side lengths for each respective colored square are equal. So we can now refer to, for example, $\bold{x_p}$ and $\bold{y_p}$ , as just $\bold{p}$ .

With this being true, the equations above can be written as:

(1) … $p + c = 5$

(2) … $r = \frac{1}{2} p$ $\rightarrow 2r=p$

(3) … $p + r = c$

These are much simpler to substitute into one another to solve for variables.

$\space$

$\large{\text{Solving equations (1), (2), and (3) for variables}}$

Let's use (2) to solve (3) for $c$ :

$2r + r = c \rightarrow c=3r$

Now that we know that $c=3r$ and $p=2r$ , we can solve (1) in terms of $r$ .

$2r + 3r = 5$

$\rightarrow 5r = 5$

$\rightarrow r=1$

So, now that we know that $r=1$ , we also know that:

From (3), where $p=2r$ , 2(1) + (1) = c)

$\rightarrow c = 3$

$p + 3 = 5 \rightarrow p=2$

Let the side lengths of the purple, blue, and red squares be $a$ , $b$ , and $c$ . We note that because $QR$ is a diameter, $\angle QPR = 90^\circ$ . Then

$\begin{aligned} \blue{PR^2} + \red{PQ^2} & = QR^2 \\ \blue{TP^2+TQ^2} + \red{PS^2+SR^2} & = 25 \\ \blue{(a-c)^2+a^2} + \red{a^2+(c+b)^2} & = 25 & \small \blue{\text{Note that }b=5-a \text{ and}} \\ (3a-5)^2 + a^2 + a^2 + (10-3a)^2 & = 25 & \small \blue{a+c = b \implies c = 5-2a} \\ 9a^2 - 30a + 25 + 2a^2 + 9a^2 - 60a + 100 & = 25 \\ 20a^2 - 90a + 100 & = 0 \\ 2a^2 - 9a + 10 & = 0 \\ (2a-5)(a-2) & = 0 & \small \blue{\text{Since }a < 2.5} \\ \implies a & = 2 \end{aligned}$

Therefore $b = 5-a = 3$ and $c = 5-2a = 1$ and the area of the three squares is $1^2 + 2^2 + 3^2 = \boxed{14}$

Let red be x, purple be y and blue be z size squares. Easy to show,

z=x+y, z+y=5 and y/(y-x)=(z+y-x)/y

Therefore, x=1, y=2 and z=3, giving answer as 14.

Let B = length of the side of the blue square, R = length of a side of the red square, and P = length of the side of the purple square. From the picture, we find:

B + P = 5

P + R = B

P + (P-R) = B

This gives three equations in three unknowns. Solving yields R = 1, B = 3, and P = 2. Hence, the sum of the areas is:

1^2 + 2^2 + 3^2 = 14