Geometry question: Find the area of blue-shared area inside this isosceles See below, looks a bit interesting, but I cannot find a solution.
I think a starting point might be the similarity of the lower white triangle and the larger triangle composed of lower blue, pink, and lower right triangle though..


Expanding-upon a comment, as requested.

Let the triangle have vertex angle $2\theta$, let $a\leq b$ be the sides of the squares, and let $t$ denote the target area. We immediately find
$$\begin{align}
a+a\cot\theta = |PQ| = b+b\tan2\theta \tag{1} \\
t = \frac12\cdot a\cdot a\cot\theta + \frac12\cdot b \cdot b\tan2\theta \tag{2}
\end{align}$$
From here, "all we have to do" is eliminate $\theta$ from $(1)$ and $(2)$. One way to do that here is first to treat the equations as a linear system in $\cot\theta$ and $\tan2\theta$; solving gives
$$\cot\theta = \frac{2t-b(a-b)}{a (a + b)} \qquad 
\tan2\theta = \frac{2t+a(a-b)}{b (a + b)} \tag{3}$$
Reciprocating $\cot\theta$ into $\tan\theta$, and substituting into the double-angle identity $\tan2\theta=2\tan\theta/(1-\tan^2\theta)$, and "simplifying" gives this relation among $t$, $a$, and $b$:
$$\begin{align}
0 
&= 8 t^3 \\
&+ 4 t^2 (a - b)(a - 2 b)  \\
&- 2 t\phantom{^2}(a^2 + b^2)(a^2 + 6 a b - b^2)  \\
&- \phantom{2}a (a - b) (a^4 + 2 a^3 b - 2 a^2 b^2 - 2 a b^3 - 3 b^4) 
\end{align} \tag{$\star$}$$
In the particular case of $a=7$ and $b=16$, $(\star)$ reduces into a factorable form and we have
$$(2t-339) (4t^2+1128t+49371) = 0 \quad\to\quad t = \frac{339}{2} \tag{$\star$'}$$
which gives our answer. $\square$

As an alternative to the above approach with $(3)$ we could re-write $(1)$  in terms of $\tan\theta$ alone:
$$(a-b)\tan^3\theta + (a+2b)\tan^2\theta-(a-b)\tan\theta-a=0 \tag{4}$$
For $a=7$ and $b=16$, this reduces to
$$(3\tan\theta-1) (3\tan^2\theta-12\tan\theta-7) =0 \tag{5}$$
Therefore (casually ignoring a positive root of the second factor),
$$\tan\theta = \frac13\qquad\to\qquad\tan2\theta=\frac{3}{4}\qquad\stackrel{\text{by (1)}}{\to}\qquad t = \frac{339}{2}\tag{6}$$
(See the edit history that also rewrites $(2)$ in terms of $\tan\theta$ alone before eliminating $\tan\theta$ from the system to get $(\star)$.)
An advantage of this approach is that $(4)$ is "more-immediate" to factor. Also, we see explicitly that $\tan\theta$ and $\tan2\theta$ are nice values. Knowing $\tan\theta=1/3$ facilitates drawing an accurate figure.
More importantly, $\tan2\theta=3/4$ reveals a non-obvious connection to the $3$-$4$-$5$ triangle. I can't help thinking that there's a clever way to detect and exploit it.

Yet-another way to proceed is to seek the distance $p:=|PQ|$.  (@Aretino's answer uses another approach to this.) Clearly,
$$\tan\theta = \frac{a}{p-a} \quad \tan2\theta=\frac{p-b}{b} \qquad\stackrel{\text{dbl ang}}{\to}\qquad p^3 - p^2(2 a + b) + 2 a^2 b  =0 \tag{7}$$
(The nicest cubic yet!) For $a=7$ and $b=16$, this reduces to
$$(p-28) (p^2-2p-56)=0 \quad\to\quad p = 28 \tag{8}$$
from which the result follows thusly:
$$t = \frac12a(p-a)+\frac12b(p-b) = \cdots = \frac{339}{2} \tag{9}$$
This solution seems like it might be the most accessible to middle-schoolers, although it, too, ignores the underlying $3$-$4$-$5$ triangle. (Perhaps I'm over-estimating the significance of that triangle!)
