PR is a diameter of the large circle, centre O. QS is a diameter of the small circle, centre O. PQ and RS are tangents to the small circle.

1. Show both triangles are right-angled

Since $PQ$ and $RS$ are tangents to the small circle, and $OQ$ and $OS$ are radii to the points of contact,
$\angle OQP=90^\circ,\qquad \angle OSR=90^\circ.$
Hence,
$\angle OQP=\angle OSR.$

2. Identify one pair of equal corresponding sides from the large circle

$OP=OR$ because they are radii of the large circle.

3. Identify another pair of equal corresponding sides and state the congruence condition

$OQ=OS$ because they are radii of the small circle.

Therefore, $\triangle OPQ \cong \triangle ORS$ by RHS.

1. Represent the relevant lengths using the given ratio

Let
$OQ: PQ: OP = 5:12:13.$
So take
$OQ=5x,\quad PQ=12x,\quad OP=13x.$
Since $OP$ is a radius of the large circle, the radius of the large circle is $13x$.

2. Find the total shaded area of the two congruent right triangles

Since $OQ \perp PQ$,
$\text{Area of }\triangle OPQ=\frac12(5x)(12x)=30x^2.$
From part (a), $\triangle OPQ \cong \triangle ORS$, so the shaded area is
$2(30x^2)=60x^2.$

3. Express the shaded area as a percentage of the large circle

Area of the large circle:
$\pi(13x)^2=169\pi x^2.$
Percentage shaded:
$\frac{60x^2}{169\pi x^2}\times 100\%=\frac{6000}{169\pi}\%$
$=11.3\%\text{, to 3 s.f.}$