The diagram shows two concentric circles with centre O. PQ is the diameter of the larger circle and RS is the diameter of the smaller circle. PS and RQ are tangents to the smaller circle.

1. State the equal right angles using the tangent-radius property.

$PS$ is tangent to the smaller circle at $S$, so $OS \perp PS$.

$RQ$ is tangent to the smaller circle at $R$, so $OR \perp RQ$.

Hence
$\angle PSO=\angle QRO=90^\circ.$

2. State the equal corresponding sides from radii of the same circles.

$OP=OQ$ since they are radii of the larger circle.

$OS=OR$ since they are radii of the smaller circle.

3. Conclude congruence using the correct congruence criterion.

In the right-angled triangles $PSO$ and $QRO$,

• $OP=OQ$ are the hypotenuses,
• $OS=OR$ are corresponding sides,
• $\angle PSO=\angle QRO=90^\circ$.

Therefore,
$\triangle PSO \cong \triangle QRO \quad \text{by RHS}.$

1. Identify the right angle and known sides in triangle $QRO$.

Since $RQ$ is a tangent to the smaller circle at $R$, $OR \perp RQ$.

So $\triangle QRO$ is right-angled at $R$.

$OQ=8\text{ cm}$ and $OR=4.36\text{ cm}$.

2. Find the length of $RQ$ using Pythagoras' theorem.

$RQ^2=OQ^2-OR^2$
$RQ^2=8^2-4.36^2$
$RQ=\sqrt{64-19.0096}$
$RQ=6.707\ldots\text{ cm}$

3. Calculate the area of triangle $QRO$.

$\text{Area of }\triangle QRO=\frac12 \times OR \times RQ$
$=\frac12 \times 4.36 \times 6.707\ldots$
$=14.621\ldots$

To $3$ significant figures, the area is $14.6\text{ cm}^2$.