该图展示了两个以 O 为圆心的同心圆。 PQ 是大圆的直径，RS 是小圆的直径。 PS 和 RQ 是小圆的切线。

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 $14.6\text{ cm}^2$.