The diagram below shows two circles, with the same centre at O. Points A, B and C lie on the bigger circle and point D lies on the smaller circle. AC is a straight line that passes through the centre of both circles. AB and BC are both tangents to the smaller circle.

1. Find the area of the larger similar triangle $ABC$.

Since $AC$ is a diameter of the bigger circle and $OC$ is a radius,
$AC=2OC.$
For similar triangles $ABC$ and $ODC$,
$\frac{\text{area of }\triangle ABC}{\text{area of }\triangle ODC}=\left(\frac{AC}{OC}\right)^2=2^2=4.$
Hence
$\text{area of }\triangle ABC=4\times 15=60\text{ cm}^2.$

2. Subtract the smaller triangle from the larger triangle to get the trapezium area.

The trapezium $ABDO$ is the remaining part of $\triangle ABC$ after removing $\triangle ODC$.
$\text{area of trapezium }ABDO=60-15=45\text{ cm}^2.$