Answer:

\(k=\dfrac{1}{18}\).

Take moments about the toppling point \(D\), using horizontal distances from \(D\).

The rectangle has area

\(2a^2\),

and its centre of mass is \(\frac{a}{2}\) to the left of \(D\).

The triangle has area

and its centre of mass is \(\frac{2a}{3}\) to the right of \(D\).

Since the lamina is uniform, the rectangle and triangle have mass ratio

\(2:1\).

Find the horizontal position \(x\) of the centre of mass of the lamina relative to \(D\), taking right as positive:

So the centre of mass of the lamina is \(\frac{a}{9}\) to the left of \(D\).

The particle at \(E\) is \(2a\) to the right of \(D\). Since the object is on the point of toppling about \(D\), the total centre of mass is vertically above \(D\). Hence the clockwise and anticlockwise moments about \(D\) balance:

Therefore

\(k=\dfrac{1}{18}\).