Review question

# When does this area within a circle have a maximum value? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9028

## Solution

If two chords $PQ$ and $PR$ on a circle of radius $1$ meet in an angle $\theta$ at $P$, for example as drawn in the diagram below,

then the largest possible area of the shaded region $RPQ$ is

1. $\theta(1 + \cos\dfrac{\theta}{2})$; $\quad$$\quad$(b) $\theta + \sin\theta$; $\quad$$\quad$(c) $\dfrac{\pi}{2}(1-\cos\theta)$;$\quad$$\quad$ (d) $\theta$.

The shaded area will be a maximum when the diagram is symmetrical.

We can see this as follows; imagine rotating $PR$ through a small angle $\alpha$ anticlockwise in the diagram above.

To keep $\theta$ the same size we need to rotate $PQ$ through $\alpha$ also, and since $PQ$ is longer than $PR$, this will cut off a greater area than rotating $PR$ did.

Thus the symmetrical arrangement maximises the shaded area.

So we need to find the shaded area for the situation above.

Remember that the area of a sector enclosing an angle $\theta$ (in radians) and with radius $r$ is $\dfrac{1}{2}r^2 \theta$.

The area of the sector we need is $\dfrac{1}{2}\times 1^2 \times 2\theta = \theta.$

Remember that the area of a triangle with sides $a$ and $b$ enclosing an angle $C$ is $\dfrac{1}{2}ab\sin C$.

The sum of the areas of the two triangles $POR$ and $POQ$ is $2\left[\dfrac{1}{2}\times 1^2 \times \sin (\pi - \theta)\right] = \sin (\pi - \theta)] = \sin \theta.$

Thus the total shaded area is $\sin \theta + \theta$, and the answer is (b).