# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | May-Jun | (P1-9709/12) | Q#2

Question

In the diagram, AYB is a semicircle with AB as diameter and OAXB is a sector of a circle with centre  O and radius r. Angle AOB = 2 radians. Find an expression, in terms of r and , for the  area of the shaded region.

Solution

It is evident from the diagram that;

First we calculate area of semi-circle AYB.

Expression for area of semi-circle with radius  is;

Therefore, we need to find radius of semicircle AYB.

We are given that AB is the diameter of the circle AYB. Hence;

It is evident from the diagram that  is an isosceles triangle with congruent sides   and base . The vertex angle of this isosceles triangle is

Now consider the diagram below to find the radius of the circle AYB.

In an isosceles triangle, the bisector of the vertex angle (the angle which is opposite to a side that might not be congruent to another side) is also perpendicular bisector of the base.

Therefore,  is an angle bisector of the  and is also perpendicular bisector of base . Hence;

Now consider the two triangles AOS and BOS.
It is evident that
and  and .

Law of Sines;

Applying law of sines to ;

Now we can find area of semi-circle AYB.

Next we calculate area of segment AXB.

It is evident from the diagram that;

Let’s first find area of circular sector AXB and then area of triangle AOB.

Expression for area of a circular sector with radius  and angle  rad is;

It is evident that sector AXB is sector of the circle with center O which has radius  and angle of circular sector is . Therefore;

Lastly, we need area of triangle AOB.

Expression for the area of a triangle for which two sides (a and b) and the included angle (C ) is  given;

For the given case, in ;

Finally;