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;
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