# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | May-Jun | (P1-9709/11) | Q#8

Question In the diagram, OAXB is a sector of a circle with centre O and radius 10 cm. The length of the chord  AB is 12 cm. The line OX passes through M, the mid-point of AB, and OX is perpendicular to  AB. The shaded region is bounded by the chord AB and by the arc of a circle with centre X and  radius XA.

i.       Show that angle AXB is 2.498 radians, correct to 3 decimal places.

ii.       Find the perimeter of the shaded region.

iii.       Find the area of the shaded region.

Solution

i.

We are required to find angle AXB.

Consider the diagram below. It is evident from the diagram that; If all three corresponding sides of two triangles are equal in length, then the triangles are congruent.

It is evident that both triangles AXM and BXM are congruent because; Both are radii of circle with center X. Both being two sections of line AB being bisected with OX at point M. Common to both triangle.

Since corresponding angles of two congruent sides all also equal; Therefore; Now we consider triangle AXM to find .

Expression for trigonometric ratio in right-triangle is; Therefore, for triangle AXM;  We need to find both AM and XM.

Let’s first find AM.

We are given that chord AB=12 and it is bisected by OX at point M. Therefore; Next we need to find XM.

It is evident from the diagram that; It is evident from the diagram that OX is radius of the circle with center O which we are given as  AX=10. Therefore, OX=10. To find MX we need to find OM.

Now consider the triangle OAM. Since OX is perpendicular bisector of AB it is a right triangle with .

Pythagorean Theorem Therefore, for triangle OAM;  We are given that OA=10 is radius of circle with center O. We have also found above that AM=6. Therefore;   Hence;   Finally we can now find the angle AXM.    Hence;   ii.

It is evident from the diagram below that;  Length of chord AB=12 is given. Therefore, we need to find length of arc AYB.

Expression for length of a circular arc with radius and angle rad is; For the arc AYB; We are given that arc AYB is of the circle with center at X. Therefore radius of circle of arc AYB; Hence; We have found in (i) that .

Now we need to find AX.

Again consider the triangle AXM.

Expression for trigonometric ratio in right-triangle is; Therefore, for triangle AXM; We have found in (i) that;  Therefore;   Now, we can find the length of arc AYB;   Finally, we can find the perimeter of shaded region;    iii.

It is evident from the diagram below that;  Let’s first find the area of sector XAYB.

Expression for area of a circular sector with radius and angle rad is; Therefore for sector XAYB;  We have found in (i) that and in (ii) that . Therefore;  Now we need to find area of triangle AXB.

Expression for the area of the triangle is; For triangle AXB;  We are given that AB=12 and we have found in (i) that XM=2. Therefore;  Hence;   