# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Feb-Mar | (P1-9709/12) | Q#9

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**Question**

**a. **

Figure 1

In Fig. 1, OAB is a sector of a circle with centre O and radius r. AX is the tangent at A to the arc AB and angle .

** i. **Show that angle .

** ii. **Find the area of the shaded segment in terms of r and .

**b. **

Figure 2

In Fig. 2, ABC is an equilateral triangle of side 4 cm. The lines AX, BX and CX are tangents to the equal circular arcs AB, BC and CA. Use the results in part (a) to find the area of the shaded region, giving your answer in terms of and .

**Solution**

**a.
**

i.

Consider the in Figure 1.

Sum of all the interior angles of a triangle is equal to radians or .

Therefore, in ;

It is evident from the Figure 1 that is an isosceles triangle with sides .

The angles opposite to equal sides of a triangle are also equal.

Therefore, in the ;

Hence;

We are given in Figure 1 that;

It is also evident from the diagram that;

We are given that;

Therefore;

We have already found that . Hence;

ii.

It is evident from the Figure 1 that;

Let us first find area of sector OAB.

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

Therefore, area of sector OAB with and angle ;

Now we find area of triangle OAB.

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

Therefore, area of triangle OAB with sides and included angle ;

Now we can find area of shaded region;

**b.
**

It is evident from the Figure 2 that;

The triangle ABC being equilateral and using results of (a), we can see that;

Therefore;

Let us first find area of triangle ABC.

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

For triangle ABC;

All sides and interior angles of an equilateral triangle are equal.

It is evident from Figure 2 that triangular ABC is an equilateral triangle with all sides;

All sides and interior angles of an equilateral triangle are equal.

Therefore;

Sum of all the interior angles of a triangle is equal to radians or .

Therefore;

Hence;

Now we find the area of sector AB.

It is pertinent to note that in Figure 2, the areas AXB, AXC and BXC are equivalents of area AXB of Figure 1, as demonstrated in diagram below.

From (a), we know that area of each of these sectors is;

We can see from the Figure 2 that . Now we need to find .

It is also evident from Figure 2 that;

Also , therefore;

Substituting ;

Therefore;

Finally;

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