# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | May-Jun | (P1-9709/13) | Q#3

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Question

The diagram shows triangle ABC which is right-angled at A. Angle radians and AC = 8  cm. The points D and E lie on BC and BA respectively. The sector ADE is part of a circle with centre  A and is such that BDC is the tangent to the arc DE at D.

i. Find the length of AD.

ii. Find the area of the shaded region.

Solution

i.

We are required to find the length of AD.

To find length of AD we consider the triangle ADC.

We are given that BDC is the tangent to the arc DE at D where A is center of the arc DE.

It is evident that AD is the radius of the sector ADE.

A radius from the center of the circle to the point of tangency is perpendicular to the tangent line.

Therefore, AD is perpendicular to straight line BDC;

Hence, triangle ADC is a right-triangle.

We know that;

Expression for   trigonometric ratio in right-triangle is;

We are given that but we need to find .

To find we consider the triangle ABC where .

We know that;

Therefore, for triangle ABC;

We are given that and , therefore;

Hence;

ii.

It is evident from the diagram that;

First, we find area of sector AED.

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

For the given sector AED;

As demonstrated in (i) AD is radius of circle/sector with center A.

It is evident from the diagram that .

Therefore, now we find .

Consider triangle ABD.

We know that;

Therefore, for triangle ABD;

As demonstrated in (i);

Since, BDC is a straight line, tangent to arc DE, therefore;

We are given that , therefore;

Hence;

Hence;

Now we can find area of sector AED;

Next, we find area of the triangle ADC.

Expression for the area of the triangle is;

For the triangle ADC.

We are given that .

We need to find DC.

Expression for  trigonometric ratio in right-triangle is;

Therefore, for triangle ADC;

As demonstrated in (i);

Therefore;

Hence;

Finally;