# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | Feb-Mar | (P1-9709/12) | Q#4

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**Question**

In the diagram, AB=AC=8 cm and angle radians. The circular arc BC has centre A, the circular arc CD has centre B and ABD is a straight line.

** i. **Show that angle radians.

** ii. **Find the perimeter of the shaded region.

**Solution**

** i.
**

It is evident from the diagram that;

Therefore;

We are given that ABD is a straight line.

Therefore, angle radians.

Now we need to find .

Consider triangle ABC.

Sum of all the interior angles of triangle equals radians or .

Therefore;

We are given that , therefore;

It is evident from the diagram that two sides of triangle ABC are equal;

Angles opposite to equal sides of a triangle are equal.

Therefore, angle opposite to equal sides must be equal.

Hence;

Now we can find the desired angle;

** ii.
**

It is evident from the diagram that;

Let’s first find circular arc CD.

Expression for length of a circular arc with radius and angle rad is;

We are given that circular arc CD has center at point B. Therefore radius of this arc is BC=BD. It is also evident that angle of this arc is .

Therefore, for the given case;

We have already found in (i) that , therefore, we only need to find the radius of arc BC.

Consider triangle ABC.

Law of Sines;

For the given case;

We have already found in (i) that;

We are also given that;

Therefore;

Using calculator;

Now we can find the length of arc CD.

Next we need to find length of arc BC.

Expression for length of a circular arc with radius and angle rad is;

We are given that circular arc BC has center at point A. Therefore radius of this arc is AC=AD. It is also evident that angle of this arc is .

Therefore, for the given case;

We are given that and , therefore;

Finally, we need length BD to find the desired perimeter.

It is evident from the diagram that BD is radius of the circular arc CD with center at point B.

It is also evident from the diagram that BC is radius of the circular arc CD with center at point B.

Therefore;

We have already found that;

Therefore;

Hence;

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