Circle - Chord Properties

The radius of the circle is given to be $$6\sqrt{2}$$ cm. We are also given that the ratio of CA to CB is 3:2. So, if we assume the value of CA to be 3x, then the value of CB becomes 2x. We are also given that $$\angle PQC=45^{0}$$ and since PCQ is an isosceles triangle, $$\angle CPQ=45^{0}$$ and $$\angle PCQ=90^{0}$$.

So, the triangle PCQ is right-angled at C. The value of PQ can be calculated using Pythagoras' theorem as,

$$PQ^2\ =\ PC^2\ +\ QC^2$$

$$PQ^2\ =\ \left(6\sqrt{\ 2}\right)^2\ +\ \left(6\sqrt{\ 2}\right)^2\ =\ 72\ +\ 72\ =\ 144$$

$$PQ\ =\ 12$$ cm

$$AQ\ =\ \dfrac{PQ}{2}\ =\ \dfrac{12}{2}\ =\ 6$$ cm

In triangle ACQ,

$$AC^2\ +\ AQ^2\ =\ CQ^2$$

$$\left(3x\right)^2\ +\ \left(6\right)^2\ =\ \left(6\sqrt{\ 2}\right)^2$$

$$9x^2\ +\ 36\ =\ 72$$

$$9x^2\ =\ 36$$

$$x^2\ =\ 4$$

$$x\ =\ 2$$

$$CB\ =\ 2x\ =\ 2\times2\ =\ 4$$ cm

In triangle BCS,

$$BC^2\ +\ BS^2\ =\ CS^2$$

$$\left(4\right)^2\ +\ \left(BS\right)^2\ =\ \left(6\sqrt{\ 2}\right)^2$$

$$16\ +\ \left(BS\right)^2\ =\ 72$$

$$BS^2\ =\ 56$$

$$BS\ =\ 2\sqrt{\ 14}$$ cm

$$RS\ =\ 2\times\ BS\ =\ 2\times\ 2\sqrt{\ 14}\ =\ 4\sqrt{\ 14}$$ cm

The area of the quadrilateral PQRS can be calculated as,

$$Area=\ \dfrac{1}{2}\times\ AB\ \times\ \left(PQ\ +\ RS\right)\ =\ \dfrac{1}{2}\times\ 10\left(12\ +\ 4\sqrt{\ 14}\right)\ =\ 20\left(3\ +\ \sqrt{\ 14}\right)cm^2$$

Hence, the correct answer is option C.

The length of the perpendicular lines drawn from the centre to each of the chords is 5 cm and 12 cm respectively. As the two chords are parallel, the distance between the two chords is 5+12 = 17 cm

There are two cases possible where the chords are on the same side of the centre or on different sides of the centre. The distance will be different in each case.