### AB and AC are two chords of a circle such that AB = 2AC. If distances of AB and AC from the centre are 3 cm and 6 cm respectively, find the area of circle. (Assume π =3)

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**Answer: **135 cm^{2}

**Step by Step Explanation: **- Take a look at the representative image below:

We are told that AB = 2AC.

Also, if the perpendicular from O to AC meets the chord at Q, then OQ = 6 cm.

Similarly, OP = 3 cm. - As the perpendicular from the centre on the chord bisects the chord, OQ bisects AC, and OP bisects AB.

From the earlier relation AB = 2AC.

Therefore, BP = 2CQ

Let us assume CQ = *x*.

Then, BP = *2x* - Now consider ΔOQC,

OC = r, the radius of the circle, and OQ = 6 cm

As, the distance of a chord from the centre is always the perpendicular distance. ΔOQC is a right-angled triangle.

By using pythagoras theorum, OQ^{2} + CQ^{2} = r^{2}

6^{2} + x^{2} = r^{2}

or, 36 + x^{2} = r^{2} ------(1) - Similarly, ΔOPB is a right-angled triangle,

OP^{2} + BP^{2} = r^{2}

3^{2} + (2x)^{2} = r^{2}

or, 9 + 4x^{2} = r^{2} ------(2) - Subtracting equation (1) from equation (2), we get:

(9 - 36) + (4x^{2} - x^{2}) = 0

or, 3x^{2} = 27

or, x^{2} = - On substituting x
^{2} = in equation (1), we get:

36 + = r^{2}

or, r^{2} = = - Therefore, area of the circle = πr
^{2} = 3 × = 135 cm^{2}