### Question 12

SSC-CGL 2018 June 4 Shift 1

The radius of a circle with O center is 10 cm, PQ and PR are the chords of 12 cm. PO cuts the chord QR at point S. What is the length of OS?

- 3.2 cm
- 2.8 cm
- 3 cm
- 2.5 cm

### Solution 1

Apply cosine formula to the isosceles $\displaystyle \Delta OPR$.

$\displaystyle 12^2 = 10^2 + 10^2 - 2\cdot 10 \cdot 10 \cos \theta$

$\displaystyle \implies \cos \theta = 0.28$

From the right triangle (why? chord bisector theorem) $\displaystyle \Delta OSR$, $\displaystyle x = 10 \cos \theta = 10 \times 0.28 = $ 2.8 cm Ans!

EXPLANATION: the circle has a radius of 10 cm. Two of its chords PQ and PR are co-terminus at P. The lengths of the two chords are each equal to 12 cm. The points P, Q, R, S and the center O of the circle form a quadrilateral. The diagonals PO and QR intersect at S. Observe that OP is a radius of the circle that is intersecting the chord QR. The angle of intersection will be 90 degree because each radius of a circle is the perpendicular bisector of each chord of the circle. Once, we establish that OS is perpendicular to QR, we are in a position to apply pythagoras theorem and other trigonometric formulae.

### Solution 2

By pythagoras to triangle OSR, $\displaystyle \overline{SR}^2 = 10^2 - x^2\text{ . . . (1)}$

By pythagoras to triangle PSR, $\displaystyle \overline{SR}^2 = 12^2 - (10 - x)^2\text{ . . . (2)}$

Equating and simplifying, we can get x = 2.8 cm Ans!

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