Question 12 SSC CGL 2018 June 4 Shift 1

Posted by Parveen(Hoven),
Aptitude Trainer and Software Developer

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Question 12 SSC-CGL 2018 June 4
[NOTE: only math questions solved]

Question: 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?

1. 3.2 cm
2. 2.8 cm
3. 3 cm
4. 2.5 cm

Method 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!

Method 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!