Try this beautiful Problem based on Chords in a Circle, Geometry, from PRMO 2017.

Chords in a Circle – PRMO 2017, Question 26

Let $A B$ and $C D$ be two parallel chords in a circle with radius 5 such that the centre $O$ lies between these chords. Suppose $A B=6, C D=8 .$ Suppose further that the area of the part of the circle lying between the chords $A B$ and $C D$ is $(m \pi+n) / k,$ where $m, n, k$ are positive integers with gcd$(m, n, k)=1$ . What is the value of $m+n+k ?$

  • $9$
  • $75$
  • $11$

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Suggested Reading

PRMO-2017, Problem 26

Pre College Mathematics

Try with Hints

First hint

Chords in a Circle

Draw OE $\perp A B$ and $O F \perp C D$

Clearly $\mathrm{EB}=\frac{\mathrm{AB}}{2}=3, \mathrm{FD}=\frac{\mathrm{CU}}{2}=4$

$\mathrm{OE}=\sqrt{5^{2}-3^{2}}=4$ and $\mathrm{OF}=\sqrt{5^{2}-4^{2}}=3$

Therefore $\Delta O E B \sim \Delta D F O$

Can you now finish the problem ……….

Second Hint

Chords in a Circle

Let $\angle \mathrm{EOB}=\angle \mathrm{ODF}=\theta,$ then

$\angle B O D=\angle A O C=180^{\circ}-\left(\theta+90^{\circ}-\theta\right)=90^{\circ}$

Now area of portion between the chords

= \(2 \times\) (area of minor sector BOD)+2 \times ar\((\triangle AOB)\)
$=2 \times \frac{\pi \times 5^{2}}{4}+2 \times \frac{1}{2} \times 6 \times 4=\frac{25 \pi}{2}+24=\frac{25 \pi+48}{2}$

Therefore $m=25, n=48$ and $k=2$

Can you finish the problem……..

Final Step

Therefore $m+n+k=75$

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