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Area of a part of circle | PRMO 2017 | Question 26

Try this beautiful problem from the Pre-RMO, 2017, Question 26, based on Area of part of circle.

Area of part of circle - Problem 26


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

  • is 107
  • is 75
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 75.

PRMO, 2017, Question 26

Higher Algebra by Hall and Knight

Try with Hints


First hint

A=2[\(\frac{1}{2} \times 25 \times \theta\)]+\(\frac{1}{2} \times 3 \times 8\)+\(\frac{1}{2} \times 4 \times 6\)

where \(\theta=[\pi-(\theta_1+\theta_2)]=[\pi-(tan^{-1}\frac{4}{3}+tan^{-1}\frac{3}{4})]\)

Area of a part of circle

Second Hint

or, \(\theta=\frac{\pi}{2}\)

or, A=24+\(\frac{25\pi}{2}\)

or, A=\(\frac{48+25\pi}{2}\)

Final Step

(m+n+k)=(48+2+25)=75

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Try this beautiful problem from the Pre-RMO, 2017, Question 26, based on Area of part of circle.

Area of part of circle - Problem 26


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

  • is 107
  • is 75
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 75.

PRMO, 2017, Question 26

Higher Algebra by Hall and Knight

Try with Hints


First hint

A=2[\(\frac{1}{2} \times 25 \times \theta\)]+\(\frac{1}{2} \times 3 \times 8\)+\(\frac{1}{2} \times 4 \times 6\)

where \(\theta=[\pi-(\theta_1+\theta_2)]=[\pi-(tan^{-1}\frac{4}{3}+tan^{-1}\frac{3}{4})]\)

Area of a part of circle

Second Hint

or, \(\theta=\frac{\pi}{2}\)

or, A=24+\(\frac{25\pi}{2}\)

or, A=\(\frac{48+25\pi}{2}\)

Final Step

(m+n+k)=(48+2+25)=75

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