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Diameter of a circle | PRMO 2019 | Question 25

Try this beautiful problem from the Pre-RMO, 2019 based on Diameter of a circle.

Diameter of a circle - PRMO 2019


A village has a circular wall around it, and the wall has four gates pointing north, southeast and west. A tree stands outside the village, 16 m north of the north gate, and it can be just seen appearing on the horizon from a point 48 m east of the south gate. Find the diameter in meters of the wall that surrounds the village.

  • is 107
  • is 48
  • is 840
  • cannot be determined from the given information

Key Concepts


Pythagoras Theorem

Equations

Integer

Check the Answer


Answer: is 48.

PRMO, 2019, Question 25

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

Let radius =r

or,\(AB=\sqrt{AO^{2}-OB^{2}}=\sqrt{(16+r)^{2}-r^{2}}\)

\(=\sqrt{256+32r}\)

or, \(AD^{2}+DC^{2}=CA^{2}\)

Diameter of a circle - figure

Second Hint

or, \(48^{2}+(2r+16)^{2}\)

\(=(48+\sqrt{256+32r})^{2}\)

or, \(r^{2}+8r=24\sqrt{256+32r}\)

or, \(r(r+8)=24(4\sqrt{2})(\sqrt{r+8})\)

Final Step

or, r=24

or, 2r=diameter=48.

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Try this beautiful problem from the Pre-RMO, 2019 based on Diameter of a circle.

Diameter of a circle - PRMO 2019


A village has a circular wall around it, and the wall has four gates pointing north, southeast and west. A tree stands outside the village, 16 m north of the north gate, and it can be just seen appearing on the horizon from a point 48 m east of the south gate. Find the diameter in meters of the wall that surrounds the village.

  • is 107
  • is 48
  • is 840
  • cannot be determined from the given information

Key Concepts


Pythagoras Theorem

Equations

Integer

Check the Answer


Answer: is 48.

PRMO, 2019, Question 25

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

Let radius =r

or,\(AB=\sqrt{AO^{2}-OB^{2}}=\sqrt{(16+r)^{2}-r^{2}}\)

\(=\sqrt{256+32r}\)

or, \(AD^{2}+DC^{2}=CA^{2}\)

Diameter of a circle - figure

Second Hint

or, \(48^{2}+(2r+16)^{2}\)

\(=(48+\sqrt{256+32r})^{2}\)

or, \(r^{2}+8r=24\sqrt{256+32r}\)

or, \(r(r+8)=24(4\sqrt{2})(\sqrt{r+8})\)

Final Step

or, r=24

or, 2r=diameter=48.

Subscribe to Cheenta at Youtube


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