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Trapezium | Geometry | PRMO-2018 | Problem 5

Try this beautiful problem from Geometry based on Trapezium.

Geometry Problem based on Trapezium | PRMO-2018 | Problem 5


Let ABCD be a trapezium in which AB||CD and AD is perpendicular on AB .suppose has an incircle which touches AB at Q and CD at P.Given that PC=36 and QB=49. Find PQ?

Trapezium
  • $64$
  • $81$
  • $84$

Key Concepts


Geometry

Trapezoid

Circle

Check the Answer


Answer:$84$

PRMO-2018, Problem 5

Pre College Mathematics

Try with Hints


Let the radius of the inner circle be r

Therefore AQ=PD=r and AD=2r

Can you now finish the problem ..........

Draw a perpendicular from C on AB at the point F

Can you finish the problem........

Trapezium figure - explanation

Let the inner circle touches BC at E

Then CE=36 (as BE & BQ are tangents)

BE=49 (as CE & PC are tangents)

Let the radius of the inner circle be r

Therefore AQ=PD=r and AD=2r

Let draw a perpendicular from C on AB at the point F

So BF=(AB-AF)=(49-36)=13

BC=85

Now in the triangle CBF we have

\(CF^2=85^2-13^2\)

\(\Rightarrow CF=84\)

Therefore CF=PQ=84

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Try this beautiful problem from Geometry based on Trapezium.

Geometry Problem based on Trapezium | PRMO-2018 | Problem 5


Let ABCD be a trapezium in which AB||CD and AD is perpendicular on AB .suppose has an incircle which touches AB at Q and CD at P.Given that PC=36 and QB=49. Find PQ?

Trapezium
  • $64$
  • $81$
  • $84$

Key Concepts


Geometry

Trapezoid

Circle

Check the Answer


Answer:$84$

PRMO-2018, Problem 5

Pre College Mathematics

Try with Hints


Let the radius of the inner circle be r

Therefore AQ=PD=r and AD=2r

Can you now finish the problem ..........

Draw a perpendicular from C on AB at the point F

Can you finish the problem........

Trapezium figure - explanation

Let the inner circle touches BC at E

Then CE=36 (as BE & BQ are tangents)

BE=49 (as CE & PC are tangents)

Let the radius of the inner circle be r

Therefore AQ=PD=r and AD=2r

Let draw a perpendicular from C on AB at the point F

So BF=(AB-AF)=(49-36)=13

BC=85

Now in the triangle CBF we have

\(CF^2=85^2-13^2\)

\(\Rightarrow CF=84\)

Therefore CF=PQ=84

Subscribe to Cheenta at Youtube


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