How Cheenta works to ensure student success?
Explore the Back-Story

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

Subscribe to Cheenta at Youtube


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


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
Math Olympiad Program
magic-wandrockethighlight