Try this beautiful problem from Geometry based on Trapezium.
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?
Geometry
Trapezoid
Circle
But try the problem first...
Answer:$84$
PRMO-2018, Problem 5
Pre College Mathematics
First hint
Let the radius of the inner circle be r
Therefore AQ=PD=r and AD=2r
Can you now finish the problem ..........
Second Hint
Draw a perpendicular from C on AB at the point F
Can you finish the problem........
Final Step
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
Try this beautiful problem from Geometry based on Trapezium.
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?
Geometry
Trapezoid
Circle
But try the problem first...
Answer:$84$
PRMO-2018, Problem 5
Pre College Mathematics
First hint
Let the radius of the inner circle be r
Therefore AQ=PD=r and AD=2r
Can you now finish the problem ..........
Second Hint
Draw a perpendicular from C on AB at the point F
Can you finish the problem........
Final Step
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