Question

a line PQ is drawn parallel to base BC of triangle ABC which meets side AB and AC at points P and Q respectively . If AP=1/3 PB. find the ratio of area of triangle APQ / Area of trapezium PBCQ​

Answer 1

Answer:

The ratio of ar(∆APQ)/ar(trapezium PBCQ) is 1:15

Step by Step explanation:

Given:

A line PQ is drawn || to base BC of ∆ ABC which meets side AB and AC at points P and Q respectively . If AP=1/3

To Find :

The ratio of ar(∆APQ)/ar(trapezium PBCQ)

Solution:

AP = 1/3 PB

AP/PB = 1/3

In ∆APQ and ∆ ABC

As PQ || BC ( corresponding angles are equal )

∠APQ = ∠ABC

∠AQP = ∠ACB

∆ APQ ~ ∆ ABC

ar ( ∆ ABC ) / ar (∆ APQ ) = AB²/AP²=4²/1²= 16:1

AP / PB = 1/3

=> AB / AP = 4/1

=> ar ( ∆ APQ)/ ar ( trapezium PBCQ )

=> ar ( ∆ APQ)/ ar ( ∆ ABC ) - ar ( ∆ APQ )

=> 1/16-1

=>1/15

=> 1:15

Answer 2

Step-by-step explanation:

HERE IS UR DIAGRAM......