Question

In the given figure, PQ || BC and AP : PB = 1 : 2. Find \[\frac{area \left(∆APQ \right)}{area \left(∆ABC \right)}\].

Answer 1

GIVEN: In the given figure PQ || BC, and AP : PB = 1 : 2

TO FIND: \[\frac{area \left( ∆ APQ \right)}{area \left( ∆ ABC \right)}\]

We know that according to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other side, then it divides the two sides in the same ratio.

Since triangle APQ and ABC are similar

Hence `(AP)/(AB)=(AQ)/(AC)=(PQ)/(BC)`

Now, it is given that `(AP)/(PB)=1/2`.

`⇒ PB =2AP`

`(AP)/(AB)=(AP)/(AP+PB)=(AP)/(AP+2AP)=1/3`

Since the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

\[\frac{Area\left( APQ \right)}{Area\left( ABC \right)} = \left( \frac{AP}{AB} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9}\]

Hence we got the result Area (ΔAPB) : Area (ΔABC) = 1 : 9