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

#### Solution

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.

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