In $$\triangle$$ABC,P is a point on BC such that BP : PC = 2 : 3 and Q is the mid point of AP. Then ar($$\triangle$$ABQ): ar($$\triangle$$ABC)is equal to:

As per the question,

$$\triangle$$ABC

P is a point on BC such that BP : PC = 2 : 3

Q is the mid point of AP,

We know that, If any triangle have the same same height, then then the ratio of there area is always equal to the ratio of the there base respectively.

In triangle ABC, $$ar(\triangle ABC)=ar(\triangle APC) +ar(\triangle APB)$$

$$\dfrac{ar(\triangle APC)}{ar(\triangle APB)}=\dfrac{AP}{PB}$$

$$\dfrac{ar(\triangle APC)}{ar(\triangle APB)}=\dfrac{2}{3}=k $$(supposed)

$$ar(\triangle ABC)=\dfrac{5}{2}ar(\triangle APB)--------(i)$$

Similarly,

In $$\triangle ABP$$,

$$ar(\triangle ABP)=ar(\triangle ABQ)+ar(\triangle BPQ)$$

$$\dfrac{ar(\triangle ABQ)}{ar(\triangle QBP)}=\dfrac{1}{1}$$

$$ar(\triangle ABQ)=ar(\triangle QBP)=\dfrac{ar(\triangle ABP)}{2}--------(ii)$$

From equation (i) and (ii)

$$\dfrac{ar(\triangle ABQ)}{ar(\triangle ABC)}=\dfrac{2}{5\times 2}=\dfrac{1}{5}$$