In $$\triangle$$ABC,P is a point on BC such that BP : PC = 3 : 4 and Q is the midpoint of AP. Then ar($$\triangle$$ABQ): ar($$\triangle$$ABC) is equal to:

Given that,

In $$\triangle$$ABC, P is a point on BC such that BP: PC = 3: 4

Q is the midpoint of AP.

We know that if two triangles have the same hight, then the ratio of the area of the triangle is always equal to the ratio of their base length.
$$ar( \triangle ABC)=ar( \triangle APB)+ar( \triangle APC)$$

But, $$ \dfrac{ar(\triangle APB)}{ar(\triangle APC)}=\dfrac{BP}{PC}$$

In $$ \triangle APB$$ and $$\triangle APC$$, both have the same height, so

$$\Rightarrow \dfrac{ar(\triangle APB)}{ar(\triangle APC)}=\dfrac{3}{4}$$

So, $$\Rightarrow ar( \triangle APB)=3k$$ and $$ar( \triangle APC)=4k$$

Now,

$$\Rightarrow ar(\triangle ABC)=ar(\triangle APB)+ar(\triangle APC)=3k+4k=7k$$------------------------(i)

Now, In $$\triangle APB$$

$$\Rightarrow ar(\triangle APB)=ar(\triangle AQB)+ar(\triangle QBP)$$

But $$ \triangle AQB$$ and $$\triangle QBP$$ have the same height,

So $$ \dfrac{ar(\triangle AQB)}{ar(\triangle QPB)}=\dfrac{1}{1}$$

$$\Rightarrow ar(\triangle AQB)=ar(\triangle QPB)$$

Hence,$$ar(\triangle ABQ)=\dfrac{ar(\triangle APB)}{2}=\dfrac{3k}{2}$$------------------------(ii)

From equation (i) and (ii)

$$\Rightarrow \dfrac{ar(\triangle ABQ)}{ar(\triangle ABC)}=\dfrac{\dfrac{3k}{2}}{7k}$$

$$\Rightarrow \dfrac{ar(\triangle ABQ)}{ar(\triangle ABC)}=\dfrac{3}{14}$$