Without the figure, it is difficult to provide a specific answer to this question. However, if we assume that the triangles ∆PQR and ∆PRX are similar, then we can use the fact that the ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding side lengths.
If we let PQ = a, QR = b, and RX = c, then we can write:
PR = PQ + QR = a + b
Using the fact that ∠PQR = ∠PRX, we can write:
∆PQR ~ ∆PRX
This means that the corresponding sides of the two triangles are proportional:
PR/PQ = RX/QR
Substituting in the values we know, we get:
(a + b)/a = c/b
Simplifying this equation, we get:
b(a + b) = ac
Expanding and rearranging, we get:
ab + b^2 = ac
Dividing both sides by a, we get:
b + (b^2)/a = c
Now, we can use the formula for the ratio of the areas of similar triangles:
(ar(∆PRX))/(ar(∆PQR)) = (RX/QR)^2
Substituting in the values we know, we get:
(ar(∆PRX))/(ar(∆PQR)) = (c/b)^2
Using the equation we derived earlier, we can substitute in c/b for (b^2)/a:
(ar(∆PRX))/(ar(∆PQR)) = (c/b)^2 = (b + (b^2)/a)^2
Expanding and simplifying, we get:
(ar(∆PRX))/(ar(∆PQR)) = (a^2 + 2ab)/(a^2 + 2ab + b^2)
Therefore, the ratio of the areas of the two triangles is (a^2 + 2ab)/(a^2 + 2ab + b^2). This is the best answer that can be provided without the specific figure.