In ΔABC, ∠C = 90°, point P and Q are on the sides AC and BC, respectively, such that AP : PC = BQ : QC = 1 : 2. Then, $$\frac{{{\text{A}}{{\text{Q}}^2} + {\text{B}}{{\text{P}}^2}}}{{{\text{A}}{{\text{B}}^2}}}$$ is equal to:
A. $$\frac{8}{3}$$
B. $$\frac{4}{3}$$
C. $$\frac{{13}}{9}$$
D. $$\frac{4}{9}$$
Answer: Option C
Solution(By Examveda Team)

Let the side of BC & CA are 3x and 3y
QC = 2x
QB = x
PC = 2y
AP = y
In triangle AQC
AQ2 = (3y)2 + (2x)2
= 9y2 + 4x2
BP2 = (3x)2 + (2y)2
= 9x2 + 4y2
AB2 = (3x)2 + (3y)2
= 9x2 + 9y2
= 9(x2 + y2)
$${\text{Now, }}\frac{{{\text{A}}{{\text{Q}}^2} + {\text{B}}{{\text{P}}^2}}}{{{\text{A}}{{\text{B}}^2}}} = \frac{{13\left( {{x^2} + {y^2}} \right)}}{{9\left( {{x^2} + {y^2}} \right)}} = \frac{{13}}{9}$$
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