In a circle, chord PQ and chord RS intersect at T such that PT :
TQ = 3 : 4 and RT : TS = 6 : 5. If PQ = p and RS = q, which of the following is true? 𝑝2 147

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**Solution:**

1. Let PT = 3k and TQ = 4k, so $p=7k\Rightarrow k=\dfrac{p}{7}$. Let RT = 6m and TS = 5m, so $q=11m\Rightarrow m=\dfrac{q}{11}$.
2. Use the chord product rule: $PT\cdot TQ = RT\cdot TS$. Substitute values: $\left(\dfrac{3p}{7}\right)\left(\dfrac{4p}{7}\right)=\left(\dfrac{6q}{11}\right)\left(\dfrac{5q}{11}\right)$.
3. Compute: $\dfrac{12p^2}{49}=\dfrac{30q^2}{121}$.
4. Cross-multiply and simplify: $12\cdot121\,p^2=30\cdot49\,q^2\Rightarrow242\,p^2=245\,q^2$.
5. Thus $p^2=\dfrac{245}{242}\,q^2$, so $p^2:q^2=245:242$. This matches option **(c)**.

FINAL_ANSWER: c

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