1. **Problem statement:** Given parallelogram PQRS with diagonals PR and QS intersecting at O, where |PQ| = 4 cm, |QR| = 7 cm, and |QO| = 3.5 cm, we need to (i) show that $\cos(\angle RSQ) = \frac{2}{7}$ and (ii) find the length |PR|. 2. **Important properties:** In a parallelogram, diagonals bisect each other, so $O$ is the midpoint of both $PR$ and $QS$. Therefore, $|QO| = |OS| = 3.5$ cm, so $|QS| = 7$ cm. 3. **Step (i): Show $\cos(\angle RSQ) = \frac{2}{7}$** - Consider triangle $RSQ$. - We know $|QR| = 7$ cm and $|QS| = 7$ cm (since $QO = OS = 3.5$ cm). - Since $PQRS$ is a parallelogram, $|RS| = |PQ| = 4$ cm. 4. Use the Law of Cosines in triangle $RSQ$ to find $\cos(\angle RSQ)$: $$ |RS|^2 = |QR|^2 + |QS|^2 - 2|QR||QS|\cos(\angle RSQ) $$ Substitute known values: $$ 4^2 = 7^2 + 7^2 - 2 \times 7 \times 7 \times \cos(\angle RSQ) $$ Simplify: $$ 16 = 49 + 49 - 98 \cos(\angle RSQ) $$ $$ 16 = 98 - 98 \cos(\angle RSQ) $$ Rearranged: $$ 98 \cos(\angle RSQ) = 98 - 16 = 82 $$ $$ \cos(\angle RSQ) = \frac{82}{98} = \frac{41}{49} $$ This is not $\frac{2}{7}$, so let's check the approach. 5. **Re-examine the problem:** The problem states $|QO| = 3.5$ cm, which is half of $QS$, so $|QS| = 7$ cm. - $|QR| = 7$ cm - $|PQ| = 4$ cm - $|RS| = |PQ| = 4$ cm 6. Since $O$ is midpoint of $PR$, $|PR| = 2|PO|$. 7. To find $\cos(\angle RSQ)$, consider vectors: - Let vector $\vec{QS}$ be along the x-axis with length 7. - Vector $\vec{QR}$ is length 7, and $\vec{RS}$ length 4. 8. Using vector approach: - Place $Q$ at origin $(0,0)$. - Then $S$ at $(7,0)$. - Since $PQRS$ is parallelogram, $R$ is at $(7,4)$ and $P$ at $(0,4)$. 9. Vector $\vec{RS} = \vec{S} - \vec{R} = (7,0) - (7,4) = (0,-4)$. - Vector $\vec{SQ} = \vec{Q} - \vec{S} = (0,0) - (7,0) = (-7,0)$. 10. Compute $\cos(\angle RSQ)$ between vectors $\vec{RS}$ and $\vec{SQ}$: $$ \cos(\theta) = \frac{\vec{RS} \cdot \vec{SQ}}{|\vec{RS}||\vec{SQ}|} $$ Calculate dot product: $$ \vec{RS} \cdot \vec{SQ} = (0)(-7) + (-4)(0) = 0 $$ Magnitudes: $$ |\vec{RS}| = 4, \quad |\vec{SQ}| = 7 $$ So: $$ \cos(\angle RSQ) = \frac{0}{4 \times 7} = 0 $$ This contradicts the problem statement. 11. **Reconsider the angle:** The problem asks for $\cos(\angle RSQ)$, which is angle at $S$ between points $R$ and $Q$. - So angle at $S$ formed by points $R$ and $Q$. 12. Vectors from $S$: - $\vec{SR} = \vec{R} - \vec{S} = (7,4) - (7,0) = (0,4)$ - $\vec{SQ} = \vec{Q} - \vec{S} = (0,0) - (7,0) = (-7,0)$ 13. Compute $\cos(\angle RSQ)$: $$ \cos(\theta) = \frac{\vec{SR} \cdot \vec{SQ}}{|\vec{SR}||\vec{SQ}|} = \frac{0 \times (-7) + 4 \times 0}{4 \times 7} = 0 $$ Again zero. 14. The problem states $\cos(\angle RSQ) = \frac{2}{7}$, so the diagram is not aligned with axes as assumed. 15. Use triangle $RSQ$ with sides: - $|RS| = 4$ - $|SQ| = 7$ - $|QR| = 7$ 16. Use Law of Cosines for angle at $S$: $$ |RQ|^2 = |RS|^2 + |SQ|^2 - 2|RS||SQ|\cos(\angle RSQ) $$ Substitute values: $$ 7^2 = 4^2 + 7^2 - 2 \times 4 \times 7 \cos(\angle RSQ) $$ $$ 49 = 16 + 49 - 56 \cos(\angle RSQ) $$ $$ 49 = 65 - 56 \cos(\angle RSQ) $$ $$ 56 \cos(\angle RSQ) = 65 - 49 = 16 $$ $$ \cos(\angle RSQ) = \frac{16}{56} = \frac{2}{7} $$ This matches the required result. 17. **Step (ii): Find |PR|** - Since $O$ is midpoint of $PR$, $|PR| = 2|PO|$. - Use triangle $POQ$ where $O$ is midpoint of $QS$. 18. Use vector approach or Pythagoras: - $|PQ| = 4$ - $|QR| = 7$ - $|QO| = 3.5$ 19. Since $O$ is midpoint of $PR$, and $QS$ is diagonal of length 7, $PR$ is the other diagonal. 20. Use parallelogram diagonal formula: $$ |PR|^2 + |QS|^2 = 2(|PQ|^2 + |QR|^2) $$ Substitute values: $$ |PR|^2 + 7^2 = 2(4^2 + 7^2) $$ $$ |PR|^2 + 49 = 2(16 + 49) = 2 \times 65 = 130 $$ $$ |PR|^2 = 130 - 49 = 81 $$ $$ |PR| = \sqrt{81} = 9 $$ **Final answers:** - (i) $\cos(\angle RSQ) = \frac{2}{7}$ - (ii) $|PR| = 9$ cm