In the diagram, SR is parallel to PQ, and SQ and PR intersect at T. Given that PT = 10cm, RT = 2,5cm, and PQ = 16cm, calculate (i) SR, (ii) the ratio of the area of triangle RTS to the area of triangle PTQ, (iii) the ratio of the area of triangle PTQ to the area of triangle QTR.

Here are the solutions to the geometry problems:

3. (b) In the diagram, SR is parallel to PQ. This means that $\triangle RTS$ is similar to $\triangle PTQ$ (by AA similarity, as alternate interior angles are equal and vertically opposite angles are equal).

(i) Calculate SR

Step 1: Identify similar triangles and their corresponding sides. Since $\triangle RTS \sim \triangle PTQ$, the ratio of corresponding sides is equal: $\frac{SR}{PQ} = \frac{RT}{TP}$

Step 2: Substitute the given values and solve for SR. Given $PT = 10$ cm, $RT = 2.5$ cm, and $PQ = 16$ cm. $\frac{SR}{16} = \frac{2.5}{10}$ $SR = 16 \times \frac{2.5}{10}$ $SR = 16 \times 0.25$ $SR = 4 \text{ cm}$ The length of SR is $\boxed{4 \text{ cm}}$.

(ii) Calculate the ratio of the area of triangle RTS to the area of triangle PTQ

Step 1: Use the property that the ratio of the areas of similar triangles is the square of the ratio of their corresponding sides. $\frac{\text{Area}(\triangle RTS)}{\text{Area}(\triangle PTQ)} = \left(\frac{RT}{TP}\right)^2$

Step 2: Substitute the ratio of the sides. $\frac{\text{Area}(\triangle RTS)}{\text{Area}(\triangle PTQ)} = \left(\frac{2.5}{10}\right)^2$ $\frac{\text{Area}(\triangle RTS)}{\text{Area}(\triangle PTQ)} = \left(\frac{1}{4}\right)^2$ $\frac{\text{Area}(\triangle RTS)}{\text{Area}(\triangle PTQ)} = \frac{1}{16}$ The ratio of the area of triangle RTS to the area of triangle PTQ is $\boxed{1:16}$.

(iii) Calculate the ratio of the area of triangle PTQ to the area of triangle QTR

Step 1: Identify that $\triangle PTQ$ and $\triangle QTR$ share the same height from vertex Q to the base PR. The ratio of their areas is therefore the ratio of their bases along the line PR. $\frac{\text{Area}(\triangle PTQ)}{\text{Area}(\triangle QTR)} = \frac{PT}{RT}$

Step 2: Substitute the given values for the bases. $\frac{\text{Area}(\triangle PTQ)}{\text{Area}(\triangle QTR)} = \frac{10}{2.5}$ $\frac{\text{Area}(\triangle PTQ)}{\text{Area}(\triangle QTR)} = 4$ The ratio of the area of triangle PTQ to the area of triangle QTR is $\boxed{4:1}$.

5. (a) In the diagram, OACB is a rectangle. X and Y are the midpoints of BC and CA respectively. Given $\vec{OA} = 2\mathbf{a}$ and $\vec{XY} = \mathbf{a} - \mathbf{b}$.

(i) Express $\vec{XC}$ in terms of $\mathbf{a}$ and/or $\mathbf{b}$

Step 1: Use the properties of a rectangle. In rectangle OACB, $\vec{BC} = \vec{OA}$. Given $\vec{OA} = 2\mathbf{a}$, so $\vec{BC} = 2\mathbf{a}$.

Step 2: Use the midpoint property. X is the midpoint of BC, so $\vec{XC} = \frac{1}{2}\vec{BC}$. $\vec{XC} = \frac{1}{2}(2\mathbf{a})$ $\vec{XC} = \mathbf{a}$ $\vec{XC} = \boxed{\mathbf{a}}$.

(ii) Express $\vec{OB}$ in terms of $\mathbf{a}$ and/or $\mathbf{b}$

Step 1: Express $\vec{XY}$ in terms of known vectors. $\vec{XY} = \vec{XC} + \vec{CY}$ We know $\vec{XC} = \mathbf{a}$. Y is the midpoint of CA, so $\vec{CY} = \frac{1}{2}\vec{CA}$. $\vec{CA} = \vec{CO} + \vec{OA}$ In a rectangle, $\vec{CO} = -\vec{OC}$ and $\vec{OA} = 2\mathbf{a}$. Also, $\vec{OC} = \vec{AB}$. Let $\vec{OC} = \mathbf{c}$. So, $\vec{CA} = -\mathbf{c} + 2\mathbf{a}$. $\vec{CY} = \frac{1}{2}(-\mathbf{c} + 2\mathbf{a}) = -\frac{1}{2}\mathbf{c} + \mathbf{a}$

Step 2: Substitute into the expression for $\vec{XY}$ and solve for $\mathbf{c}$. Given $\vec{XY} = \mathbf{a} - \mathbf{b}$. $\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\frac{1}{2}\mathbf{c} + \mathbf{a})$ $\mathbf{a} - \mathbf{b} = 2\mathbf{a} - \frac{1}{2}\mathbf{c}$ $\frac{1}{2}\mathbf{c} = 2\mathbf{a} - \mathbf{a} + \mathbf{b}$ $\frac{1}{2}\mathbf{c} = \mathbf{a} + \mathbf{b}$ $\mathbf{c} = 2(\mathbf{a} + \mathbf{b}) = 2\mathbf{a} + 2\mathbf{b}$ So, $\vec{OC} = 2\mathbf{a} + 2\mathbf{b}$.

Step 3: Calculate $\vec{OB}$. Using vector addition: $\vec{OB} = \vec{OA} + \vec{AB}$. In a rectangle, $\vec{AB} = \vec{OC}$. $\vec{OB} = \vec{OA} + \vec{OC}$ $\vec{OB} = 2\mathbf{a} + (2\mathbf{a} + 2\mathbf{b})$ $\vec{OB} = 4\mathbf{a} + 2\mathbf{b}$ $\vec{OB} = \boxed{4\mathbf{a} + 2\mathbf{b}}$.

(iii) Express $\vec{AB}$ in terms of $\mathbf{a}$ and/or $\mathbf{b}$

Step 1: Use the properties of a rectangle. In rectangle OACB, $\vec{AB} = \vec{OC}$.

Step 2: Substitute the expression for $\vec{OC}$ found in part (ii). $\vec{AB} = 2\mathbf{a} + 2\mathbf{b}$ $\vec{AB} = \boxed{2\mathbf{a} + 2\mathbf{b}}$.