1. **Problem 6:** Given that $y$ varies directly as the square of $(x+1)$, and $y=200$ when $x=4$, find the equation connecting $y$ and $x$. 2. **Step 1:** Write the direct variation formula: $$y = k(x+1)^2$$ where $k$ is the constant of proportionality. 3. **Step 2:** Use the given values $y=200$ and $x=4$ to find $k$: $$200 = k(4+1)^2 = k(5)^2 = 25k$$ $$k = \frac{200}{25} = 8$$ 4. **Step 3:** Write the equation connecting $y$ and $x$: $$y = 8(x+1)^2$$ --- 5. **Problem 7:** Given an isosceles triangle $MLN$ with height $5$ cm and angle $\angle MLN = 50^\circ$. (a) Construct the locus of point $P$ such that $\angle MPN = 50^\circ$. (b) Locate point $Q$ such that the area of $\triangle MQN$ is half that of $\triangle MLN$ and $\angle MQN = 50^\circ$. 6. **Step 1 (a):** The locus of points $P$ such that $\angle MPN = 50^\circ$ is an arc of a circle passing through points $M$ and $N$ where the angle subtended by chord $MN$ is constant at $50^\circ$. 7. **Step 2 (a):** Using a ruler and compass, draw the circle passing through $M$ and $N$ such that any point $P$ on the arc subtends $50^\circ$ at $P$. 8. **Step 3 (b):** To find point $Q$ such that area of $\triangle MQN$ is half of $\triangle MLN$ and $\angle MQN = 50^\circ$, note that the area depends on the base $MN$ and height from $Q$. 9. **Step 4 (b):** Since $\angle MQN = 50^\circ$, point $Q$ lies on the same locus as $P$ (the circle arc). 10. **Step 5 (b):** Locate $Q$ on the arc such that the perpendicular height from $Q$ to $MN$ is half the height of $5$ cm, i.e., $2.5$ cm, to ensure the area is halved. 11. **Step 6 (b):** Use ruler and compass to measure and mark $Q$ on the arc at height $2.5$ cm from $MN$. **Final answers:** - Equation connecting $y$ and $x$ is $$y = 8(x+1)^2$$ - The locus of $P$ is the circle arc subtending $50^\circ$ at chord $MN$. - Point $Q$ lies on this arc at height $2.5$ cm from $MN$ to satisfy the area condition.