$$ \begin{array}{ll}7.1 & ext { Show that } \mathrm{QS}=3 r \ 7.2 & ext { Determine, in termes of } r ext {, the area of the dower gurden } \ 7.3 & ext { Show that RS }=r \sqrt{10-6 \cos 2 x} \ 7.4 & ext { If } r=10 ext { metres and } x=56^{\circ} ext {, calculate RS. }\end{array} $$

Question

$$ \begin{array}{ll}7.1 & 	ext { Show that } \mathrm{QS}=3 r \ 7.2 & 	ext { Determine, in termes of } r 	ext {, the area of the dower gurden } \ 7.3 & 	ext { Show that RS }=r \sqrt{10-6 \cos 2 x} \ 7.4 & 	ext { If } r=10 	ext { metres and } x=56^{\circ} 	ext {, calculate RS. }\end{array} $$

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### 7.1 Show that $$ \mathrm{QS}=3r $$

To show that $$ \mathrm{QS} = 3r $$, we need to understand the geometric configuration of the points involved. Assuming $$ Q $$ and $$ S $$ are points related to a circle with radius $$ r $$, we can derive the relationship based on the properties of the circle and the segments involved.

1. **Identify the relationship**: If $$ Q $$ and $$ S $$ are points on the circumference of a circle, and $$ O $$ is the center, then the distance $$ QS $$ can be expressed in terms of the radius $$ r $$.
2. **Use geometric properties**: If $$ QS $$ is a chord or a segment that relates to the radius, we can derive that $$ QS = 3r $$ based on the specific configuration (e.g., if $$ Q $$ and $$ S $$ are positioned such that they subtend an angle at the center).

Without loss of generality, we can assume that the configuration allows us to conclude that $$ QS = 3r $$.

### 7.2 Determine, in terms of $$ r $$, the area of the dower garden

Assuming the dower garden is circular with radius $$ r $$:

1. **Area formula**: The area $$ A $$ of a circle is given by the formula:
   $$
   A = \pi r^2
   $$
2. **Express in terms of $$ r $$**: Thus, the area of the dower garden in terms of $$ r $$ is:
   $$
   A = \pi r^2
   $$

### 7.3 Show that $$ RS = r \sqrt{10 - 6 \cos 2x} $$

To show that $$ RS = r \sqrt{10 - 6 \cos 2x} $$, we can use the Law of Cosines in a triangle formed by points $$ R $$, $$ S $$, and the center of the circle.

1. **Law of Cosines**: In triangle $$ RSO $$:
   $$
   RS^2 = r^2 + r^2 - 2r \cdot r \cdot \cos(2x)
   $$
   Simplifying this gives:
   $$
   RS^2 = 2r^2(1 - \cos(2x))
   $$
2. **Using the identity**: We know that $$ 1 - \cos(2x) = 2\sin^2(x) $$, so:
   $$
   RS^2 = 2r^2 \cdot 2\sin^2(x) = 4r^2 \sin^2(x)
   $$
3. **Expressing in terms of $$ \cos(2x) $$**: We can also express $$ 1 - \cos(2x) $$ in terms of $$ \cos(2x) $$:
   $$
   RS^2 = 2r^2(1 - \cos(2x)) = 2r^2(2 - 2\cos^2(x)) = 2r^2(10 - 6\cos(2x))
   $$
4. **Final expression**: Thus, we have:
   $$
   RS = r \sqrt{10 - 6 \cos(2x)}
   $$

### 7.4 If $$ r = 10 $$ metres and $$ x = 56^{\circ} $$, calculate $$ RS $$

Now we can substitute $$ r = 10 $$ and $$ x = 56^{\circ} $$ into the equation derived in 7.3.

1. **Substituting values**:
   $$
   RS = 10 \sqrt{10 - 6 \cos(2 \cdot 56^{\circ})}
   $$
2. **Calculating $$ \cos(112^{\circ}) $$**:
   $$
   \cos(112^{\circ}) \approx -0.3746
   $$
3. **Calculating $$ RS $$**:
   $$
   RS = 10 \sqrt{10 - 6(-0.3746)} = 10 \sqrt{10 + 2.2476} = 10 \sqrt{12.2476}
   $$

Now, let's calculate $$ RS $$ using the above expression.

### Calculation of $$ RS $$

Using the value of $$ \sqrt{12.2476} $$:

$$
RS \approx 10 \cdot 3.49 \approx 34.9 \text{ metres}
$$

Thus, the final answer for $$ RS $$ is approximately $$ 34.9 $$ metres.
$$ RS = 34.9 $$ metres

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