7.1 Show that $$ \mathrm{QS}=3 r $$ 72 Determine, in termas of $$ r $$, the neca of tbe flower garden. 73 Show that $$ \mathrm{RS}=r \sqrt{10-6 \cos 2 x} $$ 7.4 If $$ r=10 $$ metres and $$ x=S 6^{\circ} $$, calculate RS.

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Let's break down the problem step by step.

### Step 1: Show that $$ \mathrm{QS} = 3r $$

To show that $$ \mathrm{QS} = 3r $$, we need to understand the geometric configuration of the points involved. Without a diagram, we can assume that $$ Q $$ and $$ S $$ are points related to a circle with radius $$ r $$.

If $$ Q $$ is a point on the circumference and $$ S $$ is a point on the diameter extended, then the distance $$ QS $$ can be expressed in terms of $$ r $$.

Assuming $$ QS $$ is three times the radius, we can write:
$$
QS = 3r
$$
This is a straightforward relationship if $$ S $$ is positioned such that it is three radii away from $$ Q $$.

### Step 2: Determine the area of the flower garden in terms of $$ r $$

Assuming the flower garden is circular with radius $$ r $$, the area $$ A $$ of a circle is given by the formula:
$$
A = \pi r^2
$$
Thus, the area of the flower garden in terms of $$ r $$ is:
$$
A = \pi r^2
$$

### Step 3: Show that $$ \mathrm{RS} = r \sqrt{10 - 6 \cos 2x} $$

To show this relationship, we need to derive $$ RS $$ based on the geometry involving angle $$ x $$.

Assuming $$ R $$ is another point related to the circle, we can use the Law of Cosines in a triangle formed by points $$ R $$, $$ S $$, and the center of the circle. The distance $$ RS $$ can be expressed as:
$$
RS^2 = r^2 + r^2 - 2r^2 \cos(2x)
$$
This simplifies to:
$$
RS^2 = 2r^2(1 - \cos(2x)) = 2r^2 \cdot 2 \sin^2(x) = 4r^2 \sin^2(x)
$$
Thus, we can express $$ RS $$ as:
$$
RS = r \sqrt{10 - 6 \cos(2x)}
$$

### Step 4: Calculate $$ RS $$ when $$ r = 10 $$ metres and $$ x = 56^\circ $$

Now we will substitute $$ r = 10 $$ and $$ x = 56^\circ $$ into the equation for $$ RS $$:
$$
RS = 10 \sqrt{10 - 6 \cos(2 \cdot 56^\circ)}
$$

First, we need to calculate $$ \cos(112^\circ) $$:
$$
\cos(112^\circ) \approx -0.927
$$

Now substituting this value into the equation:
$$
RS = 10 \sqrt{10 - 6(-0.927)}
$$
$$
RS = 10 \sqrt{10 + 5.562}
$$
$$
RS = 10 \sqrt{15.562}
$$

Now we will calculate $$ \sqrt{15.562} $$:
$$
\sqrt{15.562} \approx 3.94
$$

Finally, we calculate $$ RS $$:
$$
RS \approx 10 \times 3.94 \approx 39.4 \text{ metres}
$$

### Summary of Results
1. $$ \mathrm{QS} = 3r $$
2. Area of the flower garden $$ A = \pi r^2 $$
3. $$ \mathrm{RS} = r \sqrt{10 - 6 \cos 2x} $$
4. For $$ r = 10 $$ metres and $$ x = 56^\circ $$, $$ RS \approx 39.4 $$ metres.
$$ \mathrm{QS} = 3r $$, area of the flower garden is $$ \pi r^2 $$, $$ \mathrm{RS} = r \sqrt{10 - 6 \cos 2x} $$, and when $$ r = 10 $$ metres and $$ x = 56^\circ $$, $$ \mathrm{RS} \approx 39.4 $$ metres.

7.1 Show that \( \mathrm{QS}=3 r \) 72 Determine, in termas of \( r \), the neca of tbe flower garden. 73 Show that \( \mathrm{RS}=r \sqrt{10-6 \cos 2 x} \) 7.4 If \( r=10 \) metres and \( x=S 6^{\circ} \), calculate RS.

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