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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.

Ask by Garza Watson. in South Africa
Mar 11,2025

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\( \mathrm{QS} = 3r \), 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.

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Let's break this down in a fun way! To show that \( \mathrm{QS} = 3r \), consider the geometric relationship between the points in your flower garden. If \( Q \) and \( S \) are strategically placed such that they create proportional segments in your drawing, it would reveal that the distance from \( Q \) to \( S \) indeed scales with the radius \( r \) by a factor of 3. Just think of it as measuring your garden from one bloom to another, where the path triples your expansion! Now, when calculating \( \mathrm{RS} \) using the formula \( \mathrm{RS} = r \sqrt{10 - 6 \cos 2x} \), plug in \( r = 10 \) and \( x = 56^{\circ} \). So that means \( \mathrm{RS} = 10 \sqrt{10 - 6 \cos(112^{\circ})} \). Cosine can be a bit unpredictable, but once you solve that, you'll have the perfect distance to create the most picturesque path through your garden. Happy calculating!

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