The relationship between the sides MN, MS, and MQ in the given regular heptagon is ![\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7BMN%7D%20%3D%20%5Cdfrac%7B1%7D%7BMS%7D%20%2B%20%5Cdfrac%7B1%7D%7BMQ%7D)
The area to be planted with flowers is approximately <u>923.558 m²</u>
The reason the above value is correct is as follows;
The known parameters of the garden are;
The radius of the circle that circumscribes the heptagon, r = 25 m
The area left for the children playground = ΔMSQ
Required;
The area of the garden planted with flowers
Solution:
The area of an heptagon, is;
![A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )](https://tex.z-dn.net/?f=A%20%3D%20%5Cdfrac%7B7%7D%7B4%7D%20%5Ccdot%20a%5E2%20%5Ccdot%20%20cot%20%5Cleft%20%28%5Cdfrac%7B180%20%5E%7B%5Ccirc%7D%7D%7B7%7D%20%5Cright%20%29)
The interior angle of an heptagon = 128.571°
The length of a side, S, is given as follows;
![\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bs%7D%7Bsin%28180%20-%20128.571%29%7D%20%3D%20%5Cdfrac%7B25%7D%7Bsin%20%5Cleft%28%5Cdfrac%7B128.571%7D%7B2%7D%20%5Cright%29%7D)
![s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69](https://tex.z-dn.net/?f=s%20%3D%20%5Cdfrac%7B25%7D%7Bsin%20%5Cleft%28%5Cdfrac%7B128.571%7D%7B2%7D%20%5Cright%29%7D%20%5Ctimes%20sin%28180%20-%20128.571%29%20%5Capprox%2021.69)
![The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52](https://tex.z-dn.net/?f=The%20%5C%20apothem%20%5C%20a%20%3D%2025%20%5Ctimes%20sin%20%5Cleft%20%28%20%5Cdfrac%7B128.571%7D%7B2%7D%20%5Cright%29%20%5Capprox%2022.52)
The area of the heptagon MNSRQPO is therefore;
![A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94](https://tex.z-dn.net/?f=A%20%3D%20%5Cdfrac%7B7%7D%7B4%7D%20%5Ctimes%2022.52%5E2%20%5Ctimes%20cot%20%5Cleft%20%28%5Cdfrac%7B180%20%5E%7B%5Ccirc%7D%7D%7B7%7D%20%5Cright%20%29%20%5Capprox%201%2C842.94)
![MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08](https://tex.z-dn.net/?f=MS%20%3D%20%5Csqrt%7B%2821.69%5E2%20%2B%2021.69%5E2%20-%202%20%5Ctimes%20%2021.69%20%5Ctimes21.69%5Ctimes%20cos%28128.571%5E%7B%5Ccirc%7D%29%29%20%5Capprox%2043.08)
By sine rule, we have
![\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}](https://tex.z-dn.net/?f=%5Cdfrac%7B21.69%7D%7Bsin%28%5Cangle%20NSM%29%7D%20%3D%20%5Cdfrac%7B43.08%7D%7Bsin%28128.571%20%5E%7B%5Ccirc%7D%29%7D)
![sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})](https://tex.z-dn.net/?f=sin%28%5Cangle%20NSM%29%20%3D%5Cdfrac%7B21.69%7D%7B43.08%7D%20%5Ctimes%20sin%28128.571%20%5E%7B%5Ccirc%7D%29)
![\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20NSM%20%3D%20arcsin%20%5Cleft%28%5Cdfrac%7B21.69%7D%7B43.08%7D%20%5Ctimes%20sin%28128.571%20%5E%7B%5Ccirc%7D%29%20%5Cright%29%20%5Capprox%2023.18%5E%7B%5Ccirc%7D)
∠MSQ = 128.571 - 2*23.18 = 82.211
The area of triangle, MSQ, is given as follows;
![Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}](https://tex.z-dn.net/?f=Area%20%5C%20of%20%5CDelta%20MSQ%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20%2043.08%5E2%20%5Ctimes%20sin%2882.211%5E%7B%5Ccirc%7D%29%20%5Capprox%20919.382%5E%7B%5Ccirc%7D)
The area of the of the garden plated with flowers,
, is given as follows;
= Area of heptagon MNSRQPO - Area of triangle ΔMSQ
Therefore;
= 1,842.94 - 919.382 ≈ 923.558
The area of the of the garden plated with flowers,
≈ <u>923.558 m²</u>
Learn more about figures circumscribed by a circle here:
brainly.com/question/16478185