A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes the heptagon as shown in the diagram. The area of ΔMSQ is left for a children's playground, and the rest of the garden is planted with flowers. Find the area of the garden planted with flowers.

A garden is shaped in the form of a regular heptagon sevensided MNSRQPO A circle with center T and radius 25m circumscribes the heptagon as shown in the diagram class=

Respuesta :

The relationship between the sides MN, MS, and MQ in the given regular heptagon is [tex]\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}[/tex]

The area to be planted with flowers is approximately 923.558 m²

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;

[tex]A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )[/tex]

The interior angle of an heptagon = 128.571°

The length of a side, S, is given as follows;

[tex]\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}[/tex]

[tex]s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69[/tex]

[tex]The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52[/tex]

The area of the heptagon MNSRQPO is therefore;

[tex]A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94[/tex]

[tex]MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08[/tex]

By sine rule, we have

[tex]\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}[/tex]

[tex]sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})[/tex]

[tex]\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}[/tex]

∠MSQ = 128.571 - 2*23.18 = 82.211

The area of triangle, MSQ, is given as follows;

[tex]Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}[/tex]

The area of the of the garden plated with flowers, [tex]A_{req}[/tex], is given as follows;

[tex]A_{req}[/tex] = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

[tex]A_{req}[/tex]= 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, [tex]A_{req}[/tex] ≈ 923.558 m²

Learn more about figures circumscribed by a circle here:

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