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Naily [24]
2 years ago
5

What is the Vertex of 12 and 1

Mathematics
1 answer:
Dvinal [7]2 years ago
4 0

Answer:

Do you have an equation to find the answer?

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A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes
Alenkinab [10]

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

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 )

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)}

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

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

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

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

By sine rule, we have

\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}

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

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

∠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}

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

A_{req} = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

A_{req}= 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, A_{req} ≈ <u>923.558 m²</u>

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0
2 years ago
Suppose that Njoman rolls a fair six-sided die and a fair four-sided die S uivant
ale4655 [162]

Answer:

1/6 ,1/6, 1/24, no events A and B are not independent events.

Step-by-step explanation:

6 0
3 years ago
How to solve 18.912=2.4p
Afina-wow [57]

Answer:

p = 21.312

Step-by-step explanation:

Divide 2.4 on both sides.

p = 21.312

Hope this helped!

6 0
3 years ago
Read 2 more answers
Indicate whether the following statement is true or false: A linear function eventually exceeds a quadratic function with a posi
MA_775_DIABLO [31]

Answer:

idk

Step-by-step explanation:

idk  tbh

3 0
3 years ago
What is the area of the shaded region in the figure below ? Leave answer in terms of pi and in simplest radical form
ser-zykov [4K]

Answer:

Step-by-step explanation:

That shaded area is called a segment. To find the area of a segment within a circle, you first have to find the area of the pizza-shaped portion (called the sector), then subtract from it the area of the triangle (the sector without the shaded area forms a triangle as you can see). This difference will be the area of the segment.

The formula for the area of a sector of a circle is:

A_s=\frac{\theta}{360}*\pi r^2 where our theta is the central angle of the circle (60 degrees) and r is the radius (the square root of 3).

Filling in:

A_s=\frac{60}{360}*\pi (\sqrt{3})^2 which simplifies a bit to

A_s=\frac{1}{6}*\pi(3) which simplifies a bit further to

A_s=\frac{1}{2}\pi which of course is the same as

A_s=\frac{\pi}{2}

Now for tricky part...the area of the triangle.

We see that the central angle is 60 degrees. We also know, by the definition of a radius of a circle, that 2 of the sides of the triangle (formed by 2 radii of the circle) measure √3. If we pull that triangle out and set it to the side to work on it with the central angle at the top, we have an equilateral triangle. This is because of the Isosceles Triangle Theorem that says that if 2 sides of a triangle are congruent then the angles opposite those sides are also congruent. If the vertex angle (the angle at the top) is 60, then by the Triangle Angle-Sum theorem,

180 - 60 = 120, AND since the 2 other angles in the triangle are congruent by the Isosceles Triangle Theorem, they have to split that 120 evenly in order to be congruent. 120 / 2 = 60. This is a 60-60-60 triangle.

If we take that extracted equilateral triangle and split it straight down the middle from the vertex angle, we get a right triangle with the vertex angle half of what it was. It was 60, now it's 30. The base angles are now 90 and 60. The hypotenuse of this right triangle is the same as the radius of the circle, and the base of this right triangle is \frac{\sqrt{3} }{2}. Remember that when we split that 60-60-60 triangle down the center we split the vertex angle in half but we also split the base in half.

Using Pythagorean's Theorem we can find the height of the triangle to fill in the area formula for a triangle which is

A=\frac{1}{2}bh. There are other triangle area formulas but this is the only one that gives us the correct notation of the area so it matches one of your choices.

Finding the height value using Pythagorean's Theorem:

(\sqrt{3})^2=h^2+(\frac{\sqrt{3} }{2})^2 which simplifies to

3=h^2+\frac{3}{4} and

3-\frac{3}{4}=h^2 and

\frac{12}{4} -\frac{3}{4} =h^2 and

\frac{9}{4} =h^2

Taking the square root of both the 9 and the 4 (which are both perfect squares, thankfully!), we get that the height is 3/2. Now we can finally fill in the area formula for the triangle!

A=\frac{1}{2}(\sqrt{3})(\frac{3}{2}) which simplifies to

A=\frac{3\sqrt{3} }{4}

Therefore, the area in terms of pi for that little segment is

A_{seg}=\frac{\pi}{2}-\frac{3\sqrt{3} }{4}, choice A.

8 0
3 years ago
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