Click an item in the list or group of pictures at the bottom of the problem and, holding

7 answers
Question:

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. release your mouse button when the item is place. if you change your mind, drag the item to the trashcan. click the trashcan to clear all your answers. leave answer in exact form unless otherwise indicated. find the area of the region between a regular hexagon with sides of 6" and its inscribed circle.

Answers

  (54√3 - 27π) in²

Step-by-step explanation:


The radius of the inscribed circle is the apothem of the hexagon, so is ...

  (6 in)·sin(60°) = 3√3 in

The area of the hexagon is half the product of this and the perimeter of the hexagon (6 times the side length).

  A = (1/2)(6·6 in)(3√3 in) = 54√3 in²

The area of the circle is ...

  A = π·r² = π(3√3 in)² = 27π in²

Then the region between the hexagon and its inscribed circle will be ...

  hexagon area - circle area =

  (54√3 - 27π) in²

Area of the region = 15.03 in²

Step-by-step explanation:

Area of region between a regular hexagon with sides 6" and circle inscribed.

So Area of region = Area of regular hexagon - area of circle

Now area of regular hexagon = [tex]\frac{3a^{2} \sqrt{3}}{2}[/tex]

where a = side of the hexagon = 6"

Now area of regular hexagon = [tex]\frac{3(6^{2})\sqrt{3}}{2}=\frac{108\sqrt{3}}{2}[/tex] = 93.53 square in.

Area of circle inscribed = πr²

Here r is the radius of the circle = [tex]\sqrt{6^{2}-3^{2}}=\sqrt{36-9}[/tex]

r = 5"

So area of the inscribed circle = π(5)² = 3.14(25) = 78.5 square in.

Now area of region = 93.53 - 78.5 = 15.03 in²

You can compute the area of the region between the hexagon and the circle by subtracting:

Area of region = Area of hexagon - Area of inscribed circle

Let's find those areas.

If we draw a segment from the center of the hexagon to each of its vertex, the hexagon will be divided into 6 equal triangles. For any of those triangles, the angle in the center equals 360°/6 = 60°. Being the other angles equal, all three angles must equal 60° (because they must add up to 180°). Hence, those triangles are equilateral. The height (h) from the center divides the triangle into two equal right triangles. We know the hypotenuse is 6, because of the equilateral triangle we began with, and one of the other sides is half the side of the hexagon, 3. Applying the Pythagorean Theorem to this right triangle, we find that its height is:
 
[tex]h = \sqrt{6^2-3^2}=\sqrt{27} = 3\sqrt{3} = 5.20[/tex]

By the way, this height is also the radius of the inscribed circle, and we'll be using it later to find its area.

Now, we can find the area of each of the six triangles into which we divided the hexagon:

[tex]A = \dfrac{1}{2}\cdot 6 \cdot 3\sqrt{3} = 9\sqrt{3} = 15.59[/tex]

There are six of those triangles. So, finally, the area of the hexagon is:

Area of hexagon = 6A = [tex]54\sqrt{3} = 93.53[/tex]

Let's move on to the circle. It's easier, because we already know its radius. The area of a circle with radius r is given by:

Area of circle = [tex]\pi\cdot r^2 = \pi \cdot \left(3\sqrt{3}\right)^2 = 27\cdot \pi = 84.82[/tex]

Last, the area of the region:

Area of region = Area of hexagon - Area of circle = 93.53 - 84.82 = 8.71

Let the third side of triangle be x

Now parameter of triangle = sum of its sides

Therefore,

7a-11b=2a+b+a-9b+x

4a-4b=x(ans.)

Third side:4a-3b

Step-by-step explanation:

We know that perimeter of a triangle is sum of all three sides,i. e. x,y and z

Here in the question

x=2a+b

y=a-9b

and z=?

P=7a-11b

2a+b+a-9b+z=7a-11b

3a-8b+z=7a-11b

z=7a-11b-3a+8b

z=4a-3b

is the third side of the triangle.

Hope it helps you.

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step-by-step explanation:

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