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kumpel [21]
3 years ago
13

Area of regular polygons

Mathematics
1 answer:
Finger [1]3 years ago
8 0
Hello!

To find the area of a hexagon you do 3\frac{3 \sqrt{3} }{2} } a^{2} where a is one of the sides

Since the perimeter is 60 we can do 60/ the sides of the shape

60/6 = 10

So one side is equal to 10

you put that into the formula to get the area of a hexagon and you get 259.81

Hope this helps!
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So for this, we will be applying the Triangle Inequality Theorem, which states that the sum of 2 sides must be greater than the third side for it to be a triangle. If any inequality turns out to be false, then the set cannot be a triangle.

A+B>C\\A+C>B\\B+C>A

<h2>First Option: {6, 10, 12}</h2>

Let A = 6, B = 10, and C = 12:

6+10>12\\16>12\ \textsf{true}\\\\6+12>10\\18>10\ \textsf{true}\\\\12+10>6\\22>6\ \textsf{true}

<h2>Second Option: {5, 7, 10}</h2>

Let A = 5, B = 7, and C = 10

5+7>10\\12>10\ \textsf{true}\\\\5+10>7\\15>7\ \textsf{true}\\\\10+7>5\\17>5\ \textsf{true}

<h2>Third Option: {4, 4, 9}</h2>

Let A = 4, B = 4, and C = 9

4+4>9\\8>9\ \textsf{false}\\\\4+9>4\\13>4\ \textsf{true}\\\\4+9>4\\13>4\ \textsf{true}

<h2>Fourth Option: {2, 3, 3}</h2>

Let A = 2, B = 3, and C = 3

2+3>3\\5>3\ \textsf{true}\\\\2+3>3\\5>3\ \textsf{true}\\\\3+3>2\\6>2\ \textsf{true}

<h2>Conclusion:</h2>

Since the third option had an inequality that was false, <u>the third option cannot be a triangle.</u>

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