What is the area of the pentagon in the middle of a star inscribed in a circle of radius 10 inches.?

I hope you can convince yourself that the area of the star in total is ten times the area of triangle CAO.This means that COA is one-tenth of the way around the circle, so it is 36°. That means that EOD is 72°.

A property of circle geometry tells us that when you have two points (like A and B) on the circumference of a circle, the angle from the center (O) is twice the angle from a point on the circumference (X). So EOD = 2BXA.

This tells us that BXA is 72°/2 = 36°.

By symmetry, CAO is therefore 18°.

Since we know two angles of triangle CAO, we know the third: OCA = 180°-36°-18° = 126°.

And you are given that length AO is 10 cm.

Now with this information, we can find the area of the triangle:

At this point, you can use the Law of Sines to find the lengths of the other two sides AC and CO.

Solving this gives us lengths of 3.82 and 7.27, respectively.

Now you may be asking yourself "What good is this?" Well, if we have the sides of the scalene triangle, we can solve for the length of the bisector that would divide this into two right triangles. Once we have *this*, we can solve for only the section of the triangle that is on the *interior* of the pentagon - and there are ten such triangles, so once we have the area of that triangle, we have the area of the pentagon.

You will need to set this up as a system of two equations to determine the height that is congruent to both triangles. Let y be this height. The bottom length of 10 will be split up - but we don't know into what yet! Let's say that the shorter segment will be x, and the longer segment will then be 10 - x. Therefore, what we have is the following:

(10-x)^2+y^2=7.27^2

x^2+y^2=3.82^2

100-20x+x^2+y^2=7.27^2

y^2 = 7.27^2-100+20x-x^2

y^2 = 3.82^2-x^2

Set them equal to each other, factor, simplify, and solve.

52.85-100+20x-x^2=14.59-x^2

-47.15+20x=14.59

20x=61.74

x=3.087

Now that we have x, we can solve for y:

3.087^2+y^2=3.82^2

9.53+y^2=14.59

y^2=8.07

y=2.84

Success! Remembering that the area of a triangle is given by a=1/2bh, we have our base, x=3.087, and our height, y=2.84. Plugging these values in gives us the area of the triangle that is solely on the interior of the pentagon, a=4.38. Multiply this by ten (since this process would be repeated the entire way around the circle ten times) for a grand total for the pentagon of 43.8 in^2.

Sorry for the confusion! I did a Google search for "area of a star" and took the explanation I found a step further - if you can find the length of one of the arms, you can find the length of the triangles they create. I'll link the site I used.

Is it a hexagon or a decagon? A decagon has 10 aspects. A dodecagon has 12 aspects. in case you draw segments from the middle of the circle to the two endpoints of one area of the polygon, you create a triangle. the perspective on the middle element is 360 deg divided by ability of the variety of aspects. permit's assume 12 aspects for demonstration, so the perspective is 30 deg. good so a ways? Draw yet another section from the middle perpendicular to the area of the polygon - that bisects the perspective and the polygon area to make a top triangle. The hypotenuse of the triangle is the radius of the circle, 10". the perspective on the middle is now 15 deg. nonetheless good? sin 15 deg = opposite / hypotenuse 10 sin 15 = opposite leg opposite leg = ~2.fifty 9" nonetheless making sense? the alternative leg is in basic terms one million/2 of the area of the polygon, so double that, and you have your answer. keep in mind, if it relatively is in user-friendly terms 10 aspects, each thing else is the comparable different than the unique perspective.

AMAR SONI is wrong. He is counting the radius of the pentagon as 10 inches instead of the circle. His answer is if the points of the pentagon touch the circle. The way I understand it, you mean more like a sheriff's badge. I partailly redeem my ignorance by just now figuring it out, but supensa got there first. 43.8 in sq. is right.

Area of a triangle = (1/2)(10)(10) sin (72 degrees)

= 0.5X10X10X sin 72

= 50*sin 72..............................(i)

Area of polygon = 5*50*sin 72 = 237.7641290737875 sq inches..............................Ans