+794 Express answer in exact form. Show all work for full credit.
Find the area of one segment formed by a square with sides of 6" inscribed in a circle.
(Hint: use the ratio of 1:1:\(\sqrt2\) to find the radius of the circle.)
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Express answer in exact form. Show all work for full credit.
A regular hexagon with sides of 3" is inscribed in a circle. Find the area of a segment formed by a side of the hexagon and the circle.
(Hint: remember Corollary 1--the area of an equilateral triangle is\(\frac14s^2*\sqrt3\).)
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Find the area of the shaded portion in the equilateral triangle with sides 6. Show all work for full credit.
(Hint: Assume that the central point of each arc is its corresponding vertex.)
if you could help me out on these three questions that would help alot thanks
+128407 Find the area of one segment formed by a square with sides of 6" inscribed in a circle.
I'm not exactly sure what this is asking.....if it is asking for the area between one side of the square and the perimeter of the circle, we can proceed as follows :
To find the radius of the circle we have that
2r^2 = 36 divide both sides by 2
r^2 = 18
r = √18
So....the area we are looking for is the area of a quarter circle with a radius of √18 minus the area of an isosceles right triangle with legs of √18 and a hypotenuse of 6
So....the area is
pi (r^2) /4 - (1/2) (product of leg lengths) =
pi (√18)^2 / 4 - (1/2)(√18)(√18) =
18 ( pi /4 - 1/2) units^2
+2862 This is my HINT for the last question:
Find the area of the triangle
Find the area of the pizza sections (non-shaded)
The angles of a equilateral triangle are always 60 degrees.
That means one of the pizza sections is 1/6 of an entire circle/
Now I will leave the rest up to you.
+128407 A regular hexagon with sides of 3" is inscribed in a circle. Find the area of a segment formed by a side of the hexagon and the circle.
The circle will also have a radius of 3
The area of the circle will be pi (3)^2 = 9 pi units^2 (1)
The area of the hexagon = 6 (1/4) (3)^2 * √3 = 27√3/4 units^2 (2)
So....the area we want is just 1/6 of (1) - (2) =
[ 9pi - 27√3/4 ] / 6 =
[ 36 pi - 27√3] / 24 =
[ 12pi - 9√3] / 8 units^2 =
(3/8) [ 4pi - 3√3 ] units ^2
+794 i do not under stand this
do you see
i am on the internet
but do not have this
on the internet do you see
so do not hesitate to give me a few pointers please
or just help me out do you see
(read this as poetry)
You do have a gift, Travisio. You’ve actually found something you are worse at than math and German: Poetry.
+794 That is very kind geust
...but not to burst your bobble that was my first atempt to poety and it does not rhyme like i wanted...
+128407 For the last problem the area of the unshaded portion of the equilateral triangle is just the area of a half circle with a radius of 3 = (1/2)pi * 3^2 = (9/2)pi units^2
[ To see this.....each of the congruent unshaded sections is just the area of a 60° sector of a circle with a radius of 3 ]
So we have pi (3)^2 ( 180/360) = (1/2) pi * (3)^2 = (9/2)pi units^2
And the area of the equilateral triangle = (1/2) (6)^2 √3 / 2 = 9√3 units ^2
So....the shaded area is just
[9√3 - (9/2) pi] units^2 =
9 [ √3 - pi/2 ] units^2
