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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

 Sep 12, 2019
edited by travisio  Sep 12, 2019
 #1
avatar+103858 
+3

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

 

 

cool cool cool

 Sep 13, 2019
edited by CPhill  Sep 13, 2019
 #2
avatar+1533 
+3

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.

 Sep 13, 2019
 #3
avatar+103858 
+3

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 

 

 

cool cool cool

 Sep 13, 2019
 #4
avatar+789 
+1

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)

 Sep 13, 2019
 #10
avatar
0

You do have a gift, Travisio. You’ve actually found something you are worse at than math and German: Poetry.

Guest Sep 13, 2019
 #11
avatar+789 
+1

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...

travisio  Sep 13, 2019
 #12
avatar+104382 
+1

It is good to see you experimenting and trying things Travisio.

 

Guest is just haveing a bad day     wink

Melody  Sep 20, 2019
 #13
avatar+789 
0

thank you and welcome back

travisio  Sep 20, 2019
 #5
avatar+103858 
+1

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

 

 

cool cool cool

 Sep 13, 2019
 #6
avatar+789 
+2

thank you Chris

travisio  Sep 13, 2019
 #7
avatar+103858 
0

OK, travis.....I hope my answers were clear  !!!!

 

 

 

cool cool cool

CPhill  Sep 13, 2019
 #8
avatar+789 
+2

yes thank you they helped me sovle the next 17 questions

travisio  Sep 13, 2019
 #9
avatar+103858 
0

OK....good deal   !!!!

 

cool cool cool

CPhill  Sep 13, 2019

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