Question

A segment of a circle has a 120 arc and a chord of 8in. Find the area of the segment. No diagram.

Answer 1

Here's question 2
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:√2 to find the radius of the circle.) No diagram.

Answer 2

question 3
Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius. (Hint: A chord divides a circle into two segments. In problem 1, you found the area of the smaller segment.) No diagram.

Answer 3

Q4
Find the area of the shaded portion in the square.
(assuming the central point of the arc is the corresponding corner) Diagram posted.

Answer 4

Q5
Find the area of the shaded portion intersecting between the two circles. Diagram posted.

Answer 5

Q6
Find the area of the smaller segment whose chord is 8" long in a circle with an 8" radius. No Diagram,

Answer 6

Q7
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 1/4 s2 √3.) No diagram.

Answer 7

Q8
An equilateral triangle is inscribed in a circle with a radius of 6". Find the area of the segment cut off by one side of the triangle. No diagram.

Answer 8

Q9
Find the area of the shaded portion in the circle. Diagram posted.

Answer 9

my bad i didnt see this lol.

Answer 10

segment of a circle
the formula is \[r ^{2}(\Pi \Theta/360-\sin \Theta/2)\]
use this formula

Answer 11

Qns.1
Area of segment = area of sector -triangular portion
Chord length is 8 inches = 2 radius sin120/2
radius= 8/sqrt(3)
area of segment = (120/180) *pi*r^2 - (r^2 sin120) /2=44.68 - 9.23= 35.44

I apologize, i think you need to find exact form, but it is too tedious to simplify.

Answer 12

@triph the eq i hav given is the best 1

Answer 13

i think

Answer 14

Hmm yea your formula is actually just a rearrangement of mine. it is difficult to make out because of the way i write it.

Answer 15

qns2. a square inscribed in a circle, the diagonal of the square is the diameter of the circle. Given the side of the square is 6 inches.
The diagonal (using Pythagoras Theorem) sqrt(6^2 +6^2) = 8.5
Radius of circle = 4.25
Since the chord length is the side of the square, the central angle is 90 degrees. given all these, find area using the above equation .

Area = (@/360)pi r^2 - (r^2 sin@)/2 = (90/360)pi* 4.25^2 - (4.25^2 sin90)/2
work it out yourself thx.

Answer 16

qns1. i made a small error.
area of segment = (120/180) *pi*r^2 - (r^2 sin120) /2=44.68 - 9.23= 35.44
IT SHOULD BE
||
area of segment = (120/360) *pi*r^2 - (r^2 sin120) /2=44.68 - 9.23= 35.44

Answer 17

qns3. No idea what is the larger segment. There should be no larger segment.

Answer 18

qns4.
Shaded area in square = area of square - area of quadrant
= 6^2 - (1/4) pi 6^2

Answer 19

qns.5 shaded area = area of one segment *2

Given chord length and radius, find central angle

chord length= 2 r sin(@/2)
4 = 2 (4) sin(@/2)
@ = 60degrees

Area of segment = (@/360)pi r^2 - (r^2 sin@)/2 = (60/360) pi 4^2 - (4^2 sin60)/2

Answer 20

qns6. Don't know what is the smaller segment referring to, otherwise the method to solve is exactly similar to the above questions.

Given chord length and radius, find central angle, then substitute all into the formula to solve for area.

Answer 21

qns.7 With the chord length given to b 3 inches. The central angle is 360/6 = 60 degrees. Why because the hexagon is 6 sided. If the chord length is the side of a hexagon the central angle would be 1/6 of 360degrees.

Given chord length and central angle, find the radius.

Having theses 3 components find area with the formula used in the above questions.

Answer 22

For Q4\[A=Number Number -Number\]

Answer 23

qns.8
Given the radius of the circle is 8 inches. if the chord is one side of the triangle, the central angle is 360 /3 = 120 degrees.

Given central angle and radius, find the chord length.

Same idea, with these 3 components find area.

Answer 24

Foe Q9
\[A= Number Number \pi - Number \sqrt{Number}\]

Answer 25

I cant see what you typed. All i see is this

For Q4
A=NumberNumber−Number

Foe Q9
A=NumberNumberπ−NumberNumber−−−−−−−√

Answer 26

No you see for number 9 and 4 I guess thats how its suppose to look like

Answer 27

hmm so wat seems to be the problem. It is alright the way i see it.

Answer 28

never mind but thanks for the answers and I'll get back at you when I get stuck again.thanks.

Answer 29

Np. I will go sleep now. Im tired.