Question

if the radius of circle G is 10 and measure of angle AGB is 72, calculate the area of the sector in circle G formed by the segments AG and GB

Answer 1

@vivian_5 Are you sure angle AGB is 72 degrees?

Answer 2

yes!

Answer 3

Angle AGB is obtuse, How could it be 72 degrees? @vivian_5 I'm sure there is a mistake.

Answer 4

A) 72 pi
B) 20 pi
C) 100 pi
D) 2 pi

Answer 5

im not sure, that what the problem states. and those are the answer options

Answer 6

whats the answer?

Answer 7

@genius12 i wouldn't be so quick to jump the gun. a triangle can still be formed with that angle being acute.

could just be a drawing that re-used over and over in different editions and places

Answer 8

Hmm possibly. But the diagram should atleast adhere slightly to the information but it doesn't at all.

Answer 9

i agree. i've seen so much worse though :P lol

Answer 10

do you have a guess on what the answer might be?

Answer 11

Are we finding the area of the triangle AGB? @vivian_5

Answer 12

the area of the sector formed by the segments AG and GB, picture above

Answer 13

Yes so that would be the triangle AGB correct?

Answer 14

yeah

Answer 15

It appears to be the area inside the circle only

Answer 16

i know, whats the answer you come up with?

Answer 17

since AGB is 72 degrees, that segment there is 72/360 of the total area of the circle.

so the area of the circle is pi(r^2) = 100pi

\[100 \pi \times \frac{ 72 }{ 360 } = 20\pi\]

Answer 18

thanks!

Answer 19

glad i could help. would have helped sooner but what elbows deep in chicken wings lol. was typing slowly with pinkies

Answer 20

was*

Answer 21

can you help with one more?

Answer 22

if DEF = 97, calculate DBF.

Answer 23

Ok so by the tangent angnle theorem, we can construct the perpendicular line GE. This line forms a 90 degree angle with AEB hence GE = radius = 10 is the height of the triangle AGB. After constructing GE, we get 2 right triangles; GAE and GBE. Because we know that angle GEB is 90 degrees and is the right bisector of line AEB, then angles EGB = AEG = half of 72 = 36 degrees. Now this means that the angles GBE = GAE = 180 - (36 + 90) = 54 degrees. We can now use our trig ratios to find the base of this triangle.\[\bf \tan(54)=\frac{ GE }{ EB }=\frac{ 10 }{ EB } \implies EB = \frac{ 10 }{ \tan(54) } \]We get a same result for segment EA. Hence the area of the triangle AGB is:\[\bf Area \ of \ \triangle AGB=\frac{ GE(BE+EA) }{ 2 }=\frac{ 10(2BE) }{ 2 }=\frac{ 100 }{ \tan(54) } \approx 72.654\]

Answer 24

@Euler271 @vivian_5 OMGGGGGGGGGGG NOW I GET IT. THE QUESTION WAS ASKING FOR THE AREA OF SECTOR OF THE CIRCLE ENCLOSED BY LINES AG AND GB. I thought it was the area of the triangle AGB. Fmlllll. lols.

Answer 25

That's so damn easy no wonder it took me a while to get to the answer rofl....

Answer 26

@vivian_5 Notice that BFED is a cyclic quadrilateral and the opposite angles of a cyclic quadrilateral always add up to 180 degrees. Hence:\[\bf \angle DEF + \angle DBF=180 \implies 97+ \angle DBF = 180 \implies \angle DBF =83\]

Answer 27

@vivian_5 The above answer is 83 degrees for question 2 that you posted. For the very first question, @Euler271 is correct which is 20pi. I was thinking that you wanted the area of the triangle AGB instead of the sector of the circle enclosed by the 2 line segments lol.

Answer 28

calculate the area of JIM
HELP

Answer 29

A) 26 units squared B) 58.1 C) 52 D)116.3

Answer 30

@vivian_5 You should atleast attempt some of these questions so that you can learn how to solve if one of us isn't here. Regardless, here is how it would be worked out:

We are given that \(\bf \triangle JIM\) has sides 12, 18, 21. We are given no other information other than the side lengths and we are asked to calculate the area. There is no better situation than this to apply Heron's Formula for the area of a triangle with 3 side lengths given. You can get more information about the formula here:

www.mathsisfun.com/geometry/herons-formula.html‎

Using the formula and the 3 side lengths we get that:\[\bf s = \frac{ IJ+JM+MI }{ 2 }=\frac{ 13+18+21 }{ 2 }=26\]And now to get the area we do:\[\bf Area \ of \ \triangle JIM=\sqrt{s(s-13)(s-18)(s-21)}=\sqrt{26(13)(8)(5)}=52\sqrt{5}\]
@vivian_5

Answer 31

@Euler271