Circular Measurements (1 Viewer)

[sasquatch]

Theres a question in my book that states: "What is the area of the major segment of a circle in which the minor segment subtends a central angle of 5pi/6 and the radius is 7cm?"

For this question i applied the formula A = 1/2(ab)sinC for the major arc:

A = 1/2(7)²sin5pi/6
= 12.25 cm²

Then answer says it is 102.047 cm². I checked the area of the entire sector and it is only 64.14 cm². Did i do something wrong or is the answer wrong.

Also just underneath it another question reads: "What is the area of the minor segment cut off a circle of radius 10 cm by a chord of length 12 cm?". I got this as 16.3501 cm² which is the same as the answer in the back of the book, but it extends the question by asking:

"Check your answer by the approximate formula: area of minor segment = 2/3 (base) x (height)."

Alright, the base is 12 cm, but how the hell do you find the height of the minor segment.

 

[insert-username]

Theres a question in my book that states: "What is the area of the major segment of a circle in which the minor segment subtends a central angle of 5pi/6 and the radius is 7cm?"

For this question i applied the formula A = 1/2(ab)sinC for the major arc:

I think you're confusing arc, segment, and triangle. The formula A = 1/2(ab)sinC is used to find the area of a triangle, as far as I know, not for circle arcs. Anyway:

The area of the circle is 49pi units².

The area of the minor segment is equal to 1/2r²(@ - sin @)

= 1/2 x 7² x (5pi/6 - 0.5)

= 24.5 x (2.117993878)

= 51.89 cm²

The area of the major segment = A(circle) - A(minor segment) [since if you have a minor segment, you have a major segment]

= 49pi - 51.89

= 102.047 cm²

Thus, the book's answer is correct. I'm looking at the second question now, but for next time, make sure you're using the right formula.

Also just underneath it another question reads: "What is the area of the minor segment cut off a circle of radius 10 cm by a chord of length 12 cm?". I got this as 16.3501 cm2 which is the same as the answer in the back of the book, but it extends the question by asking:

"Check your answer by the approximate formula: area of minor segment = 2/3 (base) x (height)."

Alright, the base is 12 cm, but how the hell do you find the height of the minor segment.

Imagine a triangle with its base being the same as the minor segment with its vertex at the center of the circle. The two sides of this circle are radii and are 10 cm long. Now, draw a line from the centre of the base up to the center. You now have a right-angled triangle with sides of B, 10cm, and 6 cm. Use Pythagoras' theorem to calculate the missing side, then subtract that side from the radius and you get the height. The attached diagram should help.

A = 6cm
C = 10cm

Therefore B = 8cm (Pythargoras theorem)

Therefore height of minor segment = 2 cm

Therefore approximate area = 2/3 x 2 x 12

= 16


I_F

 

[sasquatch]

hahaha no no my problem was i didnt know what a major segment was. I was thinking the major segment was the section within a sector other then the minor segment. But i see now that the major segment is the rest of the circle apart from the minor segment, not just the sector. Thanks for that though.

 

[Templar]

Wouldn't height just be the radius in this approximation?

 

[insert-username]

Wouldn't height just be the radius in this approximation?

Nope, the segment isn't as high as the radius - see my edited-in diagram and working.


I_F

 

[sasquatch]

No, well the answer says the approximate area is 16 cm² where the exact (well to 4.d.p) is 16.3501 cm² .

Working backwards:

16 = 2/3(12)h
16 = 8h
h = 2 cm

So...i dunno then...

OH just as i write this i think i figured it out...

the height of the triangle formed by the chord and two radii arms is:

10^2 = 6^2 + h^2
h^2 = 10^2 - 6^2
.:. h = 8 cm

But the diameter is 2r = 20

Two triangles place above each other would give a combined height of 16cm, so that the remaining length is 4cm (the height of 2 sectors)

4/2 = 2cm giving the height of the single sector.

 

[SoulSearcher]

the height would be the radius of the circle subtracted by the perpendicular height of the triangle, since it is asking for the height of the minor segment

 

[sasquatch]

oh well i see we both figured it out... thanks anyway though!!!

oh well i ALSO see that there was no point doing the diameter...just the radius is sufficient...

i feel like an idiot..i do 4 unit..but i cant even do this simple 2 unit crap...

 

[sasquatch]

Hey um, just asking another question considering the graphs of trigonometric functions. y = 4cos x + 3sin x. I calculated the amplitude being 7root2 / 2 equaling 4.95 (corred to 2.d.p). The amplitude in the back of the book is given as being 5. Just wondering, is their answer due to rounding off or have i done something wrong?

By looking at the graph, i noticed the highest point would be at pi/2 so,

4cos(pi/2) + 3sin(pi/2) = 4/ root2 + 3/ root 2 = 7/ root2 = 7root2 / 2. Is this right..or did i do wrong?

 

[sasquatch]

Nevermind i found out for myself that yeah it is 5..just another question then.. How are you to calculate the amplitude?

 

[insert-username]

Nevermind i found out for myself that yeah it is 5..just another question then.. How are you to calculate the amplitude?

I don't think there's a formula to do it - you'd probably have to do a brief sketch through adding the ordinates, find the point, and work it out. I don't think 2U will give you a question like "Find the amplitude of 4 cos x + 3 sin x" - it'd be more like "Graph the function "4 cos x + 3 sin x". That's my take, anyway.


I_F

 

[sasquatch]

Well i dont have to worry bout the 2 unit test..... so i guess yeah...its just that because we graph manually its not very clear how close the curve ever gets 5. For example if the graph was y = cos x + sin x, the two graphs drawn seperately would intersect when x = pi/4 and it is evident that at that point, the y-value would be at its highest for the equation y = cos x + sin x. Hence the amplitude is 2/root 2 which equals approximately 1.4142..ect. If you did this from the graph it could seem as if the curve reaches 1.5, if you get what i mean.

 

[KeypadSDM]

Acos[x] + Bsin[x] = Csin[x + @]

Expand the right hand side, and solve the simultaneous equations. Then C is the amplitude.

Not sure if this is 2 or 3 unit, pretty sure it's 2 unit [easy enough to do]

 

[insert-username]

Ack, that's right, the transformations! I thought there may have been a link between 3² + 4² = 5², but I missed it completely. *hit head on keyboard*

It's 2-unit, I'm sure, Keypad. I remember now the looks of dismay and confusion in the class when our teacher went through it.


I_F