how to solve this? | Mathematics | Page 1

fierce pivot Jan 31, 2023, 8:21 AM

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itd be nice to know how to solve these type of questions

jade monolith Jan 31, 2023, 8:55 AM

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I have an idea. Look, we can find area of triangle BAC as BA multiplication on BC multiplication on sin(ABC) * 1/2, and perhaps it`s the answer if we prove that the small white triangle is equal to the small yellow one(in a trapezoid ADCE)

fierce pivot Jan 31, 2023, 9:03 AM

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oh! i got it now thanks!

jade monolith Jan 31, 2023, 9:04 AM

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fierce pivot oh! i got it now thanks!

No problem, you`re welcome

fierce pivot Jan 31, 2023, 9:04 AM

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wait holdon is BAC 39.6? just making sure

jade monolith Jan 31, 2023, 9:06 AM

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yes, that`s right

fierce pivot Jan 31, 2023, 9:07 AM

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woo! tyty

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i solved the rest until 14 and got stuck on 15 ahh

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im not familiar with circles

jade monolith Jan 31, 2023, 9:11 AM

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I know how to solve this, but I`m not sure that it is the most optimal

fierce pivot Jan 31, 2023, 9:12 AM

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well, if it works, it works right?
a challenge is nice so

broken coral Jan 31, 2023, 9:13 AM

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ABC is a sector I think

jade monolith Jan 31, 2023, 9:13 AM

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ok, one minute, I will open the paint

broken coral Jan 31, 2023, 9:13 AM

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Area of shaded region is area of semi circle - area of sector

fierce pivot Jan 31, 2023, 9:15 AM

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im pretty sure the area of the semi circle is 56.549

broken coral Jan 31, 2023, 9:16 AM

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56.5

fierce pivot Jan 31, 2023, 9:16 AM

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yep

broken coral Jan 31, 2023, 9:16 AM

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Then subtract sector area

fierce pivot Jan 31, 2023, 9:17 AM

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shouldnt a sector start from O?

broken coral Jan 31, 2023, 9:17 AM

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hmm

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Right sorry

fierce pivot Jan 31, 2023, 9:18 AM

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because a sector is supposed to be formed by the circles two radii

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wait holdon

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i could connect O to A
find the area if sector BOA
and subtract triangle BOA

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i would need angle BOA though

broken coral Jan 31, 2023, 9:23 AM

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Construct a line a to c

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It will be 90°

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Angle inscribed in a semi circle

fierce pivot Jan 31, 2023, 9:26 AM

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ah wait isnt it simple to just

fierce pivot Jan 31, 2023, 9:26 AM

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fierce pivot i could connect O to A find the area if sector BOA and subtract triangle BOA

do this

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angle BOA is 130 isnt it?

jade monolith Jan 31, 2023, 9:26 AM

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@fierce pivot

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if you have questions - go ahead

fierce pivot Jan 31, 2023, 9:29 AM

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ohhh i think i got it

jade monolith Jan 31, 2023, 9:30 AM

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I used formula sin(a+b)

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which is equal sin(a)cos(b)+sin(b)cos(a)

fierce pivot Jan 31, 2023, 9:31 AM

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the answer is abt 6.52 then

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if i didnt miscalculate

jade monolith Jan 31, 2023, 9:32 AM

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I never count if I can not do it

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This is why my decisions are often wrong

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joke, my decisions cannot be wrong because no one checks them

fierce pivot Jan 31, 2023, 9:36 AM

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pfft its fine

broken coral Jan 31, 2023, 9:37 AM

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I got 23.2

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Area

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Dk if correct

fierce pivot Jan 31, 2023, 9:38 AM

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idk then aha

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ill just bet on my luck

broken coral Jan 31, 2023, 9:38 AM

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It is near 23 or 24

fierce pivot Jan 31, 2023, 9:38 AM

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oo okok

jade monolith Jan 31, 2023, 9:39 AM

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broken coral I got 23.2

How did you count?

broken coral Jan 31, 2023, 9:40 AM

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My way I used angles inscribed in a semi circle

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Then construct O to A

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Sector

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Area of segment = area of sector - area of triangle

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Lemme count again

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Yes I am getting 23.2 in my method

fierce pivot Jan 31, 2023, 9:42 AM

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oh i think i understand

broken coral Jan 31, 2023, 9:43 AM

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6.52 is very small when compared to the semi circle

jade monolith Jan 31, 2023, 9:44 AM

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broken coral Area of segment = area of sector - area of triangle

I understand what you mean, now I will check the calculations

fierce pivot Jan 31, 2023, 9:44 AM

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broken coral 6.52 is very small when compared to the semi circle

yup

broken coral Jan 31, 2023, 9:45 AM

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It will be an equilateral triangle

jade monolith Jan 31, 2023, 9:45 AM

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broken coral Area of segment = area of sector - area of triangle

soo.. you got piR²/3-sqrt(3)R²/2 too

broken coral Jan 31, 2023, 9:46 AM

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Yes

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PiR^2/6

jade monolith Jan 31, 2023, 9:46 AM

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360/120

broken coral Jan 31, 2023, 9:47 AM

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60/360 ?

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Area of equilateral triangle √3/4 a^2

jade monolith Jan 31, 2023, 9:49 AM

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there is a small misunderstanding, I will now draw your method and you will tell me if I understood you correctly, okay?

broken coral Jan 31, 2023, 9:49 AM

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I think I got my mistake

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When I join A to C

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It is 90°

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By 30 60 90 theorm

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You will get the side ac

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Then you can find the area of sector

jade monolith Jan 31, 2023, 9:53 AM

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broken coral Jan 31, 2023, 9:53 AM

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no no

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See join A And c

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So angle BAC IS 90°

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Now join O to A

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so OA = 6

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Now in triangle OAC

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It is equilateral

jade monolith Jan 31, 2023, 9:55 AM

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if I understood you correctly, you will find the area of the arc AC

broken coral Jan 31, 2023, 9:55 AM

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Find we find the area of triangle then sector

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Then subtract area of triangle - area of sector

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Will give us the area of segment

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Now in Triangle BAC area = 1/2 × base × height

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By 30 60 90 theorm

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AB = 6√3

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Area of triangle ABC is 36√3

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Area of shaded region = area of semi circle - area of arc ACB

jade monolith Jan 31, 2023, 9:59 AM

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woooy men you confused your solution, I'm not saying that the solution is bad, but it's much harder to find a mistake in such I don't know exactly where your mistake is, but do you agree with this decision?:

jade monolith Jan 31, 2023, 9:59 AM

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jade monolith

.

broken coral Jan 31, 2023, 10:00 AM

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Saob means area of aob?

jade monolith Jan 31, 2023, 10:00 AM

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yes, area of triangle

broken coral Jan 31, 2023, 10:01 AM

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@jade monolith ok

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I guess

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Should I send you my work

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So you understand

jade monolith Jan 31, 2023, 10:02 AM

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sure why not

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but I may go soon

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to the lesson

broken coral Jan 31, 2023, 10:02 AM

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Ok

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I get 34.45 as the area mm

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broken coral Jan 31, 2023, 10:33 AM

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Subtract semi area from this

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56.1-34.45

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21.65

jade monolith Jan 31, 2023, 12:45 PM

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broken coral

area of ABC is equal to 31 right?

broken coral Jan 31, 2023, 12:46 PM

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No

jade monolith Jan 31, 2023, 12:46 PM

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18sqrt(3) yes

broken coral Jan 31, 2023, 12:46 PM

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Yea

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By 30 60 90 theorm we get AB √108

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Or 6√3

quartz wren Jan 31, 2023, 12:47 PM

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Oh, what are you trying to solve?

broken coral Jan 31, 2023, 12:47 PM

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Area of shaded region

jade monolith Jan 31, 2023, 12:48 PM

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quartz wren Oh, what are you trying to solve?

Just trying to found mistake

broken coral Jan 31, 2023, 12:48 PM

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broken coral

I am getting 21.65 cm^2

quartz wren Jan 31, 2023, 12:48 PM

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Ah, area of segment?

jade monolith Jan 31, 2023, 12:51 PM

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@broken coral I understand what are you doing there, but can you rewrite numbers please?

broken coral Jan 31, 2023, 12:52 PM

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Ok

quartz wren Jan 31, 2023, 12:52 PM

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Well, the area of segment of radius R and angle θ is:
S(segment) = (1/2)R^2 (θ - sin(θ))
In our case R = 6 cm, θ = 2π/3. So:
S(segment) = (1/2)(6 cm)^2 (2π/3 - sin(2π/3)) = (12π - 9√(3)) cm^2 ≈ 22.11 cm^2

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What did you get?

broken coral Jan 31, 2023, 12:55 PM

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broken coral Jan 31, 2023, 12:55 PM

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quartz wren Well, the area of segment of radius R and angle θ is: S(segment) = (1/2)R^2 (θ -...

I get 21.65 cm^2

quartz wren Jan 31, 2023, 12:56 PM

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broken coral I get 21.65 cm^2

That doesn't really help. What is the exact answer you got?

jade monolith Jan 31, 2023, 12:57 PM

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guys, do you remember problem? we found area of segment, good, not we need to subtract the area of the triangle

quartz wren Jan 31, 2023, 12:58 PM

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Wait, wait. If you've found the area of the segment, that's already the answer.

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Or did you mean the sector?

broken coral Jan 31, 2023, 12:58 PM

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I found area of shaded region

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21.65

quartz wren Jan 31, 2023, 12:58 PM

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What is the exact result?

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Rounded results can't be checked.

broken coral Jan 31, 2023, 12:58 PM

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jade monolith guys, do you remember problem? we found area of segment, good, not we need to su...

We want the area of segemnt so we subtract the area of sector - area of triangle

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@quartz wreni got the exact valueb

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21.65

quartz wren Jan 31, 2023, 12:59 PM

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That is not possible.

broken coral Jan 31, 2023, 12:59 PM

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Area of sector I got 34.45

quartz wren Jan 31, 2023, 12:59 PM

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Can you show how you got it?

broken coral Jan 31, 2023, 12:59 PM

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I did above

quartz wren Jan 31, 2023, 1:00 PM

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I don't really understand how you did it there.

broken coral Jan 31, 2023, 1:00 PM

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broken coral

.

quartz wren Jan 31, 2023, 1:00 PM

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I recommend just tackling the general case: radius R, angle θ.

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That would be better.

broken coral Jan 31, 2023, 1:01 PM

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I used 30-60-90 theorm

quartz wren Jan 31, 2023, 1:01 PM

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Well, finding the central angle is easy, that isn't really the focus.

broken coral Jan 31, 2023, 1:01 PM

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Yeah

jade monolith Jan 31, 2023, 1:01 PM

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broken coral I used 30-60-90 theorm

you can use it, and I'm sure you can go that way too, but it's easier to get confused in that way

broken coral Jan 31, 2023, 1:02 PM

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I just did what came in my mind my answer might be wrong

quartz wren Jan 31, 2023, 1:02 PM

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Well, let me show how to solve this generally, then.

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We need to find the area of a segment.
As we've established:
S(segment) = S(sector) - S(triangle)
We know:
S(sector) = (1/2)R^2 θ
S(triangle) = (1/2)R^2 sin(θ)
So:
S(segment) = S(sector) - S(triangle) = (1/2)R^2 θ - (1/2)R^2 sin(θ) = (1/2)R^2 (θ - sin(θ))

#how to solve this?