#Areas related to Circles
dim gazelleBOT
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ionic veldt
Show your work.
I actually didn't know that formula, I just used Heron's formula, but yeah.
I mean, it simplifies into that when you take a = b = c.
inland copper
Wait, Heron? But this is an equilateral triangle.
ionic veldt
Wait, Heron? But this is an equilateral triangle.
Because I didn't know the specific form of the rule for equilateral triangles.
inland copper
Because I didn't know the specific form of the rule for equilateral triangles.
Oh... Well, ok.
For an equilateral triangle with side a we have S = a^2 √(3)/4. That can be shown from the usual S = (1/2)xy sin(θ).
The region is the difference of the triangle and three identical sectors.
Hm, let me try.
inland copper
We have:
S(shaded) = S(triangle) - 3S(sector)
S(triangle) = √(3)(2r)^2/4 = √(3)r^2
S(sector) = (1/2)(π/3)r^2 = (π/6)r^2
That doesn't give a negative area.
ionic veldt
I get a positive number.
inland copper
What you you mean? The area of sector of radius r and angle θ is (1/2)θr^2.
What other formula are you using?
ionic veldt
That's just conversion. Degrees vs. radians.
inland copper
That's the same thing, you just forgot the degree sign.
Wait, hold up.
Why use 22/7?
ionic veldt
Did you actually plug it into a calculator and get something negative?
inland copper
I mean, in any case we get S(shaded) = (√(3) - π/2)r^2, no matter what formula you use.
ionic veldt
So then how do you know it's "deffo negative"?
inland copper
Oh, hold on.
We are given the area of the triangle.
No matter, not a problem.
Let S(triangle) = S. Then:
S = √(3)r^2
Then r^2 = S/√(3). So:
S(shaded) = (√(3) - π/2)(S/√(3)) = (1 - π/(2√(3)))S
Still positive, of course.
I don't think you're using that formula correctly. Just use the usual one.
Doesn't mattter.
It's the correct and valid formula.
Huh...
That teacher needs someone to talk to them, then.
That isn't ok.
The standard formula is (1/2)θr^2.
Ugh...
Fine.
Let me show with that formula, then...
S(sector) = (1/2)θr^2 = (π/360°)θr^2
So:
S(shaded) = S(triangle) - 3S(sector)
S(triangle) = √(3)(2r)^2/4 = √(3)r^2
S(sector) = (π/360°)(60°)r^2 = (π/6)r^2
Then it's the same.
What your teacher is doing is kinda stupid, to be honest. Just my opinion, though.
Oh.
Though, I'm not sure why they want you to approximate √(3) and π.
Yes?
You're welcome!
Still not sure why the approximation is needed, though...
What do you mean? √(3) is √(3).
Wait, hold on.
Where's π?
Let me calculate the answer, hold on.
Oh...
Well, why do that?
Let's find the exact answer first.
S(shaded) = (√(3) - π/2)(S/√(3)) = (√(3) - π/2)(49√(3) cm^2/√(3)) = 49(√(3) - π/2) cm^2
That's the exact answer.
We can use a calculator to approximate it: S(shaded) ≈ 7.901 cm^2
The concrete answer is the exact one, not the approximate one.
As you can see, when you first calculate it and then round it, you get the correctly rounded result.