Can You Find the Area of the BIG Circle? | Quick and Easy Tutorial

[brucelee]

I’m almost 60 years old.
The 45-45-90 triangle rule and the 30-60-90 triangle rules are two rules that I have never forgotten.
45-45-90 [side x sq.root 2 = hypotenuse]
30-60-90 [short leg x sq.root 3 = long leg] {short leg x 2 = hypotenuse}

Thank you sir.

[kenzalhunter]

I'm just curious why you approximated the equations before getting the final exact answer: A = π(7(1+√2))² = π49(3+2√2).

The earlier you approximate irrationals, the further from true the final solution gets.

[easy_s]

I wouldn't approximate any expressions like √2 until you reach the final answer for precision purposes. In this case R=7+7√2 and so the area A=πR² =π(7+7√2)² =π(147+98√2) =49π(3+2√2) which approximates 897.2 and not 896.8. Although since r is given as 7 and not 7.0 the final answer in both cases would have to be A=897.

[SergeCeyral]

the four centers of small circles build a square : L= 14 for each side. So, the length of square diagonal is 14*sqrt(2)
So the radius of the large circle is : 7 + 7*sqrt(2) = 7*(1+sqrt(2))
So the surface of large square : S = pi * 49*(1+sqrt(2))²
done

[mrminer]

Do we need to pin down an additional hypothesis of the centers of the 4 circles being at the corners of a square? They LOOK like they are, but we don't have enough geometrical info to be sure of that.

[luctheriault]

Surprisingly simple solution to an seemingly complex problem. Great explanation!! From Canada.

[nigelsw]

I used essentially the same method. Joined the four centres to form a square of side 2r. Used Pythagoras to calculate the diagonal and then added two radii to give the diameter of the larger circle.

[schoenof]

Actually it is approximately 897.2 .... you started approximating values before you got to the end and lost more precision than needed....

[quigonkenny]

As the smaller circles are arranged on a square formation, their centers form a square. Given that each is tangent both with the large circle and with two of the others, we know this is symmetrical.

If we connect each of the circle centers into a square, we see thst each side is 2r = 14. Form a diagonal of the square. This forms two isosceles right triangles, with the diagonal length being √2s = 14√2.

By observation, the diameter of the large circle is the diagonal of the square plus 2 radii of the smaller circles. The radius of the large circle is half of that, or 7√2 + 7.

Area = πr² = π(7√2+7)²
Area = π(98+98√2 + 49)
Area = 49π(3+2√2) ≈ 897.22

[suniljoshi]

Thanks.Let me understand if 8 circles are to be constructed, what will be R for bigger circle

[jarikosonen]

A=897.2 ... Looks like it collects a lot of rounding errors if rounded too early.

[ezioauditore]

I solved it by calculating the hypotenuse of the right isosceles triangle formed by the vertices A (center of small circle); B (center of large circle); C (point of tangency between two small circles) from which it is deduced that the cathetus measure 7. By adding this hypotenuse to the radius 7 I find the radius of the large circle.

[MrPaulc]

I made a smaller square linking the centre of a small circle with the centre of the large circle as a square's diagonal. This gave 7root2, so the radius of the large circle is 7 + 7root2. This allowed me to round at the end for slightly more accuracy:
((7 + 7root2)^2)*pi

[jonathancapps]

Couldn't you construct a square with the center of a small circle and the center of the big circle as opposite corners? Then R=r+✓(r²+r²). Using all four shaded circles to construct the larger square just adds needless complexity that must be undone later by dividing everything by 2.

In fact 3 of those shaded circles might be extraneous, unless they are deeded to show that the 2 larger radii tangent to one of the shaded circles form a right angle. My geometry class was nearly 30 years ago, but I feel like proving that without the other 3 circles should be possible.

[benjaminkarazi]

Hello,
This is a standard model of four circles in a circle as shown, which has a standard formula, R=r.(1+√2).
R, Large circle radius, r, Small circles radiuses
R=r.(1+√2)
A=R².π
A=[7. (1+√2)]².π
A=[7.(1+1.4142)]².π, A=897.216
Regards,

[j.r.]

Since my comment was deleted along with the previous version of this video, let me repeat that I found it simpler just to use the Pythagorean Theorem. We have a right triangle with sides of 14 and 14, so we can quickly calculate its hypotenuse, which is also the diagonal of the inner square. Then the radius of the big circle is just half the diagonal plus 7.

[johankotze]

I agree with your answer, but I fail to see how you proof the angles at the small circles' centres to be right angles, and that is crucial to you line of arguing. We must proof that the two green radii are at a right angle.

[sunilmhapankar]

hello, how did u get side of Internal square as 25? it shud be r+r = 7+7 = 14
I am getting R = 7+7√2

[aasdon]

I think its area is 897.216648 as its radius(bigger circle) is (7√2+7)👍

[Escviitash]

As the formula for a circle includes the raduis and not the diameter, it would be easier to calculate R directly by constructing a square with the two opposite coners being the origin (O) of the large cirkel and D, a square with site length r.
Then OD = sqrt(2(r^2) and R = OD + r