Can you find the Area and Perimeter of the Triangle? | (Step-by-step explanation) |

[@soniamariadasilveira7003]

Amei a questão, professor, meio difícil, mas com a sua explicação ficou fácil. Obrigada, saudações de Bauru SP

[@PreMath]

O prazer é meu! Você é incrível. Continue sorrindo👍 Amor e orações dos EUA! 😀

[@HappyFamilyOnline]

Good one 👍 Thanks for sharing 😊

[@LogicQuest]

Thank you for this question. Gratitude.

[@marioalb9726]

Side of equlateral triangle: S = 4. r + 2 . r / tan (60°/2) S = 4 . 8 + 2 . 8 / tan 30° S = 59,713 cm Perimeter of triangle: P = 3 . S P = 179,14 cm Area of equilateral triangle: A = √3/4 S² A = 1543,96 cm²

[@ybodoN]

Generalized: the side of the equilateral triangle is _2r (√3 + n − 1)_ where _r_ is the radius of the circles and _n_ is the number of circles fitted along one side of the triangle.

[@Copernicusfreud]

Yay! I solved it. I came up with the approximation of the area of the triangle using Heron's formula, which was approximately 1544 square units.

[@tombufford136]

After some thought I focused on the lower left corner angle BAC is 60 degrees and the angle to the center of the lower left circle, D, is 30 degrees from symmetry. Forming a triangle ADE where E is the intersection of a perpendicular from D and the horizontal from A. Distance AE is 8/tan(30)=13.85.Distance AB = 2*13.85+32 =59.72 ,so the perimeter =3*59.72 =179.14. The height of ABC is 0.5sqrt(3) * 59.72. using 1/2 base * height . 0.5sqrt(3)*59.72 *59.72*0.5 = area=1,544. Looking forward to the video

[@unknownidentity2846]

Here we go: . .. ... .... ..... From the center of the lower left circle three lines are drawn: one to point A and the other two to the points of intersection of this circle and the triangle. Now we have two congruent 30°-60°-90° triangles. Since the length of the shorter cathetus is the radius R of the circle, we can easily calculate the distance between A and the points of intersection of the lower left circle and the triangle: R/d = tan(30°) = sin(30°)/cos(30°) = 1/√3 ⇒ d = √3R The calculation of the remaining properties is now not difficult anymore: AB = AC = BC = 2d + 4R = 2√3R + 4R = 2(√3 + 2)R P(ABC) = 3*AB = 6(√3 + 2)R = 48(√3 + 2) A(ABC) = (√3/4)*AB² = (√3/4)*2²(√3 + 2)²R² = √3(3 + 4√3 + 4)R² = √3(7 + 4√3)R² = (12 + 7√3)R² = 64(12 + 7√3) Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@quigonkenny]

Let the center of the top circle be O, the center of the bottom right circle be P, and the center of the bottom middle circle be S. Let E and D be the points on BC that are tangent to P and O respectively, let G and F be the points on AB that are tangent with S and P respectively, and let H be the point on CA that is tangent with O. As ∠DCO = ∠DCH/2 or 30° and ∠ODC is 90°, ∆ODC and ∆CHO are congruent 30-60-90 triangles. By symmetry, the same is true for ∆BEP and ∆PFB. Triangle ∆ODC: O/H = sin 30° 8/CO = 1/2 CO = 8(2) = 16 A/H = cos 30° DC/16 = (√3)/2 DC = 8√3 As ∠PED and ∠EDO are each 90°, ED = OP = 4(8) = 32. BC = BE + ED + DC BC = 8√3 + 32 + 8√3 = 32 + 16√3 By observation, as ∠CSB is 90°, ∠SBC = 60°, and ∠BCS is 30°, ∆CSB is a 30-60-90 triangle. Triangle ∆CSB: O/H = sin 30° SB/(32+16√3) = 1/2 SB = (32+16√3)/2 = 16 + 8√3 A/H = cos 30° CS/(32+16√3) = (√3)/2 CS = (32+16√3)(√3)/2 = (48+32√3)/2 CS = 24 + 16√3) Triangle ∆ABC: Perimeter = (3)BC = 3(32+16√3) = 96 + 48√3 Area = AB(CS)/2 = (32+16√3)(24+16√3)/2 Area = (16+8√3)(24+16√3) Area = 384 + 256√3 + 192√3 + 128(3) Area = 768 + 448√3

[@williamwingo4740]

Once you have the side of the big triangle as 32 + 16 sqrt(3) you can also get the area by (1/2)(base)(altitude).. The altitude is (16 + 8 sqrt(3))(sqrt(3)) = 24 + 16 sqrt(3) so we have (1/2)(24 + 16 sqrt(2))(32 + 16 sqrt(3)) = (12 + 8 sqrt(3))(32 + 16 sqrt(3)) = 384 + 384 + (192 + 256)(sqrt(3) = 768 + 448 Sqrt(3); so it comes out the same. 🤠

[@じーちゃんねる-v4n]

Outer triangle/inner one=2+√3 perimeter=48(2+√3) area=64√3(2+√3)^2=64(12+7√3)

[@cyruschang1904]

tan30⁰ = sin30⁰ ÷ cos30⁰ = 1/2 ÷ ✓3/2 = 1/✓3 radius of each circle = r = 8 The length of each side of the triangle = 4r + 2x8✓3 = 4x8 + 2x8✓3 = 16(2 + ✓3) = 32 + 16✓3 the height of the triangle = (32 + 16✓3) ÷ 2 x ✓3 The perimeter of the triangle = 3x16(2 + ✓3) = 48 (2 + ✓3) = 96 + 48✓3 The area of the triangle = [16(2 + ✓3)] [16(2 + ✓3) ÷ 2 x ✓3 ] ÷ 2 = (2 + ✓3) (2 + ✓3) x 64 x ✓3 = (7 + 4✓3) x 64 x ✓3 = 64(12 + 7✓3)

[@wackojacko3962]

I commented on yesterday's episode no matter what 30⁰ 60⁰ 90⁰ Right Triangle you may have you need to apply the Pythagorean Theorem along with the known facts that the Hypotenuse is Twice the Shorter side opposite 30⁰and that the Longer side opposite 60⁰ is a multiple of the Shorter side times 3⅓. Very few triangles are special like this. 🙂

[@PreMath]

Excellent! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@misterenter-iz7rz]

S=8×4+2×(8×sqrt(3))=32+16sqrt(3)=16(2+sqrt(3)), therefore the perimeter is 48(2+sqrt(3))=179.14 approximately, the area is (1/2)16^2(2+sqrt(3))^2sqrt(3)/2=64(12+7sqrt(3))=1544 approximately 😊

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@giuseppemalaguti435]

È un triangolo equilatero...risulta h=24+16√3...e l=2h/tg60=16(√3+2)...perciò A=bh/2=64(7√3+12)...spero che non ci siano errori

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@misterenter-iz7rz]

First like adder.🎉

[@PreMath]

You rock!

[@DDX01]

I have solved by the very same method

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@arnavkange1487]

First time i saw so many circles inside a triangle 😂😂

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍