Can you find area of the Green Triangle? | Two circles | (Step-by-step explanation) |

[@wackojacko3962]

A very healthy and smooth solution. I absolutely love it! 🙂

[@PreMath]

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

[@aalvarma]

AD is parallel to BE, so a simple path is to define the segment AG = AD - BE = 2 - 1 = 1. AG^2 + GB^2 = AB^2. AG = 1, AB = 3, so GB = 2*2^0,5. Triangle ADC is similar to AGB, so DC is equal to 2*GB, or 4*2^0,5.

[@sarvajagannadhareddy1238]

Dear, All constructions are ingenious, exemplary, creative etc. CONGRATULATIONS !

[@marioalb9726]

A=πR² R=√(A/π) R=2cm ; r=1cm sin α = (R-r)/(R+r)=1/3 Triangle area: A = ½ b.h h = R = 2cm tan α = h/b b= h / tan α = 2/tan(asin1/3) b= 5,65685 cm A = 5,65685 cm² ( Solved √ )

[@Waldlaeufer70]

r(red) = 2 cm, r(blue) = 1cm AB = 2 + 1 = 3 1 : BC = 2 : (AB + BC) 1 : BC = 2 : (3 + BC) 2BC = BC + 3 BC = 3 AC = AB + BC = 3 + 3 = 6 DC² + r(red)² = AC² DC² + 4 = 36 DC² = 32 A(grün) = 1/2 * r(red) * DC A(grün) = 1/2 * 2 * √32 = √(16 * 2) = √16 * √2 = 4√2 cm²

[@Copernicusfreud]

That is how I solved the problem.

[@PreMath]

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

[@santiagoarosam430]

Razón de semejanza entre triángulos BEC y ADC; s=BE/AD=1/2》AC=2AB=2(2+1)=6》DC=sqrt(36-4)=4sqrt2》Verde =AD×DC/2=4sqrt2 Gracias y un saludo cordial.

[@andydaniels6363]

A handy formula for problems involving common tangents to a pair of circles: If M1, M2 and r1, r2 are the centers and radii, respectively, of two nonconcentric circles, their external similarity point is at (r2·M1 - r1·M2)/(r2 - r1). In this problem we can take M1=0, M2=BA=R+r=3, r1=R=2 and r2=r=1, so CA=6. We then obtain CD and hence also the triangle’s area via the Pythagorean theorem as in the video. Note that if you replace the subtractions in the above formula with additions you instead obtain the internal similarity point, here the point of tangency of the two circles.

[@murdock5537]

Nice! Draw a circle around B with radius 3 → AB = BD = CB = 3 → AC = 6 → ∆ DEB = ∆ BEC → sin⁡(δ) = 1/3 → cos⁡(δ) = √(1 - sin^2(δ)) = 2√2/3 → tan⁡(δ) = sin⁡(δ)/cos⁡(δ) = √2/4 = 1/CE → CE = DE = 2√2 → area ∆ ABC = (1/2)2(4√2) = 4√2 = (1/2)sin⁡(δ)6(4√2)

[@NoorAlam-jw1en]

You are great

[@PreMath]

Thanks dear ❤️

[@papanujian7758]

Very nice solution

[@PreMath]

Thanks ❤️

[@alexandrkushnir1380]

According to the video AD=R=2, BE=r=1, AD Ʇ DC, BE Ʇ EC. If each of two lines perpendicular to a third one, they are parallel to each other. So BE ‖ AD. More over BE=(1/2)*AD, Let's call a definition of midline of triangle. The midline of triangle is the line connected midpoints of two sides of triangle. It is parallel to the third side and the segment between two sides is equal a half of the third side. Backward rule says if the line is parallel to the side of triangle and cross two other sides and the segment between those two sides is equal half of the third one, so this segment belongs to the midline. BE ‖ AD and equal a half of AD, so BE is the midline. It connects midpoints of AC and CD. So AB=BC=R+r=3, AC=2AB=6. Then follow the video

[@francismoles9852]

we can deduce more quickly that AC is equal of 6 because the rapport of lenghts under BEC and ACD are of 2 for 1 also AC is equal 2* 3 equal of 6

[@misterenter-iz7rz]

Clearly it is a right angled triangle with side ratio 1×sqrt(8)×3, as the base off the triangle is 2, therefore the area is (1/2)×2×2sqrt(8)=2sqrt(8)=5.656854 approximately. 😊

[@PreMath]

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

[@bhikhubhavaghela1507]

Good explain sir❤❤❤❤❤❤🎉🎉🎉🎉🎉😢❤❤❤❤❤

[@PreMath]

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

[@arnavkange1487]

Good sense of humor today

[@PreMath]

😀

[@florianbuerzle2703]

A variation of this solution: construct the midpoint of AD and call it M. Then △ABM ≅ △BCE and BM = 2√2 by the Pythagorean theorem, and the area of △ABM is 2√2. By similarity, CD = 2 ∙ 2√2 = 4√2 and the area of △ACD is 4√2.

[@PreMath]

Thanks ❤️

[@Copernicusfreud]

Yay! I solved the problem.

[@PreMath]

Super ❤️

[@ybodoN]

If we drop a perpendicular from B to F on AD then FD = BE = 1and AF = 1. Since △ AFB ~ △ADC and AD is twice AF, AC is twice AB that is 3 ⋅ 2 = 6. So, cos  = ⅓ ⇒ sin  = ⅔√2 ⇒ area green △ADC = ½ ⋅ 2 ⋅ 6 ⋅ ⅔√2 = 4√2.

[@PreMath]

Thanks ❤️

[@tahmidulhaque414]

I had a problem where the 4 sides of a trapezium are given and we need to find the area. Please give me a solution to that.

[@TGM_12345]

Simpler AC calculation. According to Thales: AC/AD = BC/BE AC/2 = BC/1 so AC= 2BC. So AB=BC and thus AC = 6

[@Hayet-jb2sd]

Tres bien

[@BesseDenmark]

I solved it the same way! ☺ I'd like to see some interesting equations in the next videos. ☺

[@PreMath]

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

[@AmirgabYT2185]

S(ADC)=4√2≈5,66

[@giuseppemalaguti435]

Per la similitudine dei triangoli rettangoli A=4√2

[@PreMath]

Thanks ❤️

[@ybodoN]

Generalized: the area of the green triangle is _R² √(Rr) / (R − r)_ where _R_ is the radius of the red circle and _r_ is the radius of the blue circle.

[@PreMath]

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

[@quigonkenny]

Let R be rhe red radius and r be the blue. A = πR² 4π = πR² R² = 4π/π = 4 R = √4 = 2 A = πr² π = πr² r² = π/π = 1 r = √1 = 1 Let F be the point on AD wherr BF is perpendicular to AD. Since AD and BE, as radii, are perpendicular to tangent DC, DEBF is a rectangle, FD is parallel to DC, and FD = BE = 1. That means AF = 2-1 = 1. Triangle ∆AFB: a² + b² = c² 1² + FB² = 3² FB² = 9 - 1 = 8 FB = √8 = 2√2 Triangle ∆ADC: DC/AD = FB/AF DC/2 = 2√2/1 DC = 2(2√2) = 4√2 A = bh/2 = (4√2)2/2 = 4√2 cm² ≈ 5.66 cm²

[@Hayet-jb2sd]

BE egale 1

[@comdo777]

asnwer=2 isit

[@prossvay8744]

Area rectangle=4√2 I think wrong

[@gwnaveen4484]

Answer=4√2or 5.64 😅

[@PreMath]

Thanks ❤️

[@vaggelissmyrniotis2194]

First

[@PreMath]

Thanks dear ❤️

[@MathsMadeSimple101]

Plot twist: the triangle isn’t green. The illusion of the background makes you think it’s green.

[@ybodoN]

From a surrealist point of view, it is not even a triangle but only its representation 🤔