Can you find the Blue shaded equilateral triangle? | (Step-by-step explanation) |

Learn how to find the Blue shaded equilateral triangle. Important Geometry and algebra skills are also explained: Equilateral triangle; Isosceles triangle; area of a triangle formula. Step-by-step tutorial by PreMath.com
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Can you find the Blue shaded equilateral triangle? | (Step-by-step explanation) | #math #maths
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[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!!!!!!!

[@PreMath]

Thank you too❤️ You are awesome. Keep it up 👍

[@parthtomar6987]

Accurate solution with full efficiency

[@PreMath]

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

[@arthurschwieger82]

It is always fun to see the different ways to solve problems. I started with triangle ACD to get the length of CD = 6√3 (this is from the side relationship of a 30/60/90 triangle, 1-2-√3). CD=BE so I can get AB using the same relationship. AB = 6. To get the length of DE, I took 18 and subtracted AB (6) to get 12. The area of an isosceles triangle is (√3/4)*DE² = 36√3.

[@tombufford136]

At a quick glance. With Angle ADC and AC ,AD and DC are calculated.EB = DC. With angle AEB, AE and AB are calculated. ED is = AC -AB. Area = half base * height= 0.5 ED * ED*0.5*(sqrt(3).

[@MrPaulc222]

Good explanation. I didn't do the isosceles route as I worked solely with the 30-60-90 elements. Bottom right triangle has 18, 18/sqrt(3) and 36/sqrt(3) as its sides. Therefore, the upper triangle has 36/3 as its short side, meaning the equilateral has sides of 12. 12^2 - 6^2 =108 so height of sqrt(108). 6*sqrt(108) simplifies to 36*sqrt(3) so 62.35(2dp).

[@santiagoarosam430]

Ángulo A=30º+30º=60º → AC=18=DC√3 → DC=6√3 =EB → AB=6 → BC=18-6=12 =Lado del triángulo equilátero azul → Área Azul =12*6√3/2 =36√3 =62.3538 Gracias y un saludo cordial.

[@unknownidentity2846]

Very clever solution.👍 Here is my way using trigonometry: Right triangle ACD: CD/AC = tan(∠CAD) CD/18 = tan(180°−90°−60°) CD/18 = tan(30°) CD/18 = sin(30°)/cos(30°) CD/18 = (1/2)/(√3/2) CD/18 = 1/√3 CD = 18/√3 The triangles ACD and ABE are similar (30°-60°-90°): CD:AC:AD = AB:BE:AE with CD = BE CD/AC = AB/BE CD/AC = AB/CD AB = CD²/AC AB = (18/√3)²/18 AB = (18²/3)/18 AB = 18/3 = 6 Side length of the equilateral triangle DEF: s = DE = BC = AC − AB = 18 − 6 = 12 Area of the equilateral triangle DEF: A = (√3/4)s² = (√3/4)*12² = 36√3 Best regards from Germany

[@bhavesh8083]

If we take triangle ADC by tan 60 Dc = 18root 3 ,and then find the area of triangle =27 root 3

[@soniamariadasilveira7003]

Obrigada professor!

[@AmirgabYT2185]

S(tr)=36√3≈62,28

[@bigm383]

Thanks Professor!❤️😁

[@williamwingo4740]

Lots of 30-60-90 triangles in this one. All angles are in degrees. Angle ADF = 90 + 60 - 60 = 90; so triangle ADF is 30-60-90. This means DF = AD/√3. Considering triangle ADC, which is also 30-60-90: sin 60 = (√3)/2 = 18/AD ; cross-multiply and (18)(2) = (√3)(AD) = 36; so AD = 36/√3 = 36√3/3 = 12√3. So DF = (12)(√3)/(√3) = 12. Drop the altitude of triangle DEF from point F, separating the equilateral into two more 30-60-90's. The altitude is 12 sin 60 = (12)(√3)/2 = 6√3. Thus the area is (1/2)(12)(6√3) = 36√3. Carpe Diem. 🤠

[@SK-gs6xb]

AD = AC/Cos30= 36/√3 Side of the Equilateral triangle = FD = ADTan30 = 36/3= 12

[@PreMath]

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

[@Mycroft616]

We were in lock step once we knew the side length of the Blue Triangle was 12, but I used a lot more of the 30°-60°-90° Triangle identity and the observation that BEDC is a rectangle to get there. 18 = CD × 3^(1/2) CD = 18/[3^(1/2)] CD = 6 × 3^(1/2) BE = CD BE = 6 × 3^(1/2) BE = AB × 3^(1/2) 6 × 3^(1/2) = AB × 3^(1/2) AB = 6 BC = AC - AB BC = 18 - 6 BC = 12 ED = BC ED = 12 EDIT: This can also be done without trigonometry by remembering that the perpendicular bisector of an equilateral triangle creates two more 30°-60°-90° triangles and the altitude of the Blue Triangle can be solved by the identity. Or by using Heron's Theorem.

[@AndreasPfizenmaier-y7w]

Area of the equilateral triangle : a^2/4*sqrt 3. a=12

[@misterenter-iz7rz]

Let s be the side of equilateral triangle, clearly AE=ED, and angle EAD=angleEDA, so AD=root 3 s, and thus 18=(root 3 s) root 3/2=3/2 s, so s=12,therefore the answer is (1/2)12 12 root 3/2=36root 3=62.35 approximately.😊

[@PreMath]

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

[@trumpetbob15]

Interesting approach. I did it slightly differently. Triangle ABE is 30-60-90 so if AB is x, then BE is x * root 3, which is also CD given rectangle BCDE. However, CD is shorter leg of 30-60-90 triangle ACD so AC is the result CD * root 3 = x * root 3 * root 3 = 3x = 18. Thus, x = 6, BC = 18 - x = 12 = DE, which is also the side length of the shaded triangle and we get the same area after doing the same final calculations.

[@soli9mana-soli4953]

Angle in A is 60°, DAC angle is 30° so EAD = 60° - DAC = 30° EDA is 30° because complementary, so triangle AED is iscosceles and AE = ED triangles 30°,60°,90° have sides X,2X,X√ 3, so if you know one side you know all sides On ADC it is X√ 3 = 18 then DC = 18/√ 3 now you ca see that DC = EB so repeat on ABE Y,2Y,Y√ 3 Y√ 3 = 18/√ 3 Y = 6 AE = 2*Y = 12 then ED = 12 Area = 12²*√ 3 /4 = 36√ 3

[@giuseppemalaguti435]

AE=l per costruzione...teorema seni:18-l:sin30=l:sin90..l=12

[@PreMath]

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

[@murdock5537]

Nice! AC = 18; DAC = φ = 30° = BEA = FAD → CD = 18/√3 = 6√3 → AD = 12√3 → AF = 24 → EF = 12 = AD = DF → ED/2 = 6 = EM = DM → MF = 6√3 → area ∆DFE = (1/2)12(6√3) = 36√3

[@PreMath]

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

[@devondevon4366]

Answer 36 sqrt 3 or 62.35 Since 18 = sqrt 324 = sqrt 3 *sqrt 108 and triangle ACD is a 30 -60-90 triangle, then DC the sqrt 108= 6 sqrt 3. Hence AD = 12 sqrt 3 (or twice DC) Hence EB =DC 6 sqrt 3 Since triangle ABE is also a 30-60-90 right triangle AB= 6 (since EB is sqrt 3 * AB) Hence BC = (AC-AB) = 18- 6 =12 Hence the length of the equilateral triangle is 12 Area = (sqrt 3 * side^2)/4 = sqrt 3 * 12^2/4 = 36 sqrt 3 or 62.35

[@raya.pawley3563]

Thank you

[@michaelkouzmin281]

Just another solution: Let x= ED, so AB= 18-x; DC/18=tan(30) => DC=18*tan(30); BE/AB= tan(60) => BE= (18-x)*tan(60); DC=BE; 18*tan(30)=(18-x)*tan(60); x=18(1-tan(30)/tan(60)) = 18*(1-1/3) =12; Ablue = x^2*sqrt(3)/4=36*sqrt(3) = approx 62.35 sq units.

[@Istaphobic]

I wanted to challenge myself to see if I could do it without any trigonometry and presuming no knowledge of any 30-60-90 triangle side ratio relationships. Just wanted to use pure geometry. Here's what I got: Given △ACD with ∡ADC = 60° ⇒ ∡CAD = 30° (angles of a triangle), such that △ACD has angles 30, 60 and 90 and △ABE also has angles 30, 60 and 90. ∴ △ACD ≅ △ABE (similar triangles - angle, angle, angle). Let length BC = x (which is also the length of the side of the equilateral triangle as BC = ED), such that AB = AC - BC = 18 - x. Let BE = y = CD. Now, CD/AC = AB/BE, such that y/18 = (18 - x)/y, and y² = 18(18 - x). Now, ∡ADE = ∡CDE - ∡ADC = 90° - 60° = 30°. ∴ △ADE is an isosceles triangle, as ∡DAE = ∡ADE = 30° ⇒ AE = ED = x. In △AC, we have lengths AB = 18 - x, BE = y and AE = x. Using Pythagoras, AE² = AB² + BE² ⇒ (18 - x)² + y² = x². From above, we have y² = 18(18 - x), so (18 - x)² + 18(18 - x) = x². ⇒ 324 - 36x + x² + 324 - 18x = x². ⇒ 54x = 648; and ∴ x = 12. Now, we know all sides of equilateral triangle △DEF are equal and length ED = x = 12, such that ED = EF = DF = 12 (given). Drawing the height of the triangle h from point F to point G, which is orthogonal to ED, the height is the angle bisector which bisects line ED such that EG = 6. Using Pythagoras height FG = h and FG² = EF² - EG² = 144 - 36, such that FG = h = 6√3. Area △DEF = [△DEF] = ½.x.h = ½.(12).(6√3) = (6).(6√3), and ∴ [△DEF] = 36√3 units².

[@devondevon4366]

36 sqrt 3 or 63.25 It is an easy problem given that triangles ABC and ADC are 30-60- 90; hence the three sides are x, sqrt 3x, and 2x. 18= sqrt 324 = sqrt 3*sqrt 108. Therefore DC = sqrt 108 since it faces the 30 degrees angle. Hence EB also = sqrt 108 or sqrt 3^ sqrt 36 or 6 sqrt 3. Hence AB= 6 . Hence BC=12 (18-6) which means that ED= 12. Hence each side of the equilateral triangle is 12. Hence the area = sqrt 3/4 * 12^2 = sqrt 3/4*144 or sqrt 3 *144/4 = sqrt 3 * 36 Answer The formula for the area of an equilateral is sqrt 3/4 *s^2. This comes from the formulae for the area of all triangles 1/2 AB sine C = 1/2AB sine 60 (since all angles of an equilateral are 60) = 1/2 AB sqrt 3/2 ( Since Sine 60 degrees sqrt 3/2) = sqrt 3/4 *s^2

[@Copernicusfreud]

Yay! I solved the problem .

[@dadugiri3790]

Why don't you use Heron's formula for calculating area of the blue triangle after finding the length of the rectangle/the triangle as 12units?

[@ybodoN]

In the case of an equilateral triangle, even Heron's formula reduces to _¼ a² √3_ 😉

[@RajeshGupta-xz4mt]

Find the area of AED triangle

[@arnavkange1487]

Sir pls include some algebra sums side by side

[@PreMath]

Sure! Keep watching.... You are awesome. Keep it up 👍

[@mohanramachandran4550]

Very simple sir All are 30°, 60°, 90° Triangle ((18÷√(3)÷√(3))^2)√(3)÷4 = 62 .353829

[@JSSTyger]

I could be wrong because i often scribble at a 100mph but i think the answer is 36sqrt(3).

[@alster724]

Got it

[@wackojacko3962]

I've regressed from pencil and paper and solved old school with stylus and clay tablet! 🙂

[@PreMath]

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

[@thomakondaciu6417]

S=3.9

[@marioalb9726]

Height of rectangle: H = 18 / tan 60° H = 10,39 cm Base of triangle: b = H . tan 30° b = 6 cm Base of rectangle B = 18 - b B = 12 cm Area of equilateral triangle: A = √3/4 . B² A = 62,35 cm² ( Solved √ )

[@rbd77]

That was a long way around.

[@TheBengal_Tiger]

Who is the face behind this brilliance! Whats ur twitter handle ?