Can you find area of Purple triangle? | Three Squares | (Step-by-Step explanation) |

[@mathbynisharsir5586]

Very Very useful video sir 🎉

[@PreMath]

Thanks and welcome❤️

[@beta700a]

Very creative solution )) Thank you 🤩

[@PreMath]

Thanks ❤️

[@florianbuerzle2703]

By analytic geometry: Set the lower left corner of the green square as origin (0,0) of the coordinate system. The line connecting points B(2,11) and D(6,6) is then given by y = -1.25x + 13.5 with zero at x = 10.8. So we have for the side a of the triangle by the Pythagorean theorem: a² = (10.8 - 2)² + (0 - 11)² = 4961/25. So the area of the equilateral triangle is given by 1/4∙a²∙√3 = 4961/100∙√3.

[@PreMath]

Thanks ❤️

[@misterenter-iz7rz]

Tan t=5/4, sin t=5/sqrt(41)so AB=11/sin t=(11 /5)×sqrt(41), therefore the answer is 121×41/100 sqrt(3)=85.927041 approximately .😊

[@PreMath]

Thanks ❤️

[@TheEulerID]

I got a the same answer, but in a simpler way. Drop a vertical from B to meet the lowest horizontal line at G. It is clear the line GA is 11, and GF is 4. Call the length FA x, making GA 4+x. We now have two similar triangles, AGB and AFD. So, BG/GA = DF/FA or 11/(4+x) = 6/x. Multiply both sides by x(4+x) and you get 11x = 24 + 6x -> 5x = 24 -> x=24/5. That means GA = 4 + 24/5 = 44/5. We can now use Pythagoras to calculate the side AB, which I will call y. So y^2 = GA^2 + BG^2 = 11^2 + (44/5)^2 = (55/5)^2 + (44/5)^2. We can take a common factor of 11^2/5^2 and get y^2 = 11^2/5^2 x (5^2 + 4^2) = 11^2 x 41 / 5^2. The area of the purple triangle is √3 x y^2 / 2 using the standard (easily derived) formula for an equilateral triangle of side y. Substitute in for y^2 and we get Area = √3 * 121 * 21 / 100 = sqrt(3) * 4961 / 100 = 85.927 to three significant figures.

[@PreMath]

Thanks ❤️

[@Oskar5707]

after you drew the complementary angles, you could have just extended the line BE to make another bigger right triangle. It also has all the same angles, so it's similar to the upper smaller right triangle. From there, 5:11=4:x, x being the lower leg of the big right triangle. You would solve for x and solve for the side of the purple triangle using the pythagora

[@tombufford136]

At a quick glance, BD is calculated from sqrt(5^2+4^2)=sqrt(41).DA is in the same ratio as BD, DA = 6/5 * BD. BA= BD(1+6/5). The height of the purple triangle is sqrt(BA^2- 0.5*BA^2)=sqrt(1/2)BA. The area of the triangle is 1/2 base * height =1/2 BA * sqrt(1/2)*BA=sqrt(1/2)^3 * BA^2). Substituting BD .sqrt(1/2)^3* (BD(11/5))^2. BD= sqrt(41) and the area of the purple triangle is sqrt(1/2)^3 * (sqrt(41)*(11/5))^2. The area = 70.160 units. Comparing this answer with others below it seems 15 units too low.

[@PreMath]

Thanks ❤️

[@alster724]

After getting the single side, I used the basic formula for the equilateral triangle area A= s²√3/4 or A= (square area)√3/4

[@giuseppemalaguti435]

(2+3+6):(4+x)=6:x...x=24/5...l=√(11^2+(4+24/5)^2)=11√41/5...Ap=l*lsin60/2=(√3/4)121*41/25

[@PreMath]

Thanks ❤️

[@phungpham1725]

1/ Drop the height BH to the base , we have BH= 11 Notice that BE/ED = 5/4 so because of similarity BH/HA=5/4---> HA = 44/5=8.8 Let a be the side of the equilateral triangle Sq a = sq11+ sq 8.8=198.44 Area of the equilateral triangle= sqa x sqrt3/4 = 85.93 sq units

[@PreMath]

Thanks ❤️

[@zsoltszigeti758]

B point is (2;11), D point is (6;6), AB line is 5x+4y=54 => A point is (54/5;0) => side length (AB) is √((54/5-2)^2+11^2)=√198.44 => T=√3*198.44/4

[@PreMath]

Thanks ❤️

[@m.h.6470]

solution: if you continue the right side of the 4-square down to the 36-square, you get a right angle triangle. This triangle has one leg of √4 + √9 = 2 + 3 = 5 and the other leg of √36 - √4 = 6 - 2 = 4 The lower white triangle is also a right angle triangle and is mathematically similar to the first triangle. Therefore the ratios between its sides are the same as the first triangle. This second triangle has one side of √36 = 6 given. Therefore the other side is 4/5 = x/6 → x = 24/5 = 4.8. With this and Pythagoras we can calculate the entire side AB of the large purple triangle: √(4² + 5²) + √(6² + 4.8²) = √(16 + 25) + √(36 + 23.04) = √41 + √59.04 = √41 + √(144/100 * 41) = √41 + 12/10√41 = 2.2√41 ≅ 14.09 The area of an equilateral triangle is A = √3/4 * a² = √3/4 * (2.2√41)² = √3/4 * 4.84*41 = √3 * 1.21 * 41 = 49.61√3 ≅ 85.93 square units

[@PreMath]

Thanks ❤️

[@devondevon4366]

85.93 Extend the line of the yellow square through the blue to form the right triangle BPD The length of BP = 5 (sqrt 4 + sqrt 9), and the width of PD = 4 ( 1+ 3). Hence, the hypotenuse using Pythagorean BD = 6.40312 Label the triangle to the right of the green square DNA Hence, the length of DN= 6 (sqrt 36), and since it is similar to triangle BPD due to the RIGHT ANGLE of the green square ,then NA= 4.8 ( 5/4 = 6/NA, hence NA = 24/5= 4.8) Hence, the hypotenuse DA = 7.68375 Hence, the length of the equilateral (BA) = 7.68375 + 6.40312 =14.08687 Hence, the area of the purple (equilateral ) = sqrt 3/4 * 14.08687 * 14.08687 = 85.927 Answer

[@PreMath]

Thanks ❤️

[@MultiYesindeed]

Brilliant thanks

[@PreMath]

Glad you enjoyed it❤️

[@sorourhashemi3249]

Thanks.easy. I got the area through Heron formula.

[@marcgriselhubert3915]

Let's use an adapted orthonormal. B(2;11) D(6;6) VectorBD(4;-5). The equation of (BD) is: (x-2).(-5) - (y-11).(4) = 0, or -5x -4y +54 = 0. Its intersection with Ox is A(54/5;0). Then vectorBA(44/5;-11) and BA = sqrt((44/5)^2 + (-11)^2) = sqrt(4961/25) = c The area of the triangle ABC is (c^2). (sqrt(3) /4) = (4961/25). (sqrt(3)/4) = (4961. sqrt(3)) /4.

[@PreMath]

Thanks ❤️

[@LuisdeBritoCamacho]

Algebraic Geometrical Resolution, using a Cartesian Plane of Coordinates: The side of the Triangle is equal to ~ 14,0869 lu; is the Distance between Point A (10,8 ; 0) and Point B (2 ; 11) passing by Point D (6 ; 6). Using the Heron's Formula the Area of the Triangle is 85,920 su

[@PreMath]

Thanks ❤️

[@quigonkenny]

Let E be the point where BE and EA are perpendicular. Let F be the point where BF and FD are perpendicular. By observation, the side lengths of the three squares are √4 = 2 (yellow), √9 = 3 (blue), and √36 = 6 (green). By observation BF = 2+3 = 5, BE = 5+6 = 11 and FD = 6-2 = 4. Triangle ∆BEA: EA/FD = BE/BF EA/4 = 11/5 EA = 44/5 c² = a² + b² = (44/5)² + 11² BA² = 1936/25 + 121 = 4961/25 BA = √(4961/25) = (11√41)/5 Triangle ∆ACB: A = (1/2)absin θ = (1/2)[(11√41)/5]²(√3)/2 A = (4961/25)(√3)/4 = (4961√3)/100 A ≈ 85.93

[@ybodoN]

Our goal is to find the area of the purple triangle, which is equilateral, so its area is ¼ a² √3 where a² = AB². AB is the hypotenuse of ⊿ABG ~ ⊿DBE ⇒ BE/BG = DE/AG. So we can write that 5/11 = 4/AG ⇒ AG = 44/5. Then AB² = 11² + (44/5)² = 4961/25. Therefore, the area of the purple triangle is 4961/100 √3 square units.

[@PreMath]

Thanks ❤️

[@robertlynch7520]

I don't know … I just used proportionality of similar triangles to figure the whole thing out, and it worked quite easily. Let's call FA '𝒙' … Working with EB and ED, we can say: 6 / 𝒙 = 5 ÷ 4 … then cross multiplying 6 × 4 = 5𝒙 … and dividing by 5 𝒙 = 4.8 units We already know that ED is 4 units, so the baseline of that whole is (4 ⊕ 4.8 = 8.8) units. The height of the stack of boxes is 2 ⊕ 3 ⊕ 6 = 11 units. Pythagoras to the rescue, for the length of the purple △ 𝒔 = √( 11² + 8.8² ); Having that, and that the area of an equilateral triangle is (area = √3/4 ⋅ side²), then Area △ = √3/4 × √( 11² + 8.8² )² … which cancels the √ and becomes Area △ = √3/4 × (11² + 8.8²) → Area △ = 0.433013 • (121 + 77.44) Area △ = 85.927 Which is exactly the same answer, without having to actually evaluate Pythagoras at all. Seems simpler to me. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

[@PreMath]

Thanks ❤️

[@santiagoarosam430]

[(3+2)/(6-2)]=(6+3+2)/b→ b=44/5 → AB²=b²+(6+3+2)²→ AB=(11√ 41)/5 → Área ABC =AB*[(AB/2)√3]*(1/2) =(4961/100)√3 =85.93 Gracias y saludos.

[@PreMath]

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[@kimng312]

How do you know AFD is 90 degree?

[@wackojacko3962]

🙂💯👍

[@PreMath]

Thanks ❤️

[@batavuskoga]

I used the Heron's formula to calculate the triangle's area. This for the first time.

[@PreMath]

Thanks❤️

[@ybodoN]

Just like the formula A = ½ ab sin c, in the case of an equilateral triangle, Heron's formula simplifies to A = ¼ a² √3 😉

[@maxforsberg8852]

Very nice. For once my method was identical to yours. I usually over complicate my solutions.

[@PreMath]

Thanks ❤️

[@devondevon4366]

85.93

[@PreMath]

Great ❤️🌹

[@raymondruiz5839]

I believe there is another formula for area of equilateral triangle

[@devondevon4366]

Perhaps, but I am not sure. But the area of an equilateral triangle (sqrt 3)/4 a^2 can be derived from the area of a triangle formula 1/2 ab sine C. Since all sides are the same (equilateral), then 1/2 ab sine C becomes 1/2 aa sine A = 1/2 a^2 sine 60 degrees= 1/2 a^2 sine sqrt 3/2 (since sine 60 degrees = sqrt 3/2) = a^2 sine sqrt 3/4 or (sqrt 3)/4 a^2

[@tombufford136]

After watching the video and going over my previous comment/answer I have found the calculation error. The comment should be: At a quick glance, BD is calculated from sqrt(5^2+4^2)=sqrt(41).DA is in the same ratio as BD, DA = 6/5 * BD. BA= BD(1+6/5). The height of the purple triangle is sqrt(BA^2- (0.5*BA)^2)=sqrt(2/3)BA. The area of the triangle is 1/2 base * height =1/2 BA * sqrt(2/3)*BA=1/2*sqrt(2/3) * BA^2). Substituting BD .1/2*sqrt(2/3)* (BD(11/5))^2. BD= sqrt(41) and the area of the purple triangle is 1/2*sqrt(2/3) * (sqrt(41)*(11/5))^2. The area = 85.93 Units

[@PreMath]

Thanks ❤️

[@moroofoloruntola199]

There is nothing in the diagram to show that the yellow, blue and green boxes are square. The diagram is oth fully well annotated

[@Ibrahimfamilyvlog2097l]

السلام علیکم سر ویری نائس شیئرنگ سر اپ ہماری ویڈیو نہیں دیکھتے ❤❤❤❤

[@PreMath]

Thanks❤️

[@RobG1729]

You should change the caveat from "may not be" to "most definitely isn't." As excellent as your videos are, they'd be even better if diagrams were closer to scale. Visual learners may have cognitive dissonance when pictures are so inaccurate.

[@neilhunt137]

How pleasing to hear that you are at last calling the area of something as square units.....

[@PreMath]

Thanks ❤️

[@boderaner]

WTF? Why GeoGebra one-string solution was been deleted? polygon(point({2, 11}), intersect(line(point({2, 11}), point({6, 6})), line(point({0, 0}), point({2, 0}))), intersect(circle( point({2, 11}), intersect( line(point({2, 11}), point({6, 6})), line(point({0, 0}), point({2, 0})) )), circle(intersect( line(point({2, 11}), point({6, 6})), line(point({0, 0}), point({2, 0}))), point({2, 11})))) >>> 85.927