Can you find Area of the Pink Quadrilateral? | (Think outside the Box) |

Learn how to find the area of the Pink Quadrilateral. Important Geometry and Algebra skills are also explained: area of the triangle formula; Pythagorean Theorem; congruent triangles; right triangles. Step-by-step tutorial by PreMath.com
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Can you find Area of the Pink Quadrilateral? | (Think outside the Box) | #math #maths | #geometry
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[@MMmaths8800]

King of the world of Math"Premath"

[@PreMath]

You are way too generous! Thanks ❤️

[@wackojacko3962]

Once again a proof requires an auxiliary to be drawn to see this problem from a different perspective. An auxiliary line often creates congruent triangles or intersect existing lines at right angles. So the proposition @ 4:11 is used as a stepping stone to a larger result, namely the area of the Pink Quadrilateral. 🙂

[@PreMath]

Thanks for the feedback ❤️

[@soli9mana-soli4953]

Very nice solution Prof!! 👌

[@jimlocke9320]

We note that the two given lengths, 51 and 85, have a common factor 17. Designate the unit of measure as u (for units), so the lengths are 51 u and 85 u. We create a new unit U which is 17 times u. Then, 51 u = 3 U and 85 u = 5 U. To simplify the notation, leave off the U. When we are done with finding the area in units of U², we'll multiply by (17)² = 289 to convert to units of u². Doing it the hard way: We recognize that ΔCDE has a side of length 3 and hypotenuse 5, so it is a 3 - 4 - 5 right triangle and DE = 4. Construct a line through E parallel to AD and BC. It is half way between AD and BC. Drop a perpendicular from D and label the intersection F. Label the intersection with CD as point G. Drop a perpendicular from C and label the intersection H. Let DF = x. We note that ΔDEF and ΔCEH are similar and that DF = CH, therefore CH = x. Also, the distance between AD and BC is 2x and is the height of parallelogram ADCB. From Pythagoras, EH = √(3² - x²). From similarity, DF/DE = EH/CE, x/4 = (√(3² - x²))/3, which we solve for x and find x = 2.4. EH = √(3² - x²) = √(3² - (2.4)²) = 1.8 and EF = √(4² - x²) = √(4² - (2.4)²) = 3.2. FH = EF - EH = 3.2 - 1.8 = 1.4. G is the midpoint of FH, so GH = 0.7 and EG = EH + GH = 1.8 + 0.7 = 2.5 and is the median of the two bases AD and BC. So, area ADBC = (2.5)(2)(2.4) = 12 U². We multiply by 289 to get the original units squared, so area = (12)(289) = 3468 square units, as PreMath also found.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

1/ The auxiliary line you draw is so nice! So, we can have the area of the quadrilateral= 2 times that of the 3-4-5 triangle. 2/ To make it simple, just assume that the triangle DEC is simply a 3-4-5 so its area= 1/2 . 3.4= 6 3 / The area of the pink quadrilateral= 2x6x sq 17= 12x17x17=3468 sq units (:

[@PreMath]

Thanks for sharing ❤️

[@quigonkenny]

In triangle ∆DEC, CD and EC have side lengths of 85 and 51. 85 = 17(5) and 51 = 17(3). ∆DEC is thus a 17:1 ratio 3-4-5 Pythagorean triple triangle and DE = 17(4) = 68. As DE = 68, AD= 68. Extend DA and CE to intersect at F. As ∠AFE and ∠BCR are alternate interior angles (as AF and BC are parallel), ∠FEA and ∠CEB are vertical angles, and AE = EB, then ∆AFE and ∆BCE are congruent. As FE = EC, ∠FED = ∠DEC = 90°, and DE is shared, ∆FED and ∆DEC are also congruent. Since ∆AFE and ∆BCE are congruent, they have the same area, so triangle ∆CDF and quadrilateral ABCD have the same area, as one is quadrilateral AECD plus ∆AFE and the other is AECD plus ∆BCE. Triangle ∆CDF: A = bh/2 = 102(68)/2 = 102(34) = 3468 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Waldlaeufer70]

A very nice problem! I have realized that I can rotate the small triangle by 180° and that a new blue rectangular triangle is then created (DEF). What I didn't realize, however, is that the newly created triangle CDF is already the solution to the problem. Very nice! Thanks for sharing!

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thank you too ❤️

[@DB-lg5sq]

MERCI BEAUCOUP POUR VOTRE EFFORT S(ABCD)=2S(CDE) =51×68=3468

[@maxforsberg8852]

As per usual solved without trigonometry whereas I, also as per usual, used a ton of trigonometry to solve this. I did spot the hidden 3-4-5 triangle, scaled up by a factor 17, right away though.

[@LuisdeBritoCamacho]

Hello everybody!! 1) DE = 68 lin un My Intuition says that : 2) If I fold Triangle [ADE] with Axis being Line DE, and cover part of Triangle [CDE]. 3) If I fold Triangle [BCE] with Axis being Line CE, and cover part of Triangle [CDE]. 4) All Triangle [CDE] will be covered. 5) So, my Intuitive Based Answer is that Pink Quadrilateral Area is 51 * 68 = 3.468 Square Units. Twice the Area of Triangle [CDE].

[@MegaSuperEnrique]

Did we calculate √4624 without a calculator?

[@hongningsuen1348]

85 = 5 x 17 and 51 = 3 x 17 hence DE = 4 x 17 = 68

[@MegaSuperEnrique]

@@hongningsuen1348 yes, I watched the video. Did you?

[@dirklutz2818]

yes. First divide 4624 with 70; Answer=66.0... So the square root must be between 70 and 66.0. Then try the average, which is 68 . And 4624 divided with 68 gives 68, which is, of course, the square root.

[@mauriziograndi1750]

There is always a starting point don’t worry is not difficult. Start with the sin 51/85 once you know this angle you will know the adjacent, then with an angle and a side you will solve the next triangle and then you near the solution. The author should give the starting point instead to sit in the chair.

[@surveer176]

In this case area of triangle is half of the area of quadrilateral

[@michaelkouzmin281]

I dare say your solution is based/implied on the statement that AD is parallel to CB (i.e. ABCD is a trapezoid) but it is not mentioned in the terms of the problem and ABCD is called a quadrilateral. Ooops

[@caryross2693]

I had the same issue, but have now learned that the small arrows drawn on the two line segments indicate that they are parallel.

[@misterenter-iz7rz]

No idea at all, except I know DA=DE=17×4=68.......😢😢😢

[@PreMath]

Thanks for the feedback ❤️