Can you find Area of the Pink Quadrilateral? | (Think outside the Box) |
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- เผยแพร่เมื่อ 14 พ.ค. 2024
- 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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King of the world of Math"Premath"
You are way too generous!
Thanks ❤️
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. 🙂
Thanks for the feedback ❤️
Very nice solution Prof!! 👌
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.
Excellent!
Thanks for sharing ❤️
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 (:
Thanks for sharing ❤️
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
Excellent!
Thanks for sharing ❤️
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!
Thank you!
You are very welcome!
Thank you too ❤️
MERCI BEAUCOUP POUR VOTRE EFFORT
S(ABCD)=2S(CDE)
=51×68=3468
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].
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.
Did we calculate √4624 without a calculator?
85 = 5 x 17 and 51 = 3 x 17 hence DE = 4 x 17 = 68
@@hongningsuen1348 yes, I watched the video. Did you?
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.
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.
In this case area of triangle is half of the area of quadrilateral
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
I had the same issue, but have now learned that the small arrows drawn on the two line segments indicate that they are parallel.
No idea at all, except I know DA=DE=17×4=68.......😢😢😢
Thanks for the feedback ❤️