Can you find the area of the Pink Trapezoid? | Trapezoid | (Trapezium) |
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- Опубликовано: 15 май 2024
- Learn how to find the area of the Pink shaded Trapezoid. Important Geometry and algebra skills are also explained: Trapezoid; Trapezium; Trapezoid area formula; Trigonometry; Triangle area formula. Step-by-step tutorial by PreMath.com
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Can you find the area of the Pink Trapezoid? | Trapezoid | (Trapezium) | #math #maths | #geometry
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Every time I see your Premath examples, my mind starts running, learning and reinforcing.
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△ADB is a special 30°-60°-90° right triangle.
AD = (AB)/2
= 6/2
= 3
By observation, trapezoid ABCD is a right trapezoid.
Draw a segment thru D and a point E on base AB such that it is a height of the trapezoid.
This segment divides trapezoid ABCD into a rectangle, BCDE, and a right triangle, △AED.
△AED is also a special 30°-60°-90° right triangle.
AE = (AD)/2
= 3/2
= 1.5
DE = (AE)√3
= (3/2)√3
= (3√3)/2
Apply the Geometric Mean (Altitude) Theorem on △ADB.
(DE)² = AE * BE
[(3√3)/2]² = 1.5 * BE
1.5(BE) = 6.75
BE = 4.5
By the Parallelogram Opposite Sides Theorem (BCDE is a rectangle), BC = (3√3)/2 & CD = 4.5.
A = [(a + b)/2]h
= 1/2 * (4.5 + 6) * (3√3)/2
= (63√3)/8
So, the area of the pink trapezoid is (63√3)/8 square units (exact), or about 13.64 square units (approximation).
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AD=3 DB=3√3
BC=3√3/2 CD=9/2
Pink Trapezoid area :
(6+9/2)＊3√3/2＊1/2=63√3/8
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In triangle ADB we have AD = AB.cos(60°)= 6.(1/2) = 3. We now note H the orthogonal projection of D on (AB).
In triangle ADH we have DH = AD.sin(60°) = (3.sqrt(3))/2, and this is the height of the trapezoïd ABCD.
Also in triangle ADH we have AH = AD.cos(60°) = 3.(1/2) = 3/2, so DC = HB = AB - AH = 6 - 3/2 = 9/2.
The area of the trapezoïd ABCD is ((DC + AB)/2). DH = ((9/2 + 6)/2). (3.sqrt(3))/2 = (21/4).(3.sqrt(3))/2 = (63.sqrt(3))/8.
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Another formula for the area of a 30°-60°-90° special right triangle is A = (h²)(√3)/8, where h is the length of the hypotenuse. Method 2 can be simplified by using this formula. For ΔABD, the hypotenuse has length 6, so its area is (6)²(√3)/8 = (36)(√3)/8. For ΔBCD, the hypotenuse is 3√3, so its area is (3√3)²(√3)/8 = (27)(√3)/8. Summing the areas, we get (36)(√3)/8 + (27)(√3)/8 = (63)(√3)/8 square units, as PreMath also found.
Two other useful formulas for area, where the hypotenuse is known and designated h, not to be confused with height also being designated h, are: For the 45°-45°-90° special right triangle, A = (h²)/4. For the 15°-75°-90° right triangle, A = (h²)/8. (The 15°-75°-90° right triangle is not considered "special" in geometry, but appears frequently enough in problems that students should make a "special" effort to know its formulas.)
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This is awesome, many thanks, Sir!
φ = 30°; ∎ABCD → AB = AE + BE → sin(AED) = 1; DAB = 2φ → EDA = ABD = φ →
sin(φ) = 1/2 → cos(φ) = √3/2 → AD = 3 → AE = 3/2 → DE = 3√3/2 →
CD = BE = 6 - 3/2 = 9/2 → area ∎ABCD = (1/2)(6 + 9/2)(3√3/2) = 63√3/8
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Calculated in my head:
A = (6 + 4.5) / 2 * 3/2 * √3 = 5.25 * 3/2 * √3 = 15.75 / 2 * √3 = 7.875 * √3 (= 13.64) square units
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You could also obtain the value of the diagonal of the trapezoid through the right triangle 30°, 60°, 90°, where the hypotenuse is 2a, one side is a and the other is the root of 3, that is, a^(1/3).
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🔺 ABD is a 30-60-90 triangle.
The hypotenuse given.
Hence AD will be known.
Now we may drop a perpendicular DP on AB.
APD will be a 30-60-90 triangle.
As AD is now known, we may find PD and AP.
DP=BC
PB=CD=AB-AP(now AP and AB are known)
Required area =
Rectangle PBCD +🔺 APD= PB*BC +1/2*AP*PD
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Today I have time for some math:
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The right triangle ABD is a 30°-60°-90° triangle, so we can conclude:
AD = AB/2 = 6/2 = 3
BD = √3*AD = 3*√3 = 3√3
Since ∠CBD=90°−∠ABD=90°−30°=60°, the right triangle BCD is also a 30°-60°-90° triangle:
BC = BD/2 = 3√3/2 = (3/2)√3
CD = BC*√3 = (3/2)√3*√3 = 9/2
Now we can calculate the area of the trapezoid in two different ways:
A(ABCD)
= A(ABD) + A(BCD)
= (1/2)*AD*BD + (1/2)*BC*CD
= (1/2)*3*3√3 + (1/2)*(3/2)√3*9/2
= (9/2)√3 + (27/8)√3
= (36/8)√3 + (27/8)√3
= (63/8)√3
A(ABCD)
= (1/2)*(AB + CD)*BC
= (1/2)*(6 + 9/2)*(3/2)√3
= (12/2 + 9/2)*(3/4)√3
= (21/2)*(3/4)√3
= (63/8)√3 ✓
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AB=6→ AD=6/2=3→ CB=3√3/2→ Área ABCD =(6 -3/4)*3√3/2 =13,6399 ud².
Gracias y saludos.
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You're welcome. A hug from Porto Alegre, RS, Brazil.

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Extended CD
Drop perpendicular AP on extended CD
Area of rectangle ABCP -area of 🔺 APD will be the answer.
(AP = BC and PD =6-4.5=1.5)
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3rd method is the best! Watching an artist master at work solving. ..namely you sir! Have a nice day. 🙂
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1/ We have two halves of two equilateral triangles.
Area of a equilateral triangle of which the side= a:
Area = sqa .sqrt3/4
So Area of ABD= 36.sqrt3/8
Area of DBC=27sqrt3/8
Area of the trapezoid= 63.sqrt3/8
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Thank you! 30-60-90 for the win!
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S(ABD) = (3 × 3√3)/2. ▲BCD ~ ▲ABD, K = (√3)/2, K² = 3/4.
S(ABCD) = S(ABD) × (1 + K²) = [(3 × 3√3)/2] × (7/4) = (63√3)/8.
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"Super Duper " 👍
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The triangle adb is a special triangle 90-60-30 so the ypothenus is 6 , ad is 3✓3 and db is 3.
Now dbc is another special triangle 90-60-30 , so if db is 3 the other measures are dc = 1.5 and bc = 1.5✓3
The first area of triangle is 3✓3 x 3 /2 = 27✓3/2
the second area is 3/2√3 x 3/2 /2 = 9√3/8
the sum is 9√3/8 + 27√3/2 = 111√3/8
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S=63√3/8≈13,64
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Both methods are easy!!!!! 🎉🎉🎉🎉🎉
Sin 60 = DB/AB = DB/6.
DB = 6 sin 60 = 5.196.
BDC = ABD = 30 degrees.
Sin 30 = BC/DB = BC/ 5.196.
BC = 0.5 x 5.196 = 2.598.
Area of left triangle = 1/2 x AB x DB x sin 30.
0.5 x 6 x 5.196 x 0.5 = 7.794.
Area of right triangle = 1/2 x BC x DB x sin 60.
0.5 x 2.598 x 5.196 x 0.8660 = 5.845.
Total area = 7.794 + 5.845 = 13.64.
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Not difficult（1/2)(6+9/2)×3sqrt(3)/2=(63/8)sqrt(3)😊.
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Challenge:
Find the Sides of a Right Triangle (a ; b ; c; and : a < b < c) whose Numerical Value of the Area is equal to Seven Times the Numerical Value of the Perimeter : "A = 7 * P" : 7 * (a + b + c) = ((a * b) / 2). Prove your answer for Positive Integer Solutions.
This could probably be answered using Heron's formula, since that uses a, b, c for perimeter and area. However, Euclid's formula, where
a = m^2-n^2;
b = 2*m*n;
c = m^2+n^2
(m>n>0)
also works.
P = a+b+c
P = (m^2-n^2) + (2*m*n) + (m^2+n^2)
P = 2*m^2+2*m*n
P = 2*m*(m+n)
7P = 14*m*(m+n)
A = (m^2-n^2) * (2*m*n) / 2
A = (m+n)*(m-n)*m*n
Since A = 7P, A/7P = 1
So,
((m+n)*(m-n)*m*n) / (14*m*(m+n)) = 1
or
(m-n)*n = 14
To check:
use n=7, so m=2+7 = 9 [not many choices here, m and n must be integers and (m-n) and n must be factors of 14]
a = 9^2-7^2 = 81-49 = 32
b = 2*9*7 = 126
c = 9^2+7^2 = 81+49 = 130
P = 32+126+130 = 288
7P = 7*288 = 2016
A = 32*126/2 = 2016
Although the triangle sides could be scaled by an integer (say k), the relationship between P and A would fail as the perimeter would be scaled by k but the area by k^2.
Thanks, that was a good exercise.
@@Slimmo_09 thank you.
First, this is not an Exercise or an Amusement. This is Numbers Theory!! Where you study the Properties of Right Triangles and Primeval Pythagorean Triples.
I am going to show you that your answer is not a Primeval Pythagorean Triple (PTT). The Solution must be given in terms that (a ; b ; c) have no Common Factor between them. They must be Primes between them.
In your case, your solution is the Pythagorean Triple (32 ; 126 ; 130) is a Derivative of (16 ; 63 ; 65).
If you use this PPT what do we find?
A = 16 * 63 / 2 ; A = 504
P = (16 + 63 + 65) ; P = 144
7 * P = 7 * 144 = 1.008 ; 7 * P = 1.008
1.008 / 504 = 2 ; the very same Factor you use
As you can easily see the DPT (32 ; 126 ; 130) is ((2 * 16) ; (2 * 63) ; (2 * 65))
With the PPT (16 ; 63 ; 65) the Relationship between the Area and the Perimeter is not a Positive Integer:
A = 504
P = 144
A / P = 504 / 144 ; A / P = 252 / 72 ; A / P = 126 / 36 ; A / P = 63 / 18 ; A / P = 21 / 6 ; A / P = 7 / 2 ; A / P = 3,5
Multiplying (A / P) by 2 what do we get? Et Voilà !! The Number 7!!
My Solution is (29 ; 420 ; 421), as you can easily see, don't share any Common Factor. And they have the Property asked : A = 7 * P ; 6.090 = 7 * 870
Thank you again!!
😁😁😁😁😁😁😁
@@LuisdeBritoCamacho What I showed was only to satisfy the conditions as then asked. Since 14 has four factor pairs, there will be four triangles that satisfy the condition that A=7P. I just chose one to test the calculations, and it coincidentally satisfied the condition that ab, but it is a PPT.
3. (7,2) n=7, m-n=2, m=9 (a, b, c) = (32, 126, 130)
a>b but not a PPT.
4. (14,1) n=14, m-n=1, m=15 (a, b, c) = (29, 420, 421)
a>b and is a PPT.
All of these triangles have 7*(a+b+c) = a*b/2.
I thought you might have been onto something here with the new PPT condition, as you'll notice only the final factor pair produce a triangle that satisfies all conditions. Testing with A=5P also gave the final factor pair the only one to satisfy the conditions: n=10, m-n=1, m=11 (a, b, c) = (21, 220, 221). However, testing with A=6P gave two suitable triangles:
(4,3) n=4, m-n=3, m=7 (a, b, c) = (33, 56, 65);
(12,1) n=12, m-n=1, m=13 (a, b, c) = (25, 312, 313).
Again, the final factor pair produces a satisfactory triangle. Notice that in each of these cases the final factor pair produces a triangle with c-b=1. Perhaps you need to add this as another condition.
Consider the general case of triangles that satisfy A=kP. The significance of being able to select just the final (or the first if that had worked) factor pair for calculations is that there is no need to factor. Set n=k*2, m=n+1, and calculate a,b,c. Right there is a practical application of your work - for one triangle this makes no difference, but computing 1 million such triangles and not having to factor k*2 is a big time saving. Of course if you need all PPT triangles that satisfy the equation then this won't work and factoring is the choice.
@@Slimmo_09, thank you for sharing your knowledge.
It seems that you are the only one, here, who has the same interest in the Relationships between Primeval Pythagorean Triples and Right Triangles.
A PPT is : (a ; b ; c) and a DPT is : (a*k ; b*k ; c*k)
For instance : (3 ; 4 ; 5) wich is not only a PPT but the Fundamental Primeval Pythagorean Triple (which Sides are (a ; (a + 1) ; (a + 2)), has infinite many "Sons" (so to speak); DPT : (3*k ; 4*k; 5*k).
There is only one FPPT : (3 ; 4 ; 5).
Let's assume that there are infinite many PT that satisfy the assumption (n ; (n + 1) ; (n + 2)) in the Field of Positive Integers.
n^2 + (n + 1)^2 = (n + 2)^2
n^2 + n^2 + 2n + 1 = n^2 + 4n + 4
n^2 - 2n - 3 = 0
n = -1 or n = 3
So, one must conclude, beyond any reasonable doubt in "The Uniqueness of the Fundamental Primeval Pythagorean Triple".
Thank you so much for your interest in Numbers Theory!!
@@LuisdeBritoCamacho Another fact you no doubt already know. A (3,4,5) triangle will not satisfy any triangle where the area has to be a multiple of the perimeter, as posed by your challenge. However, if you reformulated the challenge to have the perimeter as a multiple of the area, then a (3,4,5) satisfies P=2A. You also need to emphasize the (5,12,13) more, since it is one of two triangles, and the only PPT, where A=P.
Let's go!! 😅😅😅😅😅😅😅
1) AB = 6 and AB^2 = 36
2) sin(60º) = (sqrt(3) / 2) ; sin(60º) ~ 0,86603
3) cos(60º) = 1 / 2 = 0,5
4) BD = 6 * sin(60º) ; BD = 6 * (sqrt(3) / 2) ; BD = (3 * sqrt(3)) lin un ; BD ~ 5,2 lin un
5) AD = 6 * cos(60º) ; AD = 6 * (1 / 2 ) = 3 lin un
6) DD' // BC and DD' = BC
7) DD' = 3 * sin(60º) ; DD' = (3 * sqrt(3) / 2) lin un ; BC = DD' ~ 2,6 lin un
8) BC = (3*sqrt(3)/2) lin un ; BC ~ 2,6 lin un
9) DC^2 = BD^2 - BC^2 ; DC^2 ~ 20,28 ; DC ~ 4,5 lin un
10) M = 6 lin un
11) m = 4,5 lin un
12) h = 2,6 lin un
10 TPA = (M + m / 2) * h ; TPA = ((6 + 4,5) / 2) * 2,6 ; TPA = (10,5 / 2) * 2,6 ; TPA ~ 13,65 lin un
11) My Best Answer is Trapezoid Pink Area (TPA) is approx. equal to 13,65 Square Units.
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The rain brought a lot of destruction to our state, especially here in the capital.
(9-2,25)^0 5*5,25
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First comment and first like, can you pin it?
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