Can you find area of the Trapezoid ABCD? | Trapezoid | Trapezium | #math #maths

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Can you find area of the Trapezoid ABCD? | Trapezoid | Trapezium | #math #maths

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[jonchester]

Another day, another wonderful mathematical mystery to solve from PreMath! This is the way I love to start my day. Thx much as always. 🙂

[marcgriselhubert]

Let's note a = AD, b = DC and let H be the orthogonal projection of C on (AB).
Angle HBC is 180° - 135° = 45°, so triangle HBC is right angle and isosceles and HB = HC = AD = a and AB = a + b.
We use the Pythagorean theorem in right angle triangle ADC: a^2 + b^2 = 15^2 = 225, and also in BAD: (a+b)^2 + a^2 = 19^2 = 361
Let's develop, we have then a^2 + b^2 = 225 and 2.(a^2) + 2.a.b + b^2 = 361. At second equation we substract the first equation, we obtain: a^2 + 2.a.b = 136.
The area of the trapezoid is ((DC + AB)/2). AD = ((b+ a+b)/2).a = (a +2.b).(a/2) = (a^2 + 2.a.b):2, so it is 136/2 = 68.

[visualjewels]

So nice math tutorial, stay blessed my dear friend ❤❤❤👍

[mumps]

I love it…solving h(x+y)/2 without determining actual values for x, y, or h! I don’t know how people’s brains work this way, but I’m glad they do!

[Abby-hisf]

The angle 135 was given for a good reason! Thanks again

[jimlocke]

Very clever! At 6:10, we have 3 equations and 3 unknowns, so we could solve for the 3 unknowns and then use them in the formula for area of a trapezoid. However, you compute a numeric value for the expression h(x + y), without the individual values of h, x, and y. That is all you need to compute the area. Nicely done!

When I tried to compute the individual values, I wound up with a fourth order, even orders only, equation for h. So, with only even orders, I could reduce it to a quadratic equation and solve. I did not carry all the way through with the calculation, which would have meant doing involved algebra, very carefully to avoid making errors. However, I have full confidence that the problem can be solved that way.

[ChuzzleFriends]

Trapezoid Area Formula = h[(a + b)/2]
Extend segment DC to point E. Then draw segment BE such that it is perpendicular to segment DE.
Thus, we get a rectangle ABED. This splits the exterior angle into two angles.
(I'll label a point on the extended side outside segment AB as point F. Large exterior angle becomes ∠CBF)
∠EBF is a right angle. So, m∠CBE = 45°.
Because ∠E is also a right angle by definition of rectangles, m∠BCE = 45°. This makes △BEC an isosceles right triangle (Special 45°-45°-90°). Segment CE has the length of the height of the trapezoid.
The Parallelogram Opposite Sides Theorem proves opposite sides of all parallelograms are congruent. Label segment CD as a (from the formula).
Then, the length of segment AB, base b = a + h.
The area of the trapezoid becomes: h[(2a + h)/2]
a² + h² = 15² = 225
(a + h)² + h² = 19²
a² + 2ah + h² + h² = 361
h² + 2ah + 225 = 361
h² + 2ah = 136
h(2a + h) = 136
h[(2a + h)/2] = 68 (This is the formula!)
So, the area of trapezoid ABCD is 68 square units.

[santiagoarosam]

Llamamos E a la proyección ortogonal de C sobre AB → El ángulo EBC =180-135=45º → DC=b=AE ; AD=h=CE=EB →→
h²+b²=15² ; h²+(b+h)²=19² = h²+b²+h²+2bh =(h+b)² + (h²+2bh) =(15²) + (2*ABCD) =19² →→ ABCD=(19²-15²)/2 =68
Interesante acertijo. Gracias y un saludo cordial.

[gulshanjoshi]

Managed to workout the area =h/2 (2a+h)=68. Then also managed to work out DC=a=15, AD=h=4 and AB=a+h=19. Area = (15+19)×4/2=68. Found it challenging. Thanks.

[hongningsuen]

When there are 3 variables, search for 3 independent simultaneous equations relating these 3 variable to ensure existence of a solution. In this case, there are 3 equations:
X^2 + h^2 = 19^2, Y^2 + h^2 = 15^2, Y = X - h (by Pythagoras theorem and equal sides of isosceles triangle). So a solution is ensured. The problem is algebraic solution for the variables is complicated. The algebraic solution for individual variables is not necessary here when geometric intuition is used. Substitute X - h for Y into equation 2 to give (X - h)^2 + h^2 = 15^2. Subtraction of equation 2 from equation 1 gives 2Xh - h^2 = 136. Note that Xh is area of a rectangle when the missing triangle of area h^2/2 is added to the trapezium. Hence area of the trapezium = Xh - X^2/2. 2Xh - h^2 = 2(Xh - h^2/2) = 136. Hence area of trapezium = 136/2 = 68.

[phungpham]

Alternative approach:
Label CD= a, BC= b and CE=h (the height of the trapezoid)
The area of the trapezoid= Area of the rectangle ADCE + area of the right isosceles triangle CEB = h.a + sqh/2
Consider the triangle ABD: sq h = sq 19 - sq b= sq 19- sq(a+h)= sq19 - ( sqa+sqh + 2ah)
Because (sqa+sqh)= sq sq h = sq19 - (sq15 +2ah)
----> sqh +2ah = sq19-sq15
sqh/2 + ah= (sq19-sq15)/2
Area of the trapezoid= (sq19-sq15)/2= 68 sq units

[johnjones]

7:10 Ah, the part that eluded me. I noticed h = x - y, but I didn't think of substitution. Instead, I tried to solve for x and y and got a valid algebraic result, but it was a geometrically impossible result (19-15)(19+15). So I was stumped.

I have since graphed all the information and found x ≈ 18.547 and y ≈ 14.422.

I wonder what the exact values are...

[jan-willemreens]

... Good day, I guess I was in the mood for a long math/arithmetic adventure this time (lol) ... I extended DC to new point E to create rectangle ABED ... Triangle CEB is an isosceles 45-45-90 triangle, so I AD I = I CE I = I EB I = X ... now applying Pythagoras as follows ... SQRT(15^2 - X^2) = I DC I and now using I DB I = 19 as follows ... 19^2 - X^2 = [ SQRT(15^2 - X^2) + X ]^2 ... after a few tedious algebraic steps I obtained the equation ... 5X^4 - 1172X^2 + 136^2 = 0 [ X^2 = T ] .... 5T^2 - 1172T + 136^2 = 0 .... applying Quadratic Formula ... T1 = (1172 + SQRT(D))/10 = approx. 217.383 (rejected) v T2 = (1172 - SQRT(D))/10 = approx. 17.017 , where D = 1003664 ... again after some more algebra I obtained the most logic outcome for X ... X = approx. 4.125 & I AB I = approx. 18.547 & I DC I = approx. 14.422 .... so finally A(Tr.) = ( 18.547 + 14.422 )/2 * 4.125 = approx. 68 u^2 ... a very hard way, so only fools like me took this arithmetically tedious path for once (LOL) ... thanks for your much better solution strategy (lol) ... best regards, .... pfff ... Jan-W

[devondevon]

68
Let label DC = m
DA =p
Draw a perpendicular line from AD to C forming a 45 -45-90 right
triangle SBC
SC=DA = p
SB= DC =m
Hence, AB = m+ p
Hence, the area of the trapezoid is (m+ m + p)*p*1/2 =
(2m + p)p/2 = (2mp + p^2 )/2 THIS WILL GIVE THE AREA OF THE TRAPEZOID
Using PYthagorean, then
m^2 + p^2 =15^2
(m + p)^2 + p^2 = 19^2
m^2 + p^2 + 2mp + p^2 = 361
15^2 + 2mp + p^2 =361 [substitute the value for m^2 + p^2 = 15^2]
2mp + p^2 = 361- 225
2mp +p^2 = 136
(2mp + p^2)/2 = 68

[StephenRayWesley]

(15)^2°=225° (19)^2= 361° (225°+361°)= 586° (90°+90°+135°+45°) =360° (586°-360°)=√226 √2^√11 3^2 1^√11^1 3^2 √1^√1^√1 3^2 (x+2x-3)

[prossvay]

Area of the trapezoid=68 square units. ❤❤❤ Thanks

[LuisdeBritoCamacho]

My solution is very simple:
x = DC ; y = AD ; (x + y) = AB, just because CB is the diagonal of a square [CC'BB'] and CB' = C'B = AD.
Solving a System of Nonlinear equations with 2 Unknowns:
1) x^2 + y^2 = 15^2 ; x^2 + y^2 = 225
2) (x + y)^2 + y^2 = 19^2 ; (x + y)^2 + y^2 = 361
There are only one set of integer Solutions, wich is : x ~ 14, 422 and y = 4, 125
Now, B ~ 18, 545 ; b ~ 14, 422 ; h ~ 4, 125
A = [(B + b) * h/2] = (18, 545 + 14, 422) * (4, 125 / 2) = 32, 967 * 2, 0625 = 67, 99444375
Answer:
The Area of Trapezoid [ABCD] equals approx. 68 square units.

[parthtomar]

🎉🎉🎉
Nice solution sir but it's a class 10th level question

[XR]

2 x (AC + DB) = 68 Big coincidence. 🤔

[wackojacko]

Yom! That's Mennonite for Yah Mon! 🙂