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Can you find area of the Purple shaded Trapezoid? | (Perimeter) | #math #maths | #geometry

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[imetroangola]

To briefly check this question, if you encountered it in a test, the first thing would be to see if the right triangle EBC has a solution (6, 8, 10).

Let's see:

If y=5, then 5=45 -5x→5x=40, then

x=8. We conclude that in fact ∆EBC has a solution (6, 8, 10), where y=5 and x=8. Like this,

AB=2×8 - 3=13, DC=3×8 - 3=21 and AD=y+1=5+1=6.

The area of ​​the trapezoid S, is given by:

S=(AB + CD)×AD/2

S=(13+21)×6/2=34×3

*S=102 square units*

[prossvay]

Perimeter of the trapezoid=50
2x-3+y+1+3x-3+10=50
5x+y=50++6-1-10=45
So 5x+y=45 (1)
Connect B to E (E on CD and BE right CD)
In ∆BCE
CE^2+BE^2=BC^2
CE=CD-AB=3x-3-2x+3=x
BC=10
BE=y+1 ; BC=10
x^2+(y+1)^2=10^2
x^2+y^2+2y+1=100
x^2+y^2+2y=99 (2)
(1) y=45-5x
(2) x^2+(45-5x)^2+2(45-5x)=99
So x=8 ; x=126/13 rejected
so y=45-40=5
2x-3=16-3=13
3x-3=24-3=21
y+1=5+1=6
Purple area=1/2(13+21)(6)=102 square units.❤❤❤

[unknownidentity]

Let's find the area:
.
..
...
....


From the known perimeter P we can conclude:

P = 50
AB + BC + CD + DA = 50
(2x − 3) + 10 + (3x − 3) + (y + 1) = 50
5x + y + 5 = 50
5x = 45 − y
⇒ x = 9 − y/5

Now let's add point E on CD such that ABED is a rectangle. In this case BCE is a right triangle and we can apply the Pythagorean theorem:

BE² + CE² = BC²
DA² + CE² = BC²
DA² + (CD − DE)² = BC²
DA² + (CD − AB)² = BC²
(y + 1)² + [(3x − 3) − (2x − 3)]² = 10²
(y + 1)² + x² = 10²
(y + 1)² + (9 − y/5)² = 10²
y² + 2y + 1 + 81 − (18/5)y + y²/25 = 100
y² + y²/25 + 2y − (18/5)y + 82 = 100
y² + y²/25 + (10/5)y − (18/5)y − 18 = 0
y² + y²/25 − (8/5)y − 18 = 0
25y² + y² − 40y − 450 = 0
26y² − 40y − 450 = 0

y = {40 ± √[40² − 4*26*(−450)]}/(2*26)
y = [40 ± √(1600 + 46800)]/52
y = (40 ± √48400)/52
y = (40 ± 220)/52

Since the length of DA (y+1) must be positive, the only useful solution is:

y = (40 + 220)/52 = 260/52 = 5
x = 9 − y/5 = 9 − 5/5 = 9 − 1 = 8

AB = 2x − 3 = 2*8 − 3 = 16 − 3 = 13
CD = 3x − 3 = 3*8 − 3 = 24 − 3 = 21
DA = y + 1 = 5 + 1 = 6

Now we are able to calculate the area of the trapezoid:

A(ABCD) = (1/2)*(AB + CD)*DA = (1/2)*(13 + 21)*6 = 3*34 = 102

Best regards from Germany

[georgebliss]

Not sure why AD is stipulated to be y + 1 because it could be designated as any unknown letter as per normal proceedure, and makes no difference to the answer.
e.g let AD = y.
y = 50 - (2x -3) - 10 - (3x -3).
y = 50 - 2x + 3 - 10 - 3x + 3.
y = 46 - 5x.
Then substitute for y in triangle EBC where y^2 + x^2 = 100.

[nandisaand]

Why didnt you do the y-substitution BEFORE expanding the binomial? Wouldve been a little less work

[MrPaulc]

P = 50,
5x + y + 5 = 50 which can be 5x + y = 45 and maybe an equation (1).
Drop a vertical from B to a new point E.
Spli DC into 2x - 3 and x.
(x^2) + (y+1)^2 = 100 perhaps an equation 2
I will expand the two equations:
(1) 5x + y = 45
(2) x^2 + y^2 + 2y + 1 = 100
Simplify (2) slightly with x^2 + y^2 + 2y = 99
Rearrange (1) to y = 45 - 5x
Now rewrite (2):
x^2 + (45 - 5x)^2 + 90 - 10x = 99
Second rewrite: x^2 + 2025 - 450x + 25x^2 + 90 - 10x = 99
26x^2 - 460x + 2115 = 99
26x^2 - 460x + 2016 = 0
Reduce: 13x^2 - 230x + 1008 = 0
That looks quite a kludge, but I'll test with the quadratic formula.
(230+or-sqrt(52, 900 - 4*13*1008))/26
(230+or-sqrt(484))/26
(230+or-22)/26
252/26 or 208/26. 208 looks the more likely as 208/26 is an integer (8), but that isn't proof.
For the moment, assume x = 8. That would make BCE a 6, 8, 10 if y = 5.
Try with those:
Base 6 (AD) and average height (13 + 21)/2 = 17.
The most likely solution is 102 un^2.
Now backtrack and test the other x value (252/26).
Assume x = 126/13, which looks much harder.
100 - (126/13)^2 = (y + 1)^2

100 - (15, 876/169) = (y + 1)^2

that would give x as about 6.06. Call it 6 for easier working.
y^2 + 2y + 1 = 6 so y^2 + 2y - 5 = 0
-2+or-sqrt(4 - 4*-5))/2
y=-2+or-sqrt(24))/2
y = -2+or-2*sqrt(6))/2
y = -1 + sqrt(6)
I will stop there, but it looks likethere might be two valid solutions.
Actually, ignore the last few lines because the perimeter wouldnt be 50 LOL

[santiagoarosam]

Perímetro =50=2x-3+3x-3+y+1+10→ y=45-5x→ y+1=46-5x → (46-5x)²+x²=10²→ x=8→ y=5→ AD=6 ; AB=13, DC=21 → Área ABCD =(21+13)*6/2 =102 ud².
Gracias y un saludo.

[himo]

y+5x-5=40 y=45-5x
3x-3-(2x-3)=x x²+(y+1)²=10² x²+(46-5x)²=100 x²+2116-460x+25x²=100
26x²-460x+2016=0 13x²-230x+1008=0 (13x-126)(x-8)=0 x=126/13 is rejected , x=8
y=5 y+1=5+1=6 2x-3=13 3x-3=21

Purple Trapezoid area = (13+21)*6/2 =102

[sergioaiex]

Solution:

Perimeter (P) = 50
P = (3x - 3) + (2x - 3) + (y + 1) + 10
(3x - 3) + (2x - 3) + (y + 1) + 10 = 50
5x + y + 5 = 50
y = 45 - 5x ... ¹

Pythagorean Theorem

a² + b² = c²

a = (3x - 3) - (2x - 3)
a = x

b = y + 1
b = (45 - 5x) + 1
b = 46 - 5x

c = 10

x² + (46 - 5x)² = 10²
x² + (2116 - 460x + 25x²) = 100
26x² - 460x + 2.016 = 0

x = [460 ± √ (211.600 - 209.664)]/52
x = (460 ± √1936)/52
x = (460 ± 44)/52
x = 115 ± 11/13
x' = 126/13 Rejected

x" = 104/13
x" = 8 Accepted

Let's replace in our equation 1

y = 45 - 5x
y = 45 - 5 (8)
y = 5

In Final Step, let's calculate Purple Trapezoid Area

A = ½ h (m + n)

h = 6
m = 21
n = 13

A = ½ h (m + n)
A = ½ 6 (21 + 13)

A = 102 Square Units

[Birol]

My way of solution ▶
P(ABCD)= 50
[AB]= 2x-3
[BC]= 10
[CD]= 3x-3
[DA]= y+1
⇒
2x-3+10+3x-3+y+1= 50
5x+y= 45
y=

b) E ∈ [DC] and [BE] ⊥ [EC]
by considering the right triangle ΔECB, we have:
[EC]= [DC]-[AB]
[EC]= 3x-3-(2x-3)
[EC]= x
[CB]= 10
[BE]= [DA]
[BE]= y+1
by applying the Pythagorean theorem we get:
[BE]²+[EC]²= [CB]²
(y+1)²+x²= 10²
y²+2y+1+x²= 100
x²+y²+2y=

c) By considering equation-1 and eqaution-2, and inserting the y value into the 2nd equation, we get :
x²+y²+2y= 99
y= 45-5x
⇒
x²+(45-5x)²+2*(45-5x)= 99
x²+2025-450x+25x²+90-10x= 99
26x²-460x+2016=0
dividing both sides by 2
13x²-230x+1008
Δ= (230)²-4*13*1008
Δ= 484
√Δ= 22

x₁= (230+22)/2*13
x₁= 252/26
x₁= 126/13
for the value x= 126/13
y= 45-5x
y= 45-5*126/13
y= - 45/13 ❌
y < 0 which is not possible !

x₂=( 230-22)/2*13
x₂= 8
⇒
x= 8 length units
y= 45-5*8
y= 5 length units

d) the area of the Trapezoid, A(ABCD)
A(ABCD)= [(2x-3)+(3x-3)]*(y+1)/2
A(ABCD)= [(2*8-3)+(3*8-3)]*(5+1)/2
A(ABCD)= [13+21]*(5+1)/2
A(ABCD)= 102 square units

[alster]

The factoring was tricky but the others were straightforward

[michaeldoerr]

The answer is x = 5 and y = 5. I really liked that you showed how to apply divisibility on justifying how to factor and group a quadratic eqn with large numbers. And this is probably good practice to eliminate x or y upon substitution!!!

[lilfuglsangstergaard]

Focus on triangle BEC. It is a special 90° triangle with side relation 3:4:5.
Side BE=10 is given and it is 2*5 then because of relationship on side length:
Side CE=2*4=8 and as in video this side is =x, so x=8
and Side BE=2*3=6 is = y+1, (so y=5) but we only need (y+1) thus it is the hight of trepezoid we need for computing area.
Area = [h*(a+b)]/2, h=6, a=side AE = 2X-3=2*8-3=13 and b=side DE = 3X-3=3*8-3=21. Then area=[6*(13+21)]/2= 102 😎
Then

[andryvokubadra]

My version

1:20 - 2:18
(2x-3) + (3x-3) + 10 + (y + 1) = 50
5x - 46 = -(y + 1)
46 - 5x = y + 1

6:05 - 8:48 10² - (3x-3-2x+3)² = (y + 1)²
100 - (x)² = (46 - 5x)²
100 - x² = 25x² - 460x + 2116
0 = 26x² -460x + 2016
x = 8 and 126/13

Ofcourse y & area values going to be found 😊😊😊

[ShelleyChristineSmith]

Always remember with these that they are annoyingly NEVER drawn to scale.

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :

01) Perimeter = 50 lin un

02) BC = 10 lin un

03) (2X - 3) + (3X - 3) + (Y + 1) = 50 - 10 ; (2X - 3) + (3X - 3) + (Y + 1) = 40 ; 5X + Y - 5 = 40 ; 5X + Y = 45

04) B'C = (3X - 3) - (2X - 3) ; B'C = (3X - 2X) - (-3 + 3) ; B'C = X lin un

05) AD = BB' = (Y +1) lin un

06) System of Two Equations (Linear and Nonlinear) with two Unknowns :

a) 5X + Y = 45
b) (Y + 1)^2 + X^2 = 100

07) Positive Integer Solutions :

08) X = 8 and Y = 5

09) AB = 13 lin un ; DC = 21 lin un ; AD = 6 lin un

10) Area = (13 + 21) * (6 / 2) ; Area = 34 * 3 ; Area = 102 sq un


Therefore,


OUR BEST ANSWER :

Purple Trapezoid Area equal 102 Square Units.

[wasimahmad-tc]

21+13=34÷2=17×6=102area x=8 y=5 answare