Can you find area of the Yellow Trapezoid? | (Triangle) |
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- Опубликовано: 7 фев 2025
- Learn how to find the area of the yellow shaded Trapezoid. Important Geometry and Algebra skills are also explained: Trapezoid; Trapezium; Heron's Formula; area of the trapezoid formula; area of the triangle formula. Step-by-step tutorial by PreMath.com
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Very interesting, professor. I didn't know that I can use the formula to calculate the area of trapezoid even if the length of the side from the figure are different (7 and 8). Thank you for the video, it really helps.
You are welcome!
Thanks for the feedback ❤️🙏
This construction is a mathematical impossibility.
I solved the problem, but hit a curious impossibility in the process. Drop a perpendicular down from points A and B. This creates two triangles with height h and two different hypotenuses equalling 7 and 8. The base of each of these triangles can be labelled x and 3-x. Using the pythagorean theorem, we get equation (1) x^2+h^2 = 49 and (3-x)^2+h^2 = 64 or equation (2) 9-6x+x^2+h^2 = 64. Let's subtract the first equation (1) from the second equation (2), which produces 9-6x = 15, which simplifies to 6x = -6, or x= -1, which is a geometric impossibility given the diagram above. Yet plugging this value back into equations (1) and (2) correctly produces the value h = sqrt48 = 4*sqrt3, and the area of the trapezoid can be correctly calculated. Further, you can use this value of x= -1 to calculate the area of the trapezoid in a piecemeal fashion by adding up the area of the rectangle in the middle (4*sqrt3*6), the area of the triangle on the right side ((4*sqrt3*4)/2) and SUBTRACTING the area of the triangle on left ((-4*sqrt3)/2). This totals up to 30*sqrt3. This would mean that the line with value 7 runs inward and strikes the base at a value of x=-1 to the perpendicular, creating a "negative triangle" that needs to be subtracted from the total area. Or is there a mistake somewhere in my reasoning? Mr. Premath, please weigh in!
Your are correct. The drawing is misleading. The line with value 7 runs inward.
The diagram can't exist! Suppose we draw a circle of radius 7 centred on D, and a circle of radius 8 centred on C. Let height h vary.
AB = 9 - (sqrt(7^2-h^2) + sqrt(8^2-h^2))
With h=0, AB = 9-7-8 = -6, clearly not a solution. However the right hand side is an increasing function of h. Let's see if we can get AB=6. The maximum for real AB is when h=7, in which case
AB = 9 - sqrt(64-49)
= 9 - sqrt(15)
~ 5.13
Hence, the diagram can't be drawn. Others point out that it can be drawn if, in effect, x is negative, as you found.
x=-1 means that the trapezium is drawn wrongly. ie point D is located 1 unit to the right of the foot of h.
I got the same as you.
Solving the equation gives me -1 for x ....
PreMath qualified "this figure may not be %100 true to the scale." In fact angle D is obtuse.
(From diagram at 4:48) If h = 4*SQRT(3), then using Pyth theorem, FC^2 + (4*SQRT(3))^2 = 8^2 ==> FC^2 = 64 - 48 ==> FC = 4 ==> this is impossible because DF would be 9-4=5, but DF is obviously > 6
I calculated the height in almost the same way as shown in the video. Then I checked my results in the way you did, and I was also confused. But it is possible to construct a trapezoid with the given side lengths. In this case it is a so called obtuse trapezoid (point A is located on the left side of point D, not on the right side).
Best regards from Germany
Mathematics King is Only and Only PreMath, Thank You for Giving Us Knowlege
Kevinmathscience?
In spanish language, trapezoid can't have parallel sides !!!!
Only trapezium can have !!!
"Trapezium" is not equal to "trapezoid" in Spanish
There is a great confusion about this !!!
This trapezoid is not possible with these dimensions.
I think you might be right. See my earlier post where I was unable to draw it.
@@abstragic4216 At 6:26 the right triangle with a hypotenuse of 8 and a height of 4√3 would need a base of 4.
I used an obtuse trapezoid calculator to construct this figure. The side lengths and the height are as given in the diagram, but the side AD is obtuse, and angle ADC is equal to 98.21°. Further, angle DCB = 60°, angle CBA = 120° and angle BAD = 81.79°. That is an accurate construction that yields h= 4*sqrt3 and an area of 30*sqrt3.
This is obtuse trapezoid
h2 + FC2 = 64 then FC = 4 and 4 + 6 = 10 > DC
*_Solution: IS IT POSSIBLE TO CONSTRUCTION OF THIS TRAPEZOID?_*
Let AM and BN be the height of the trapezoid drawn from vertices A and B, such that AM=BN=h. Furthermore, let DM=x and CN=y.
AMNB is a rectangle , therefore
y + x + 6 = 9 → *y + x = 3.*
By Pythagoras, in ∆BNC and ∆ADM, we have, respectively:
64 = h² + y² and 49 = h² + x², hence,
64 - 49 = h² + y² - h² - x² = y² - x²
y² - x² = 15
(y + x ) (y - x) = 15
since y + x = 3, then
3 (y - x) = 15 → y - x = 5.
Now, y + x = 3 and y - x = 5, therefore
y = 4 and x = -1. *_What is impossible!!!_*
I run to the same problem
I used a different way to solve the problem and got the same result as you:
Let H and K the orthogonal projections of A and B on CD, then DH=x, HK=6 and KC=3-x. Pythagoras: h²=7²-x² = 8²- (3-x)² and x=-1!!
@WernHerr Ok! I wanted to detail the problem more, but the important thing is to verify that it is not possible to construct the trapezoid.
0:04 @imetroangola17 Read my reply to this problem. I hit the same issue. The trapezoid can be constructed, but the line with the value of 7 would need to tilt in the same direction as the line with the value of 8. The bottom of the trapezoid would still be 9, but point D would be one unit to the right of point A, while point C would be 4 units to the right of point B. So the trapezoid would not be convex the way it is depicted in the diagram. This is the only way for the trapezoid to be drawn.
@imetroangola17 The trapezoid cannot be drawn the way it is in the diagram. If I use an X,Y coordinate system and put point A at (0,4*sqrt3), then point B is at (6,4*sqrt3), point D is at (1,0) and point C is at (10,0). This would work geometrically and would also provide the right area of the trapezoid.
Be H and K the orthogonal projections of A and B on (CD) and h = AH = BK. In triangle ADH: DH^2 = AD^2 - AH^2 = 49 -h^2, so DH = sqrt(49 -h^2)
In the same way, we get KC = sqrt(64 -h^2). As DC = DH + HK + KC we have: 9 = sqrt(49 -h^2) + 6 + sqrt(64 -h^2), or sqrt(49 - h^2) + sqrt(64 - h^2) = 3
We square: 49 - h^2 + 64 - h^2 +2.sqrt((49 -h^2).(64 - h^2)) = 9, or sqrt((49 - u).(64 - u)) = u - 52, with u = h^2. Once again we square:
3136 - 49.u - 64.u +u^2 = u^2 - 104.u +2704, or -113.u + 3136 = -104.u + 2704, and then u = 48 and h = sqrt(48) = 4.sqrt(3)
The area of the trapezoïd is h.((AB + DC)/2) = (4.sqrt(3)).((6+9)/2) = 30.sqrt(3)
Draw vertical to DC from A and B which will meet DC at E and F respectively. Using Pythagoras theorem we can find h from triangle ADE and BFC Where DE DE is x and FC is 3-x.
Thank you!
You're welcome!
Thanks for the feedback ❤️🙏
"The drawing may not be 100% true to the scale!" As several viewers have found, if you drop a perpendicular from A to CD, point D is to the RIGHT of the intersection by 1 unit. The drawing inaccurately implies that it is on the left. If you drop a perpendicular from B to CD, point C is 4 units to the right of the intersection. Points C and D remain 9 units apart. If the base of the triangle formed on the left is x, the base of the triangle on the right is 3 + x. The two perpendiculars, AB and a line segment of length 6 between the perpendiculars' intersections with extended CD form a rectangle of area 6h, where h is the distance between AB and CD. Add the area of the triangle on the right and deduct the area of the triangle on the left to get the area of the trapezoid. Or, just use the trapezoid area formula with bases AB and CD and height h.
Thanks for the feedback ❤️🙏
Focusing on PreMath's ΔBCE, construct a half circle with end points B and C (center at the midpoint of BC) and point E on the same side of BC as the half circle. Construct CF of length 3 where F is on the half circle. Construct BF. ΔBFC must be a right triangle because
I used an obtuse trapezoid calculator to construct this figure. The side lengths and the height are as given in the diagram, but the side AD is obtuse, and angle ADC is equal to 98.21°. Further, angle DCB = 60°, angle CBA = 120° and angle BAD = 81.79°. That is an accurate construction that yields h= 4*sqrt3 and an area of 30*sqrt3.
Let FC=x;
BC²-FC²=BF²=AG²=AD²-DG²;
BC²-FC²=AD²-DG²;
8²-x²=7²-(9-x-6)²;
64-x²=49-(3-x)²;
64-x²=49-9+6x-x²;
24=6x;
x=4
So the triangle BCF has a double hypotenuse of the smaller leg, that is, it is a classic 30°-60°90° right triangle, therefore:
h=BF=FC*√3=4√3
The rest is trivial
The dimension of the trapezoid is not possible.
We can draw two vertical lines from the upper-right and upper-left corners to the base. Both lines will correspond to the height h. We can then apply Pythagorean formula and write
√(7^2 - h^2) + √(8^2 - h^2) = 3
√(7^2 - h^2) = 3 - √(8^2 - h^2)
Square both sides
(7^2 - h^2) = 3^2 + (8^2 - h^2) - 6√(8^2 - h^2) 7^2 - 8^2 - 3^2 = - 6√(8^2 - h^2)
-24 = -6√(8^2 - h^2)
4 = √(8^2 - h^2)
h^2 = 64 - 16 = 48, so h = 4√3
So far so good.
But if you write
√(8^2 - h^2) = 3 - √(7^2 - h^2)
After squaring both sides, you get
(8^2 - h^2) = 3^2 + (7^2 - h^2) - 6√(7^2 - h^2) 8^2 - 7^2 - 3^2 = - 6√(8^2 - h^2)
6 = - 6√(8^2 - h^2) this is not possible
If you plug in h = 4√3 to
√(7^2 - h^2) + √(8^2 - h^2)
√(7^2 - 48) + √(8^2 - 48) = √(1) + √(16) = 5, not 3
exactly.
Let's find the area:
.
..
...
....
.....
First of all we add point E on CD such that ABCE is a parallelogram. Now we calculate the area of the triangle ADE by applying Heron's formula:
AD = 7
AE = BC = 8
DE = CD − CE = CD − AB = 9 − 6 = 3
s = (AD + AE + DE)/2 = (7 + 8 + 3)/2 = 18/2 = 9
A(ADE) = √[s*(s−AD)*(s−AE)*(s−DE)] = √[9*(9−7)*(9−8)*(9−3)] = √(9*2*1*6) = √108 = 6√3
The height h of the trapezoid ABCD has the same length then the height of the triangle ADE according to its base DE:
A(ADE) = (1/2)*DE*h(DE) = (1/2)*DE*h ⇒ h = 2*A(ADE)/DE = 2*6√3/3 = 4√3
Now we are able to calculate the area of the yellow trapezoid:
A(ABCD) = (1/2)*(AB + CD)*h = (1/2)*(6 + 9)*4√3 = 30√3
Remark: This trapezoid does not have the shape shown in the diagram. In fact, it is an obtuse trapezoid.
Best regards from Germany
Trazamos una paralela a BC por A y obtenemos el triángulo AED---> ED=9-6=3---> Área del triángulo AED =√(9*6*2*1) =6√3---> Distancia de AB a DC =h=2*6√3/3 =4√3 ---> Área ABCD =AED+ABCE =6√3 +(6*4√3) =30√3 u².
Gracias y un saludo cordial.
It only works if AB ≠ 6 but AB = 4 → CE = 5 → CBE = α →
cos(α) = 11/14 → sin(α) = 5√3/14 → h = 4√3 →
area ABCD = (1/2)(4 + 9)4√3 = 26√3
Professor's answer is correct, but the shape of the figure is not as shown. Drop a vertical line from A and mark the intersection of that line with DC as E, and drop a vertical line from B and mark the intersection of that line with DC as F.
Define a as the length of DE. Then the length of FC is 3-a. If you now apply pythagorean to AED and BCF you get two equations in a and the height h. Combine the two equations to eliminate h and solve for a and you get a=-1. So point D is actually down and right of A, not down and left of A. But the area I calculated this way is still the same as professor's answer.
One thing that is interesting in this problem is that typically in geometry you reject lengths that come out negative, but in this case a negative dimension actually works.
At 6:30 in the video, we have h= 4*sqrt3 in a right triangle with the hypotenuse equal to 8. According to the pythagorean theorem, FC must be equal to 4. But just before this point, at 6:25 in the video, EC is shown as equal to 3. Interesting that our instructor gets the overall right answer with a flawed figure and values that cannot be true. This problem has been an adventure....
Thank you for another good one.
I used Pythagoras instead of Heron's thus:
Construct parallel line BE and drop perpendicular BF as per given solution.
Let FC = x.
Then EF = 3-x
Using Pythagoras in triangles BEF and BCF respectively :
h^2 = 7^2-(3-x)^2
and h^2 = 8^2 - x^2
therefore 7^2-(3-x)^2 = 8^2 - x^2
this simplifies to x = 4
which means h^2 = 8^2 - 4^2 = 48
therefore h = sqrt(48) = 4*sqrt(3)
And yellow area = h*(9+6)/2 = 30*sqrt(3)
Amusingly, this agrees with the given answer, but I haven't figured out how to draw it to scale, because x being 4 is greater than EC (which is 3)!
49 - X^2 = 64 - (3-X)^2
X = 1, h = 48^.5 = ~6.93
S = h × (9 + 6)/2 = ~52
Move AD so that point A is on point B.
AB, C, D is a triangle with sides 7,8,3.
By Heron's, area is sqrt((9)(1)(2)(6)) = sqrt(108) so 2*sqrt(27).
Base is 3. Area is 2*sqrt(27)
(1/2)base*height = area, so area/((1/2)base) = height
(2*sqrt(27))/(3/2) equivalent to (2*sqrt(27)) * (2/3)
height = (4*sqrt(27))/3
sqrt(27) is 3*sqrt(3), therefore,
(12*sqrt(3))/3 which = 4*sqrt(3).
h = 4*sqrt(3)
4*sqrt(3) * (9+6)/2
4*sqrt(3) * 15/2 = 30*sqrt(3) un^2 approx 51.96 un^2
Unusually, I see we did it pretty much the same way :)
Find height (h).
Drop a perpendicular from B to a point I will call E.
CE is x
8^2 - x^2 = h^2
The problem is quite right...no mistake...area is available !
√(49-h^2)+6+√(64-h^2)=9, h=4√3.
To correct this problem, the slope of line AD should be reversed.
To be clear: the triangle ebc can have that h and area, but dec is bent, not a straight line.
Thanks for the feedback ❤️🙏
Assalamualaikum Sir
How do you do Sir?
Mahin From Bangladesh.
Bom dia Mestre
Trapezium formula=1/2 a.(h1+h2)//
???
Sir while finding the area of trapezoid in the last how A = 9 and B = 6???
If height is sqr(48) then you get indeed :
- One triangle with 90 ° and dimensions : 1, 7 and sqr(48)
- Another triangle with 90 * and dimensions : 4 , 8 and sqr(48)
But then you get a lenght of DC of 1 + 6 + 4 ....
This exercice is wrong
This dimensions are not possible for trapezium.
Thanks for the feedback ❤️🙏
No Solution Possible with these lengths.
Thanks for the feedback ❤️🙏
@@PreMath Thank Very Much for your daily work in the name of Mathematical Sciences.
Solution is possible if you allow angle D to be obtuse.
@@LuisdeBritoCamacho
Many thanks for your kind words😀
You are very generous❤
u have to delete this video sir
lengths r wrong
Drawing is impossible. You can delete this exercice