Can you calculate area of the Trapezoid? | (Triangles) | #math #maths | #geometry

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Can you calculate area of the Trapezoid? | (Triangles) | #math #maths | #geometry

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[charlesmitchell]

Nice problem. Haven’t followed you in a while and I miss your lessons. YouTube Has a way of scattering your interest all over creation! Thanks for the video.

[cyruschang]

h = height
2^2 - a^2 = 3^2 - (3 - a)^2 = h^2
4 - a^2 - 9 + 9 - 6a + a^2 = 0
4 - 6a = 0
a = 2/3
h = √(32/9) = (4√2)/3
area = (1 + 4)((4√2)/3)/2 = (10√2)/3

[davidellis]

I found a different solution. Dropping perpendiculars from A and B to CD divides CD into three segments with lengths x, 1 and 3-x. Then the required altitude h can be found by applying the Pythagorean Theorem to the two right triangles: h^2=2^2-x^2=3^2-(3-x)^2. The x^2 terms cancel out, leaving a linear equation with solution x=2/3. Then h^2=32/9, so h=sqrt(32/9)=4*sqrt(2)/3, and the area of the trapezoid is (h/2)(4+1), which is 10*sqrt(2)/3. Which approach do readers prefer?

[nandisaand]

@ 7:30:
Assign (X) to segment EF.
Assign (3-X) to DF.
X²+H²=2²
(X-3)²+H²=3²
Two eqns, 2 varibles. Solve and you find X=⅔.
Plug back into eqn above, and you find H=4•Sqrt(2)/3

[highlyeducatedtrucker]

Once you have the 3, 3, 2 isosceles triangle, you don't need to find its area. Area of a Trapezoid is (1/2)(a+b)(h). a and b are given, and you can find h from the isosceles triangle as you demonstrated in the video.

[jimlocke]

The area of ΔADF can, alternatively, be found using Heron's formula. The sides a, b, c are 2, 3, and 3. The semi perimeter, s, is (a + b + c)/2 = (2 + 3 + 3)/2 = 4. Area = √(s(s - a)(s - b)(s - c)) = √((4)(2)(1)(1) = √8 = 2√2. There is also Bretschneider′s formula for quadrilaterals, which would find the area directly from the lengths of the 4 sides and cos(Θ/2), where Θ is the sum of either 2 opposite angles. We have the lengths of all 4 sides. However, it is probably more straightforward to use PreMath's approach instead finding cos(Θ/2).

[scottdort]

Once you divide up the trapezoid into a triangle and a parallelogram, you can sum of those areas separately. The area of the triangle is going to be equal to 2*sqrt2.
The area of the parallelogram is going to be 1*2*cosx. Cosx = 2*sqrt2/3. Therefore, the area of the parallelogram is going to be 4/3*sqrt2.
If you have these two areas together, the sum is 10/3*sqrt2 just Premath previously stated.

[johnbrennan]

Use Heron’s formula on triangle DFA to get area = 2sqroot2. Area also = 1/2(3). h. So h=(4root2)/3 etc

[bakrantz]

Add line connecting AF. Use Heron's Formula to calculate area of triangle ADF as 2 root 2. Calculate height of trapezoid which is same as height of triangle ADF using the base DF of 3 and the area from Herons as (4 root 2)/3. Calculate area of trapezoid as 1/2h(a+b) which equals 20/6 root 2.

[TurquoizeGoldscraper]

I never thought of breaking it up to make an isosceles triangle like that. I broke it up into the center rectangle and two right triangles and used Pythagoras to figure out the height.

[andrepiotrowski]

Enlarge the isosceles triangle so that the side length is 4 and use the ray theorem to calculate the areas!
A = A_4 - A_1 = A_3 * ((4/3)^2 - (1/3)^2) = A_3 * 15/9 = 5/3 * sqrt(8) = 10/3 * sqrt(2)

[ChuzzleFriends]

Draw a segment thru point A and a point E on base CD such that it is parallel to leg BC.
This forms a parallelogram ABCE (since segment CE is on base CD, so it is parallel to base AB).
By the Parallelogram Opposite Sides Theorem, AE = 2 & CE = 1.
So, DE = CD - CE = 4 - 1 = 3.
So, △ADE is an isosceles triangle.
Draw a segment thru point D and the midpoint F of segment AE. This is an altitude of △ADE. It should intersect with segment AE at a right angle.
So, AF = EF = 1. Apply the Pythagorean Theorem on △AFD.
a² + b² = c²
1² + (DF)² = 3²
(DF)² + 1 = 9
(DF)² = 8
DF = √8
= 2√2
Find the area of △ADE.
A = (bh)/2
= 1/2 * 2 * 2√2
= 2√2
Draw another altitude of △ADE. This altitude should intersect point A and a point G on segment DE. It is the height of the trapezoid (and the parallelogram on that statement).
Use the area formula to find this height AG.
A = (bh)/2
2√2 = 1/2 * 3 * AG
2√2 = 3/2 * AG
AG = 2√2 * 2/3
= (4√2)/3
Find the area of parallelogram ABCE.
A = bh
= 1 * (4√2)/3
= (4√2)/3
Trapezoid ABCD Area = △ADE Area + Parallelogram ABCE Area
= 2√2 + (4√2)/3
= (6√2)/3 + (4√2)/3
= (10√2)/3
So, the area of the yellow trapezoid is (10√2)/3 square units (exact), or about 4.71 square units (approximation).

[sorourhashemi]

Thanks challenging. Connect B to CD as perpendicular and mark it as M . so ∆BMC and mark MD as a. MC=4-a. ===>BM^2=4-(4-a) and connect A to CD as well. So BM^2= 9- (a-1)===>a=10/3 and MB ( height)= 1.888, area

[yalchingedikgedik]

Very well
Thanks Sir
That’s wonderful method of solve
Thanks PreMath
Good luck with respects ❤❤

[quigonkenny]

Draw EA, in such a way that E is the point on CD where EA is parallel with BC. As AB is already parallel with CE, then ABCE is parallelogram, and CD = AB = 1, EA = BC = 2, and ED = CD-CE = 4-1 = 3. Let ∠EDA = θ.

In triangle ∆EDA, by the law of cosines:

cosθ = (ED²+DA²-AE²)/(2ED(DA))
cosθ = (3²+3²-2²)/(2(3)(3))
cosθ = (9+9-4)/18
cosθ = 14/18 = 7/9

sin²θ + cos²θ = 1
sin²θ + (7/9)² = 1
sin²θ + 49/81 = 1
sin²θ = 1 - 49/81 = 32/81
sinθ = √(32/81) = (4√2)/9

Draw AF, where F is the point on ED where AF and ED are perpendicular. Let AF = h.

sinθ = AF/DA
(4√2)/9 = h/3
h = 3(4√2)/9 = (4√2)/3

Trapezoid ABCD:
A = h(a+b)/2
A = ((4√2)/3)(1+4)/2
A = (2√2)5/3
[ A = (10√2)/3 ≈ 4.714 sq units ]

[MrPaulc]

I got the same answer with a different method. I extended AB with lines x and y to make a rectangle. I made two equations by Pythagoras and compared them to calculate x, y, and h. I then subtracted the two triangles that I made from the rectangle area. It is a valid method but I felt yours was more compact and a bit cleaner.

[michaelstahl]

Great solution again, Sire . I didn`t expect that way. I propose to draw a vertical line from A to CD and from point B to CD. I call it point F . The distance between E and F ist 1 LU.
The distance between F and C may be x units. The distance DE is 4 - 1 - x = 3 - x . When I use the theorem of Pythagoras on the triangles ADE and BFC so I got for x = 2/3.
So h = 4 * Square ( 2 ) / 3 and the area is 10 * square ( 2) / 3.

[bandarusatyanandachary]

When 3 sides of a triangle are given we can find its area by using simple formula square root of s(s -a)(s - b)(s - c)

[alster]

Tricky at first but easy as it progresses

[ellenmortensen]

Im so gratefull for your channel. Learning to put in to formula has been great. Done all of your tutorials.
But now im looking for some tutorials being a bit more difficult. Can you provide.