Nice problem. Haven’t followed you in a while and I miss your lessons. TH-cam Has a way of scattering your interest all over creation! Thanks for the video.
Yes, sometimes TH-cam also gives recommendations to me for videos with pretty girls showing clothing fashion 😅 But I always come back to math and geometry, I promise!
@ 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
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?
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.
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).
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.
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
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.
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.
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.
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 ]
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.
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.
Copiamos AB sobre D y obtenemos un triángulo isósceles DBA, cuya altura por A es √(3²-1²)=2√2= Área DBA---> Altura por D =4√2/3---> Área ABCD =[(4+1)/2]*(4√2/3) =10√2/3. Gracias y saludos
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).
Bom dia Mestre Fiz um pouco diferente do Sr dividi o Trapézio em 2 triângulos retângulos fiz Pitágoras e igualei as alturas. Daí só foi usar (B+b)xh)/2. Obrigado pela aula
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)
STEP-BY-STEP RESOLUTION PROPOSAL : 01) DC = X + 1 + Y ; X + 1 + Y = 4 ; X + Y = 3 02) X + Y = 3 03) h^2 = 9 - X^2 04) h^2 = 4 - Y^2 05) 9 - X^2 = 4 - Y^2 06) System of Two Nonlinear Equations with Two Unknowns : a) X + Y = 3 b) X^2 - Y^2 = 5 07) Solutions : a) X = (7 / 3) lin un b) Y = (2 / 3) lin un 08) Check : (7 / 3) + (2 / 3) = (9 / 3) = 3 09) h^2 = 9 - (49 / 9) ; h = (81 - 49) / 9 ; h = (32 / 9) 10) h^2 = 4 - (4 / 9) ; h = (36 - 4) / 9 ; h = (32 / 9) 11) h = (sqrt(32) / 3) lin un 12) h = (4*sqrt(2) / 3) lin un 13) (h / 2) = (2*sqrt(2) / 3) 14) TA = 5 * (2*sqrt(2) / 3) sq un 15) TA = [(10*sqrt(2)) / 3] sq un 16) TA ~ 4,714 sq un Therefore, OUR BEST ANSWER : Trapezoid Area equal [(10*sqrt(2)) / 3] Square Units or approx. equal to 4,714 Square Units.
My way of solution ▶ By the given trapezoid (ABCD) E ∈ [AC] and [DE] ⊥ [AE] and F ∈ [AC] and [FC] ⊥ [BF] [AB]= [EF]= 1 [DE]= x [FC]= 4-1-x [FC]= 3-x step-I) Let's apply the Pythagorean theorem for the right triangle ΔDEA : [DE]² + [EA]²= [AD]² [DE]= x [EA]= h [AD]= 3 ⇒ x² + h²= 3² x²+h²= 9...........Eq-1) step-2) Applying the Pythagorean theorem for the right triangle ΔFCB : [BF]²+[FC]²= [CB]² [BF]= h [FC]= 3-x [CB]= 2 ⇒ h² + (3-x)²= 2² h²+(3-x)²= 4.......Eq-2) if we get h² from the equation-1 and put this value in equation-2, we get: x²+h²= 9 h²= 9-x² ⇒ 9-x² + (3-x)²= 4 9-x²+9-6x+x² = 4 14= 6x x= 7/3 h= √9-x² h= √(9 - 49/9) h= 4√2/3 A(ABCD)= (4+1)*h/2 h= 4√2/3 A(ABCD)= 5*4√2/6 A(ABCD)= 10√2/3 A(ABCD) ≈ 4,714 square units
Solution: Let's label 2 points, one perpendicular to point C, point R, and other perpendicular to point D, point S Let's label BR = x RC = h In ∆ BCR, applying Pythagorean Theorem: x² + h² = 2² x² + h² = 4 ... ¹ In ∆ ADS, applying Pythagorean Theorem after label AS AS = 4 - (1 + x) AS = 3 - x (3 - x)² + h² = 3² 9 - 6x + x² + h² = 9 x² + h² = 6x ... ² Replacing ¹ in ² 4 = 6x x = 4/6 x = 2/3 Replacing x = 2/3 in equation ¹ (2/3)² + h² = 4 4/9 + h² = 4 (×9) 4 + 9h² = 36 9h² = 32 h² = 32/9 h = 4√2/3 Trapezoid Area A = ½ h (a + b) A = ½ 4√2/3 (1 + 4) A = 10√2/3 Square Units ✅ A ≈ 4,7140 Square Units ✅
Let's find the area: . .. ... .... ..... Let's add the points E and F on CD such that ABEF is a rectangle. In this case the triangles ADF and BCE are both right triangles and we can apply the Pythagorean theorem. With x=CE and y=AF=BE we obtain: AD² = AF² + DF² BC² = BE² + CE² AD² = AF² + (CD − EF − CE)² BC² = BE² + CE² AD² = AF² + (CD − AB − CE)² BC² = BE² + CE² 3² = y² + (4 − 1 − x)² 2² = y² + x² 9 = y² + (3 − x)² 4 = y² + x² 9 − 4 = (3 − x)² − x² 5 = 9 − 6x + x² − x² 6x = 4 ⇒ x = 4/6 = 2/3 ⇒ y² = 4 − x² = 4 − (2/3)² = 4 − 4/9 = 36/9 − 4/9 = 32/9 ⇒ y = √(32/9) = 4√2/3 Since y=AF=BE is the height of the trapezoid, we are able to calculate its area: A(ABCD) = (1/2)*(AB + CD)*y = (1/2)*(1 + 4)*(4√2/3) = 10√2/3 Best regards from Germany
Nice problem. Haven’t followed you in a while and I miss your lessons. TH-cam Has a way of scattering your interest all over creation! Thanks for the video.
I'm glad you liked the video! I’m always trying to keep things interesting, but it’s good to hear from you again.
Thanks for the feedback ❤️🙏
@@PreMath I followed your videos all the time . Sometimes I have no other way to solve a problem. Be sure you`re great !
Yes, sometimes TH-cam also gives recommendations to me for videos with pretty girls showing clothing fashion 😅
But I always come back to math and geometry, I promise!
@ 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
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Thanks for sharing ❤️
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
Excellent!
Thanks for sharing ❤️
@@PreMath Thank YOU
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?
This is the approach I took. I find the math is simpler this way.
your approach is better. thanks for sharing
Thank you!
You are very welcome! Thanks for the feedback ❤️
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.
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Thanks for the feedback ❤️
Tricky at first but easy as it progresses
Very well
Thanks Sir
That’s wonderful method of solve
Thanks PreMath
Good luck with respects ❤❤
I am so glad you liked it! Thanks for the feedback ❤️🙏
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).
Excellent!
Thanks for sharing ❤️
S=10√2/3≈4,717≈4,72
Excellent!
Thanks for sharing ❤️
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.
Keep watching..
Thanks for the feedback ❤️
S(ADF) =(3/4)S(ADC) = 2√2
S(ADC) = (4/3)S(ADF).
S(ABCD) = (5/3)S(ADF) = (10√2)/3.
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
You are very welcome!
Thanks for the feedback ❤️
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.
Excellent!
Thanks for sharing ❤️
3×3-1÷1=8squrooth=2.8284+1.94=4.7674
2.8284+1.9703=4.798769area
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.
Excellent!
Thanks for sharing ❤️
Use Heron’s formula on triangle DFA to get area = 2sqroot2. Area also = 1/2(3). h. So h=(4root2)/3 etc
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Thanks for sharing ❤️
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.
Thanks for the feedback ❤️
Strange we can think about things that don't exist. ....more to come! 😊
😀
Thanks for the feedback ❤️
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 ]
Excellent!
Thanks for sharing ❤️
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.
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.
Excellent!
Thanks for sharing ❤️
Copiamos AB sobre D y obtenemos un triángulo isósceles DBA, cuya altura por A es √(3²-1²)=2√2= Área DBA---> Altura por D =4√2/3---> Área ABCD =[(4+1)/2]*(4√2/3) =10√2/3.
Gracias y saludos
Excellent!
You are very welcome!
Thanks for sharing ❤️🙏
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).
Hello 'lm so happy to when of yours follower's I'm from Morocco
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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)
Bom dia Mestre
Fiz um pouco diferente do Sr dividi o Trapézio em 2 triângulos retângulos fiz Pitágoras e igualei as alturas. Daí só foi usar (B+b)xh)/2.
Obrigado pela aula
Excellent!
You are very welcome!
Thanks for sharing ❤️
And area is 4.714
Excellent!
Thanks for sharing ❤️
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)
Area of the trapezoid=1/2(1+4)(4√2/3)=10√2/3 square units.❤❤❤
Excellent!
Thanks for sharing ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) DC = X + 1 + Y ; X + 1 + Y = 4 ; X + Y = 3
02) X + Y = 3
03) h^2 = 9 - X^2
04) h^2 = 4 - Y^2
05) 9 - X^2 = 4 - Y^2
06) System of Two Nonlinear Equations with Two Unknowns :
a) X + Y = 3
b) X^2 - Y^2 = 5
07) Solutions :
a) X = (7 / 3) lin un
b) Y = (2 / 3) lin un
08) Check : (7 / 3) + (2 / 3) = (9 / 3) = 3
09) h^2 = 9 - (49 / 9) ; h = (81 - 49) / 9 ; h = (32 / 9)
10) h^2 = 4 - (4 / 9) ; h = (36 - 4) / 9 ; h = (32 / 9)
11) h = (sqrt(32) / 3) lin un
12) h = (4*sqrt(2) / 3) lin un
13) (h / 2) = (2*sqrt(2) / 3)
14) TA = 5 * (2*sqrt(2) / 3) sq un
15) TA = [(10*sqrt(2)) / 3] sq un
16) TA ~ 4,714 sq un
Therefore,
OUR BEST ANSWER :
Trapezoid Area equal [(10*sqrt(2)) / 3] Square Units or approx. equal to 4,714 Square Units.
Excellent!
Thanks for sharing ❤️
My way of solution ▶
By the given trapezoid (ABCD)
E ∈ [AC] and [DE] ⊥ [AE] and
F ∈ [AC] and [FC] ⊥ [BF]
[AB]= [EF]= 1
[DE]= x
[FC]= 4-1-x
[FC]= 3-x
step-I) Let's apply the Pythagorean theorem for the right triangle ΔDEA :
[DE]² + [EA]²= [AD]²
[DE]= x
[EA]= h
[AD]= 3
⇒
x² + h²= 3²
x²+h²= 9...........Eq-1)
step-2) Applying the Pythagorean theorem for the right triangle ΔFCB :
[BF]²+[FC]²= [CB]²
[BF]= h
[FC]= 3-x
[CB]= 2
⇒
h² + (3-x)²= 2²
h²+(3-x)²= 4.......Eq-2)
if we get h² from the equation-1 and put this value in equation-2, we get:
x²+h²= 9
h²= 9-x²
⇒
9-x² + (3-x)²= 4
9-x²+9-6x+x² = 4
14= 6x
x= 7/3
h= √9-x²
h= √(9 - 49/9)
h= 4√2/3
A(ABCD)= (4+1)*h/2
h= 4√2/3
A(ABCD)= 5*4√2/6
A(ABCD)= 10√2/3
A(ABCD) ≈ 4,714 square units
Solution:
Let's label 2 points, one perpendicular to point C, point R, and other perpendicular to point D, point S
Let's label
BR = x
RC = h
In ∆ BCR, applying Pythagorean Theorem:
x² + h² = 2²
x² + h² = 4 ... ¹
In ∆ ADS, applying Pythagorean Theorem after label AS
AS = 4 - (1 + x)
AS = 3 - x
(3 - x)² + h² = 3²
9 - 6x + x² + h² = 9
x² + h² = 6x ... ²
Replacing ¹ in ²
4 = 6x
x = 4/6
x = 2/3
Replacing x = 2/3 in equation ¹
(2/3)² + h² = 4
4/9 + h² = 4 (×9)
4 + 9h² = 36
9h² = 32
h² = 32/9
h = 4√2/3
Trapezoid Area
A = ½ h (a + b)
A = ½ 4√2/3 (1 + 4)
A = 10√2/3 Square Units ✅
A ≈ 4,7140 Square Units ✅
Excellent!
Thanks for sharing ❤️
Let's find the area:
.
..
...
....
.....
Let's add the points E and F on CD such that ABEF is a rectangle. In this case the triangles ADF and BCE are both right triangles and we can apply the Pythagorean theorem. With x=CE and y=AF=BE we obtain:
AD² = AF² + DF²
BC² = BE² + CE²
AD² = AF² + (CD − EF − CE)²
BC² = BE² + CE²
AD² = AF² + (CD − AB − CE)²
BC² = BE² + CE²
3² = y² + (4 − 1 − x)²
2² = y² + x²
9 = y² + (3 − x)²
4 = y² + x²
9 − 4 = (3 − x)² − x²
5 = 9 − 6x + x² − x²
6x = 4
⇒ x = 4/6 = 2/3
⇒ y² = 4 − x² = 4 − (2/3)² = 4 − 4/9 = 36/9 − 4/9 = 32/9 ⇒ y = √(32/9) = 4√2/3
Since y=AF=BE is the height of the trapezoid, we are able to calculate its area:
A(ABCD) = (1/2)*(AB + CD)*y = (1/2)*(1 + 4)*(4√2/3) = 10√2/3
Best regards from Germany
I solved it this way, as well.
Done this way too !
Excellent!
Thanks for sharing ❤️