0:54 Extend AE to meet extended DC at F. ∆AEB is Congruent to ∆ECF by ASA, with CE=3 and EF =4. In the Right DAF, AD=6 Drop a perpendicular AQ on DF. By Similarity of ∆DAF with ∆AQF, AQ=4*6/10=12/5 AREA=10*12/5=24
*Solução:* Prolongue o segmento AC até encontrar com o prolongamento do segmento DC no ponto F. Os triângulos ∆ABE e ∆ECF são congruentes, pois: •∠EAD=∠CFE (AB//DC) • BE=EF •∠BEA=∠CEF(opostos pelo vértice) Assim , EF=4 e CF=3, por Pitágoras no ∆ADF, vamos ter AD=6. A altura h do trapézio é a mesma altura do ∆ADF, daí h×10=6×8 →h=24/5. A área S do trapézio é dada por: S=(3+7)24/10 *S=24 unidades quadradas*
The area is 24 units square. At the 2:10 mark, I simply thought that you could have concluded that since there is only one Pythagorean triple that has a side length of 4, the other lengths are 5 and 3. Also I NEEDED this review of what happens if HL similarity requires a 90° angle and a common angle: the first two letters are the first pair and the first and third letters are the second pair. I think that there should be a mnemonic considering how familit these rules regarding HL similarity are!!! I hope that I am going somewhere!!!
At 5:45, we can, alternatively, find the value of x using the Δ area formula. If AF = 3 is considered the base of ΔAEF, AE = 4 is the height and area = (1/2)(3)(4) = 6. If EF = 5 is the base, then AP = x is the height and the area is unchanged, so 6 = (1/2)(5)(x) and x = 12/5, as Math Booster also found at 7:37. Proceed from there.
Let F be the midpoint of AD. Draw EF. Draw AM, where M is the point on CD where AM is parallel to BC. Let N be the point of intersection between AM and EF. As AM and BC are parallel, and as AB, EF, and CD are parallel, and as BE = EC = x, then ABEN and NECM are congruent parallelograms and EN = CM = AB = 3 and AN = NM = x. As ∠ANF and ∠AMD are corresponding angles and thus congruent, and as ∠FAN = ∠DAM, then ∆NFA and ∆MDA are similar triangles. NF/AN = MD/AM NF/x = (7-3)/2x NF = (x)4/2x = 4/2 = 2 EF = EN + NF = 3 + 2 = 5 In triangle ∆FAE, as AE = 4 and EF = 5, then ∆FAE must be a 3:4:5 Pythagorean triple right triangle and FA = 3. As F is the midpoint of DA, then DF = FA = 3. Drop a perpendicular from A to G on CD. Let H be the point on intersection between EF and AG. As EF and CD are parallel, then ∠EHA = ∠CGA = 90°. As ∠EHA = ∠FAE = 90° and ∠AEF = ∠AEH, then ∆EHA and ∆FAE are similar triangles. HA/AE = FA/EF HA/4 = 3/5 HA = (4)3/5 = 12/5 As ∠AHF = ∠AGD = 90° and ∠FAH = ∠DAG, then ∆AHF and ∆AGD are similar triangles. AG/DA = AH/FA AG/6 = (12/5)/3 = 4/5 AG = (6)4/5 = 24/5 Trapezoid ABCD: A = h(a+b)/2 = (24/5)(3+7)/2 A = (12/5)10 = (12)2 = 24 sq units
@@ritwikgupta3655 so is it always true like in general that in a trapezoid when we draw any line parallel to the two bases of a trapezoid will always divide altitude of a trapezoid in the same ratio as it divides the two non parallel bases ? Here the line is a median so it is dividing both the non parallel bases and altitudes in equal halves .. Am I correct?
0:54 Extend AE to meet extended DC at F. ∆AEB is Congruent to ∆ECF by ASA, with CE=3 and EF =4.
In the Right DAF, AD=6
Drop a perpendicular AQ on DF.
By Similarity of ∆DAF with ∆AQF, AQ=4*6/10=12/5
AREA=10*12/5=24
*Solução:*
Prolongue o segmento AC até encontrar com o prolongamento do segmento DC no ponto F.
Os triângulos ∆ABE e ∆ECF são congruentes, pois:
•∠EAD=∠CFE (AB//DC)
• BE=EF
•∠BEA=∠CEF(opostos pelo vértice)
Assim , EF=4 e CF=3, por Pitágoras no ∆ADF, vamos ter AD=6.
A altura h do trapézio é a mesma altura do ∆ADF, daí
h×10=6×8 →h=24/5.
A área S do trapézio é dada por:
S=(3+7)24/10
*S=24 unidades quadradas*
梯形中線=5
左腰=6
三角形ADE=4*6/2=12
梯形面積=12*2=24
The area is 24 units square. At the 2:10 mark, I simply thought that you could have concluded that since there is only one Pythagorean triple that has a side length of 4, the other lengths are 5 and 3. Also I NEEDED this review of what happens if HL similarity requires a 90° angle and a common angle: the first two letters are the first pair and the first and third letters are the second pair. I think that there should be a mnemonic considering how familit these rules regarding HL similarity are!!! I hope that I am going somewhere!!!
At 5:45, we can, alternatively, find the value of x using the Δ area formula. If AF = 3 is considered the base of ΔAEF, AE = 4 is the height and area = (1/2)(3)(4) = 6. If EF = 5 is the base, then AP = x is the height and the area is unchanged, so 6 = (1/2)(5)(x) and x = 12/5, as Math Booster also found at 7:37. Proceed from there.
Agreed again! But he loves similar triangles.
Let F be the midpoint of AD. Draw EF. Draw AM, where M is the point on CD where AM is parallel to BC. Let N be the point of intersection between AM and EF. As AM and BC are parallel, and as AB, EF, and CD are parallel, and as BE = EC = x, then ABEN and NECM are congruent parallelograms and EN = CM = AB = 3 and AN = NM = x.
As ∠ANF and ∠AMD are corresponding angles and thus congruent, and as ∠FAN = ∠DAM, then ∆NFA and ∆MDA are similar triangles.
NF/AN = MD/AM
NF/x = (7-3)/2x
NF = (x)4/2x = 4/2 = 2
EF = EN + NF = 3 + 2 = 5
In triangle ∆FAE, as AE = 4 and EF = 5, then ∆FAE must be a 3:4:5 Pythagorean triple right triangle and FA = 3. As F is the midpoint of DA, then DF = FA = 3.
Drop a perpendicular from A to G on CD. Let H be the point on intersection between EF and AG. As EF and CD are parallel, then ∠EHA = ∠CGA = 90°.
As ∠EHA = ∠FAE = 90° and ∠AEF = ∠AEH, then ∆EHA and ∆FAE are similar triangles.
HA/AE = FA/EF
HA/4 = 3/5
HA = (4)3/5 = 12/5
As ∠AHF = ∠AGD = 90° and ∠FAH = ∠DAG, then ∆AHF and ∆AGD are similar triangles.
AG/DA = AH/FA
AG/6 = (12/5)/3 = 4/5
AG = (6)4/5 = 24/5
Trapezoid ABCD:
A = h(a+b)/2 = (24/5)(3+7)/2
A = (12/5)10 = (12)2 = 24 sq units
Can you explain me in the video how did he do AP/PQ= BE/EC?
(3)^2(4)^2(7)^2={9+16+49}=74 {45°A+45°B+90°C+90°D}=270°ABCD/74=3.48ABCD 3.2^24 3.2^12 3.2^2^6 3.1^1^2^3 1.1^2^3 2^3 (ABCD ➖ 3ABCD+2).
Uau! Que questão show. Bonita, muito bonita. Parabéns pela escolha!!! Brasil - Outubro de 2024.
หาค่า x จากสามเหลี่ยมบนที่เป็นสามเหลี่ยมมุมฉากยาว3 หน่วย 4 หน่วยและ 5 หน่วยก็ได้ครับไวด้วย
(3+7)/2=5 √[5²-4²]=3 3*4/2=5*h/2 h=2.4 2.4+2.4=4.8
area of trapezoid ABCD = (3+7)*4.8/2 = 24
How did you do AP/PQ= BE/EC ? What did you apply?
Because these are line segments between parallel lines
@@ritwikgupta3655 so is it always true like in general that in a trapezoid when we draw any line parallel to the two bases of a trapezoid will always divide altitude of a trapezoid in the same ratio as it divides the two non parallel bases ? Here the line is a median so it is dividing both the non parallel bases and altitudes in equal halves .. Am I correct?
AE×DC=F; ΔABE=ΔFCE (ASA)
FE=AE=4; AF=4+4=8
FC=AB=3; FD=7+3=10
ΔAFD(3/4/5)2; AD=3•2=6
[ABCD]=[AFD]=½8•6=24 sq.un. 😁
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