<a href="#" class="seekto" data-time="54">0:54</a> 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 <a href="#" class="seekto" data-time="130">2:10</a> 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 <a href="#" class="seekto" data-time="345">5:45</a>, 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 <a href="#" class="seekto" data-time="457">7:37</a>. 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 <a href="#" class="seekto" data-time="184">3:4</a>: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?
