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The known points D and E are on the sides AB and AC of the triangle ABC, ab = 10cm, AC = 6cm, ad = 2cm=__ Triangle ade is similar to triangle ABC Since there is no drawing, there will be two situations I have one more question to answer
There are two
1. When AD / AB = AE / AC, AE = 1.2
2. When AD / AC = AE / AB, AE = 10 / 3
AD.AE They are the center line and height of the triangle ABC, BC = 6cm, AE = 4cm
S ⊿ ABC = 4 × 6 / 2 = 12 (cm? 2) s ⊿ abd = s ⊿ ADC = 6 (cm? 2) [s ⊿ ade cannot be determined! ∈ [0., + ∞)]
It is known that △ ABC is similar to △ AED, D is the midpoint of AB, AB is equal to 10cm, AC = 6cm, and the length of AE is calculated
∵ D is the midpoint of ab
∴AD=1/2AB=5
∵△ABC∽△AED
∴AD/AC=AE/AB
5 / 6 = AE / 10
6AE=5×10
AE=50/6
As shown in the figure, given the triangle ABC ∽ triangle ade, AE = 5, AC = 9, de = 6, ∠ a = 70 ° and ∠ B = 40 degrees, find the degree of 1. ﹤ C and ﹤ AED, 2. The length of BC As shown in the figure, the known triangle ABC ∽ triangle ade, AE = 5, AC = 9, de = 6, ﹤ a = 70 °, and ﹤ B = 40 degrees, seek 1. Degree of ∠ C and ∠ AED 2. Length of BC
.∠C=180°-70°-40°=70°
From triangle ABC to triangle ade
∠AED=∠C=70°
AE/AC=DE/BC
5 / 9 = 6 / BC, then BC = 54 / 5
Given the triangle ABC ~ triangle AED, where angle B = angle AED, ad = 5cm, bd7cm, AC = 8cm, find the length of AE and the similarity ratio of triangle ABC and triangle AED
Triangle ABC ~ triangle AED, AB / AE = AC / ad AE = ab × AD / AC AE = (7 + 5) 5 / 8 = 7.5
The similarity ratio of the triangle is equal to the ratio of the corresponding edge = AC / ad = 8 / 5
As shown in the figure, in △ ABC, AD is 2 times of AC, AE is 3 times of AB, and the area of △ ABC is several times of △ AED?
It should be one sixth. LZ did not give the graph. I guess AD and AC are in the same line, AE and ab are in the same line. Then extend ad to f to make AF = 3aC. Then LZ should know how to do it
As shown in the figure, D and E in △ ABC are points on BC, BD = ad, AE = EC, ∠ ade = 80 ° and ∠ AED = 66 ° to find the degrees of △ ABC angles
Because ad = BD, so ∠ abd = ∠ bad = 40 °
Similarly, ∠ ace = ∠ EAC = 33 °
∠BAC=∠BAD+∠EAC+∠DAE(=180°-∠ADE-∠AED)=107°
(1) AB = 4, AE = 3, BC = 5, find the circumference of △ Abe, (2) AB = 4, AC = 3, △ Abe and △ ace perimeter (3) What is the relationship between △ Abe and △ ace areas in 2
Because AE is the center line of BC, be = CE = BC / 2 = 5 / 2 = 2.5, so the circumference of triangle Abe = AB + be + AE = 4 + 3 + 2.5 = 9.5; 2) because AB = 4, BC = 5, AC = 3, then AB ^ 2 + AC ^ 2 = 16 + 9 = 25 = 5 ^ 2 = BC ^ 2, according to the Pythagorean theorem, the triangle ABC is a right triangle, ∠ BAC = 90
It is known that AE is the median line of RT △ ABC, ab = 6, AC = 8. ① find s △ Abe; ② find the circumference difference of △ ace and △ Abe
AB = 6, AC = 8, so BC = 10
So ace = 2 AE = 1
AC = 10 so be = CE = 5
Perimeter difference = 5 + 8 + 5-5-5-6 = 2
As shown in the figure, in △ ABC, AE is the center line on the edge of BC. We know that AC = 4cm, BC = 3cm, AE = 5cm, and the circumference of △ ace of the ball
Because AE is the midpoint on the edge of BC, BC = 3cm, so EC = 1.5cm,
Because AC = 4 cm, AE = 5 cm, so △ ace = EC + AC + AE = 1.5 + 4 + 5 = 10.5 (CM)
RELATED INFORMATIONS
- 1. As shown in the figure, it is known that AE is the center line of triangle ABC, and the perimeter of triangle Abe is 2cm more than that of triangle ace, if AB + A If AB + AC = 18cm, request AB, AC length
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