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As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC, points E and F are on AB and AC respectively, and AE = AF, it is proved that de = DF
0
Finding AE ratio EC of BD: DC 2:3 AF: DF 4:1 in triangular ABC
DG ∥ AC, BF to g,
Because BD / DC = 2 / 3
CE/DG=CB/BD
If BD / BC = 2 / (2 + 3), then BC = (5 / 2) BD
∴CE=(5/2)DG
AE/DG=AF/FD=4/1
∴AE=4DG
∴AE/EC=4DG/[(5/2)DG]=8/5
As shown in the figure, ∠ 1 = ∠ 2, ∠ EDC = ∠ BAC, AE = AF, ∠ B = 60 ° in △ ABC, then the line segments AF, BF, AE, EC, ad, BD, DC and DF in the graph with the same length as de are______ Article
Connect Fe to ad to o,
The △ AFE is an isosceles triangle
∵∠1=∠2，
If Ao ⊥ EF, and fo = OE, DF = De
∵∠EDC=∠BAC，
∴△ABC∽△EDC，
∵∠ABC=60°，
Ψ Dec = 60 °, AED = 120 °, AFD = 120 °,
The △ FBD is an equilateral triangle
∴BF=BD=DF=DE．
Therefore, there are three lines equal to the length of de
(please note: when ∠ BAC = 60 °, all the other 7 segments except AD are equal to the length of De)
So the answer is: 3
As shown in the figure, in the triangle ABC, BD: DC = 1:2, e is the midpoint of AD. if the area of triangle ABC is 120 square centimeters, what is the area of shadow part?
BD: DC = 1:2, so the triangle abd = 12, the triangle ADC = 13, the triangle ABC = 120 × 13 = 40 (square centimeter), e is the midpoint of AD, so the triangle AEC = triangle Dec = 12 triangle ADC, thus we can get: Triangle abd = triangle Dec = triangle AEC = 40 (square centimeter), the crossing point E is eg ‖ BC, and the intersection AB is ab
As shown in the figure: the area of triangle ABC is 70 square centimeter, BD = CD = 6 cm, ∠ C = 45 ° to find out the area of shadow part
Area of triangle abd = area of triangle ADC = 70 △ 2 = 35 (square centimeter),
Because BD = CD = 6 cm, ∠ C = 45 °, the area of the triangle DCE is 9 (square centimeter),
Shadow area = 35-9 = 26 (square centimeter),
A: the shadow area is 26 square centimeters
Given that the area of the triangle ABC is 12 square centimeters, calculate the area of the shadow part?
Let the right sides of the isosceles right triangle ABC be a, 12a2 = 12, A2 = 24; the area of sector abd: 18 π A2 = 3 π = 9.42 (square centimeter), the area of blank part BCD: 12-9.42 = 2.58 (square centimeter), the area of semicircle: 12 π (a △ 2) 2 = 12 × 3.14 × 14a2 = 18 × 3.14 × 24 = 9.42 (square)
As shown in the figure, the area of triangle ABC is 50cm2, twice the area of parallelogram efcd, and the area of shadow part in the figure is______ cm2．
50÷2÷2，
=25÷2，
=5 (square centimeter);
A: the shadow area is 12.5 square centimeters
So the answer is: 12.5
In the triangle ABC of the right figure, DC = 2bd, CE = 3aE, the area of shadow part is 20 square centimeter, and the area of triangle ABC is______ Centimeter
According to "DC = 2bd, CE = 3aE", s △ ade = 1
4S△ADC，S△ADC=2
3S△ABC，S△ADE=1
6S△ABC，
So the area of the triangle ABC is 20 ÷ 1
6 = 120 (square centimeter),
A: the area of triangle ABC is 120 square centimeters
So the answer is: 120 square centimeter
In the triangle ABC of the right figure, DC = 2bd, CE = 3aE, the area of shadow part is 20 square centimeter, and the area of triangle ABC is______ Centimeter
According to "DC = 2bd, CE = 3aE", s △ ade = 1
4S△ADC，S△ADC=2
3S△ABC，S△ADE=1
6S△ABC，
So the area of the triangle ABC is 20 ÷ 1
6 = 120 (square centimeter),
A: the area of triangle ABC is 120 square centimeters
So the answer is: 120 square centimeter
As shown in the figure, translate the right triangle ABC along the BC direction to get the triangle def. If AB = 8cm, be = 4cm, DH = 3cm, calculate the area of the shadow part Although the characters are ugly, I can still see them clearly. The black part is the shadow part. Please give me the process. I want to complete it. It's better to use ∵
370190792：
∵ ∵ ABC translation to obtain △ def
∴DE=AB=8cm
S△ABC=S△DEF
∴S△ABC-S△HEC=S△DEF-S△HEC
That is, s right angle trapezoid abeh = s shadow
∵DE=8cm,DH=3cm
∴HE=DE-DH=8-3=5cm
ν s shadow = s right angled trapezoid abeh
=1/2(AB+HE)×BE
=1/2(8+5)×4
=(1/2)×13×4
=26cm²
The area of the shadow part is 26 cm 2
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