If the circumference of the parallelogram ABCD is 50 cm and ab is 3 cm shorter than CD, then CD,

The perimeter of the parallelogram ABCD is 50cm 2 (CD + AD) = 50 and cd-ad = 3. The solution is: CD = 14, ad = 11, CD = 14cm, ad = 11cm

It is known that the circumference of the parallelogram ABCD is 42cm, ab = 2BC. Find the length of each side of the parallelogram ABCD

If the graph is known to be a parallelogram, then AB = CD, ad can be assumed= BC.AB +BC = 42 / 2 = 21cm, because AB = 2BC, so AB + BC = BC * 3 = 21, BC = 7cm, then ad = BC = 7cm, ab = CD = 14cm

It is known that the perimeter of the parallelogram ABCD is equal to 60, and ab: BC = 1:2, find the length of each side of the parallelogram ABCD

Because the quadrilateral ABCD is a parallelogram
So AB = CD, BC = ad
Let AB = x, then CD = x, BC = ad = 2Ab = 2x
So x + X + 2x + 2x = 60
So x = 10
So AB = CD = 10, BC = ad = 20
A: the sides of parallelogram ABCD are 10, 10, 20 and 20

If the circumference of straight parallelogram ABCD is 36cm, the sum of diagonals is 24cm, AB: BC = 4:5, what is the circumference of triangle ABC

In the parallelogram ABCD, AB: ad = 2:3, if the perimeter is 20cm, BC =?

AB: ad = 2:3, ab = 2 / 3 ad, perimeter = 2 (AB + AD) = 10 / 3 ad = 20, the solution is ad = 6, so BC = ad = 6

As shown in the figure, in the parallelogram ABCD, the crossing point a is AE ⊥ BC, the perpendicular foot is e, connecting De, and F is a point on the line De, and ∠ AFE = ∠ B. If AB = 4, ad = 33, AE = 3, then the length of AF is______ ．

In RT △ ade, de = ad2 + AE2 = 6; ∵ ad ∥ BC, ab ∥ CD, ∵ AE ⊥ BC, ∵ AE ⊥ ad, ∵ ad = 33, AE = 3; in RT △ ade, de = ad2 + AE2 = 6; ∵ ad ∥ BC, ab ∥ CD, ∫ ADF = ∠ CED, ∠ B + ∠ C = 180 °, ∫ AFE + ∠ AFD = 180, ∠ AFE = ∠ B, ∫ AFD = ∠ C

In the parallelogram ABCD, the crossing point a is AE ⊥ BC, the perpendicular foot is e, connecting De, and F is a point on the line De, and ∠ AFE = ∠ B. find △ ADF ∽ Dec

If ABCD is a parallelogram, then ∠ B ＋ C = 180 degrees, and ∠ AFE = ∠ B, ∠ AFE ＋ AFD = 180 degrees
So ∠ AFD ＋ C
AE ⊥ BC, that is, AE ⊥ ad, then ⊥ AED ＋ ade = 90 degrees, and ⊥ AED ＋ DCE = 90 degrees
Therefore, ADE = DCE
It can be obtained as follows: △ ADF ∽ Dec

In the parallelogram ABCD, the crossing point a is AE ⊥ BC, the perpendicular foot is e, connecting De, and F is a point on the line De, and ∠ AFE = ∠ B. prove △ ADF ∽ Dec
If AB = 4, ad = 3, radical 3, AE = 3, find the length of AF

A quadrilateral ABCD is a parallelogram
∴ AD//EC
∠ADE=∠DEC ①
And ∠ AFE = ∠ B
∠AFD=180°-∠AFE
∠C=180°-∠B
∴ ∠AFD=∠C ②
Also, EDC = DAF
From (1), (2) and (3)
△ADF∽△DEC
From △ ADF ∽ Dec
AF:DC=AD:DE ④
And AE ⊥ BC
The ead is a right triangle
There is Pythagorean law
DE=√(3√3)^2+3^2=6
And ab = DC = 4, ad = 3, radical 3
From 4
AF:4=3√3:6
AF=3√3*4/6=12√3/6=2√3
That is AF = 2 √ 3

As shown in the figure, in the parallelogram ABCD, the crossing point a is AE vertical BC, the perpendicular foot is e, connecting De, f is a point on the line De, and the angle AFE = angle B. (1) verification
(1) Proof: Triangle ADF is similar to triangle Dec.

∠B+∠DCE=180
∠AFE+∠AFD=180
Angle AFE = angle B
Therefore, DCE = AFD
AD//BC
∠ADF=∠DEC
So triangle ADF is similar to triangle Dec

In the parallelogram ABCD, the crossing point a e is perpendicular to BC, the perpendicular foot is e. connect De, f is a point on the line De, and ∠ AFE = ∠ B
(1) Prove that triangle ADF is similar to triangle Dec
(2) If AB = 4, ad = 3, radical 3, AE = 3, find the length of AF

It is proved that: first, because ad is parallel to BC, so ∠ ADF = ∠ Dec; then, ∠ AFE = ∠ B, and ∠ AFD + ∠ AFE = 180,
Therefore, AFD = DCE: so similar

If the circumference of the parallelogram ABCD is 50 cm and ab is 3 cm shorter than CD, then CD,