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As shown in the figure, in the isosceles trapezoid ABCD, ad ‖ BC, ad = 3, BC = 5. AC and BD intersect at point O, and ∠ BOC = 60 °. If AB = CD = x, then the value of X is______ ．
As shown in the figure, through point D, make de ⊥ BC, ∵ ad ∥ BC, ad = 3, BC = 5. ∠ BOC = 60 °, the ∥ BOC and ∥ AOD are equilateral triangles, OB = OC = 5, OA = od = 3, ∥ BD = 8, ∥ be = 4, CE = 1, ∥ de = 43. In RT △ DCE, we can get DC = 7, that is, x = 7
The rectangle ABCD is folded along the straight line BD, so that point C falls at point C ', BC intersects ad at point E, ad = 8, ab = 4, find the area of △ BDE
I can't transfer pictures. I'm sorry,
Prove that △ Abe ≌ △ c'de (AAS)
So AE = c'e
Let c'e = x, then de = 8-x, c'd = 4
Pythagorean theorem, the solution is x = 3,8-x = 5
So s △ BDE = (AB * de) / 2
=(4*5)/2
=10
Fold the rectangle ABCD along the diagonal BD so that point a falls to point F. if AB = 4, ad = 8, what is the area of the overlapping triangle BDE?
Because the folding can get BF = AB = 4, angle f = angle c = 90 degrees
Simultaneous angle bef = angle Dec
So △ bef ≌ △ Dec
So be = de
Let CE = x, then be = de = 8-x
We can get (8-x) ^ 2 = 4 ^ 2 + x ^ 2
The solution is x = 3, be = 5
The triangle area is (1 / 2) be × CD = 5 × 4 △ 2 = 10
As shown in the figure, fold the rectangle ABCD along the straight line BD so that the point C falls at C ', and BC' intersects with E. ad = 8, ab = 4, be = 5, and find the area of triangle c'ed
When the angle BCD is lifted up, BC will intersect with E
According to the conditions:
RCH '= RCH = 90 degrees
c'D=DC=AB=4,c'B=CB=8
So c'e = c'b-be = 8-5 = 3
So s △ c'ed = 1 / 2 * c'e * c'd = 1 / 2 * 3 * 4 = 6
When ad = 8, ab = 4, the area of triangle ABF is calculated
Because be = De, if AE = x, then be = 8-x
So AB ^ 2 + AE ^ 2 = be ^ 2
If x = 3, then AE = 3, be = 5, then the height of be is equal to 12 / 5
And BF = BC = 8, so the area of triangle ABF is 48 / 5
In the parallelogram ABCD, the angles DAB = 60 degrees, ab = 2, ad = 4 fold the triangle CBD to the triangle EBD along BD, and make the plane EBD perpendicular to the plane abd
In the parallelogram ABCD, the angles DAB = 60 degrees, ab = 2, ad = 4 fold the triangle CBD to the triangle EBD along BD, and make the plane EBD perpendicular to the plane abd
Verification, AB vertical de
Finding the side area of triangular pyramid e-abd
In the triangle abd, using the angle DAB = 60 degrees, ab = 2, ad = 4, it is not difficult to get BD = 2 times root 3 and ab vertical BD, plane EBD, vertical plane abd, EF vertical plane abd, EF vertical ab. in the plane bed, ABAB vertical BD has EF vertical ab. so AB vertical plane EBD, so AB vertical de!
As for the side area, it doesn't count
Calculate the area skillfully, as shown in the figure rectangle ABCD, ab = 4, ad = 6, e is on the midpoint of AB, BF + 2FC, calculate the area of triangle EFD
Boss, why don't you have a picture? And "BF + 2FC" should be "BF = 2FC"
Triangle EFD = rectangle ABCD triangle ade triangle EFB triangle FDC = 24-6-4-4 = 10
Guess it should be
In the AA parallelogram ABCD, the angles DAB = 60 degrees, ab = 2, ad = 4 fold the triangle CBD along BD to the position of the triangle EBD, so that the plane EDB is perpendicular to the plane
Finding the side area of e-abd
Because, DAB = 60 °, ab = 2, ad = 4
So ∠ abd = ∠ BDC = ∠ BDE = 90 °
Because plane EBD ⊥ plane abd
So the de ⊥ plane abd
So ab ⊥ de
Known: as shown in the figure, in rectangular ABCD, ab = 3, BC = 4, fold △ BCD along the straight line of BD, so that point C falls on point F, if BF intersects ad at e, find the length of AE
Let AE = x, de = 4-x, according to Pythagorean theorem, we can get: be = x2 + 9, so EF = bf-be = bc-be = 4-x2 + 9. ∵ AEB ∵ def, ∵ aeef = Bede. ∵ AB = 3, BC = 4, ∵ x = 78
As shown in the figure, in rectangular ABCD, points E and F are on edge AB and CD respectively, BF / / De, if ad = 12cm,
AB = 7cm, and AE: EB = 5:2, then the shadow is a rectangular part. What is the area of ebfd? 30 points will be awarded! Otherwise, no points! No picture, please draw it yourself!
Because ABCD is rectangular, ad is parallel and equal to BC, AB is parallel and equal to DC, because AB = 7cm and AE: EB = 5:2, AE = 5cm, EB = 2cm, ED is parallel to BF, so ebfd is parallelogram, that is EB = DF = 2cm, CF = 5cm, so s triangle AED = s triangle CFB = 1 / 2 * 7 * 5 = 17.5cm ^ 2, so s yin = s
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