In rectangle ABCD, ab = 4cm, ad = 3cm, fold the rectangle along the straight line AC, and point B falls at point E, connecting de. what is aced in four sides? Please explain the reason and calculate its area! Very urgent, this is today's home work, tomorrow to hand in, online 30 minutes, please hurry up, good answer bonus!

In rectangle ABCD, ab = 4cm, ad = 3cm, fold the rectangle along the straight line AC, and point B falls at point E, connecting de. what is aced in four sides? Please explain the reason and calculate its area! Very urgent, this is today's home work, tomorrow to hand in, online 30 minutes, please hurry up, good answer bonus!

Extending CE, ad, intersecting point O, (forming triangle OCA) it is proved that △ AEC ≌ △ CDA is SSS. (I will not describe it here) ≌ ECA = ≌ DAC, CE = ad (congruent triangle correspondence...) ∵∠ ECA = ∠ DAC ∵ co = Ao (equilateral to equilateral) ∵ co = Ao CE = ad ∵ OE = OD ∵ OED = ODE (equilateral to equilateral)

As shown in the figure, fold the rectangular piece of paper ABCD, first fold out the crease BD, and then fold so that the ad side coincides with the diagonal BD, and get the crease dg. if AB = 2, BC = 1, then the length of Ag is___ ．

According to the title: ab = 2, ad = BC = 1, in the RT △ abd, BD = AB2 + ad2 = 4 + 1 = 5. Passing point G is GH ⊥ BD, the vertical foot is h, from the folding, we can know: △ AGD ≌ △ HGD, ad = BC = 1, set the length of Ag as X, Hg = Ag = x, Hg = Ag = x, BG = 2-x, BG = 2-2-2-x, bh5-5-1 = 5-1 in the RT △ abd, in the RT △ bGH, from the Pythagorean theorem, we get BG2 = BH2 = BH2 + Hg2, (2-x) 2, (2-x) 2 (2-x) 2 (2) (2-x) 2 = (2) (2) (2-x) 2 = (5-x) 2 = (5-1) (5-1) (5-1) 2 (5-1) 2) (2 = (5 12

As shown in the figure, fold the rectangular piece of paper ABCD, first fold the diagonal BD, and then fold, so that the ad edge coincides with BD, and get the crease DG, if AB = 8, BC = 6, find the length of Ag

In RT △ abd, ab = 8, ad = BC = 6, BD = AB2 + ad2 = 82 + 62 = 10. From the properties of folding, △ ADG ≌ △ a'dg, ｜ a'd = ad = 6, a'g = AG, ｜ a'B = bd-a'd = 10-6 = 4, let Ag = x, then a'g = Ag = x, BG = 8-x. in RT △ a'bg, X2 + 42 = (8-x) 2, the solution is x = 3, that is, Ag = 3

As shown in the figure, fold the rectangular piece of paper ABCD, first fold out the crease BD, and then fold so that the ad side coincides with the diagonal BD, and get the crease dg. if AB = 2, BC = 1, then the length of Ag is___ ．

According to the meaning of the title: ab = 2, ad = BC = 1, in RT △ abd, BD = AB2 + ad2 = 4 + 1 = 5. Go through point G for GH ⊥ BD, the foot is h. from the folding, we can know: △ AGD ≌ △ HGD, ≌ ad = DH = 1, let the length of Ag be x, Hg = Ag = x, BG = 2-x, BH = 5-1. In RT △ bGH, we get BG2 = BH2 + Hg2, (2-x) 2 by Pythagorean theorem

As shown in the figure, in rectangular ABCD, ab = 4, ad = 8, ABC is folded along diagonal AC to get AEC. If EC intersects ad with F, FD is obtained

Let FD = x, AF = y, then AF + FD = x + y = ad = 8. (1) from the angle AEF = angle FDC = 90 degrees, AFE = cfdae = DC = 4 (AB = AE after folding along the diagonal AC), we can get the congruence of AFE and FDC, and we can get EF = FD = X. because AEF is a right triangle, we can get X2 (the square of EF) + 16 (the square of AE) = Y2 (AF

The quadrilateral ABCD is rectangle, the quadrilateral aced is isosceles trapezoid, AC ‖ De, ad = CE

∵ the diagonals of isosceles trapezoid are equal and equally divided, so ∠ ACD = ∠ CAE,
≌ △ AEC ≌ △ CDA (known ad = CE, AC common edge) ≌ △ ABC {symmetry axis of rectangular diagonal system}

The two right sides of a right triangle are 10cm, 8cm high and 5cm long

Solution: according to the two methods of calculating the area of right triangle, the length of oblique side can be obtained. The area of right triangle = one right side x the other right side △ 2, the area of right triangle = oblique side x the height of oblique side △ 2
Let the length of hypotenuse be a
Right triangle area = 10 cm × 8 cm △ 2 = a × 5 cm △ 2
The solution of the equation is a = 16 cm
A: the length of the hypotenuse is 16 cm
However, it is found that the sum of the squares of the two right angles is less than the square of the hypotenuse, which does not conform to the Pythagorean theorem

If the two right sides of a right triangle are 10cm and 5cm long, then the median line on the hypotenuse of the right triangle is? C
There is another: if the area of a square is 27, then the side length of the square is

Hypotenuse = radical (10 ^ 2 + 5 ^ 2) = 5 radical 5
Because the center line on the hypotenuse of a right triangle is half of the hypotenuse
So the center line on the hypotenuse = (1 / 2) * hypotenuse = (5 root sign 5) / 2
The side length of a square is root 27 = 3 root 3

If the height and the length of the middle line on the hypotenuse of a right triangle are 5cm and 6cm respectively, then the area of the right triangle is 5cm______ ．

The length of the middle line on the hypotenuse of the right triangle is 6cm, ∫ the length of the hypotenuse is 12cm, ∫ the height on the hypotenuse of the right triangle is 5cm, ∫ the area of the right triangle = 12 × 12 × 5 = 30cm2

If the height and the length of the middle line on the hypotenuse of a right triangle are 5cm and 6cm respectively, then the area of the right triangle is 5cm______ ．

The length of the middle line on the hypotenuse of the right triangle is 6cm, ∫ the length of the hypotenuse is 12cm, ∫ the height on the hypotenuse of the right triangle is 5cm, ∫ the area of the right triangle = 12 × 12 × 5 = 30cm2