If the upper and bottom of a trapezoid is increased by 5 cm, it becomes a parallelogram. If the upper and bottom are reduced by 7 cm, it becomes a triangle, and the area is reduced by 21 square cm, Original area of trapezoid

It shows that the upper bottom of the original trapezoid is 7, and the lower bottom is 7 + 5 = 12. The area reduction of 21 is a triangle with the bottom of 7cm
First find the height of trapezoid, 21x2 divided by 7 = 6cm
(7 + 12) X6 divided by 2
=19x6 divided by 2
=57

Triangle AOD, triangle AOB, triangle COD are four congruent triangles, and quadrilateral ABCD is a diamond
Triangle AOD, triangle AOB, triangle COD are four congruent triangles. Is quadrilateral ABCD a diamond?
Think for yourself! As usual, I can't take pictures with my mobile phone

Because the diagonals of the quadrilateral ABCD intersect at O and △ AOB, △ cob, △ cod, △ AOD are four congruent right triangles,
Because the diagonals of the quadrilateral ABCD intersect at O, the diagonals are equally divided
So ad = AB = BC = CD
So the quadrilateral ABCD is a diamond

In diamond ABCD, how many isosceles triangles and right triangles do the diagonals AC and BD intersect at point o

Four isosceles triangles: △ ABC, △ abd, △ ADC, △ BCD
Four right triangles: RT △ AOB, RT △ AOD, RT △ BOC, RT △ cod

As shown in the figure, ad in trapezoidal ABCD is parallel to BC, the diagonal intersects o, the area of triangle AOD is 20, and the area of triangle ABO is 30

∵ s △ abd = s △ ACD ∵ s △ ABO = s △ CDO = 30 ∵ s △ CDO ∶ s △ BCO = 2 ∶ 3 ∵ s △ BCO = 30 × 3 / 2 = 45 ∵ area of trapezoidal ABCD = 20 + 30 + 30 + 45 = 125 question: why s △ CDO ∶ s △ BCO = 2 ∶ 3 question: why s △ CDO ∶ s △ BCO = 2 ∶ 3

Quadrilateral ABCD, diagonal AC, BD intersect o, if △ AOB ≌ △ AOD ≌ △ cod, then quadrilateral ABCD is what special quadrilateral? Tell me your reason

Square
prove
Because △ AOB ≌ AOD ≌ cod
So Ao = Bo = co = do
So Ao + CO = Bo + do
So AC = BD
And because △ AOB ≌ △ AOD
So AB = ad
According to the fact that two diagonals are equal, a set of edges are equal
So it is proved that the quadrilateral is a square

In the parallelogram ABCD, AC and BD intersect at the point O. what kind of motion does the triangle AOD go through to get the triangle cob

Rotate 180 ° on the plane with 0 as the center, and then rotate 180 ° on the plane perpendicular to the paper plane

If s △ AOB = 4 and s △ cod = 9, the minimum area of quadrilateral ABCD is ()
A. 21B. 25C. 26D. 36

Let the distance from point a to edge BD be H. as shown in the figure, in any quadrilateral ABCD, s △ AOB = 4, s △ cod = 9; ∵ s △ AOD = 12od · h, s △ AOB = 12ob · H = 4, ∵ s △ AOD = OD · 4ob = 4 × odob, s △ BOC = ob · 9od = 9 × obod; let odob = x, then s △ AOD = 4x, s △ BOC = 9x; ∵ s quadrilateral ABCD =

If s △ AOB = 4 and s △ cod = 9, the minimum area of quadrilateral ABCD is ()
A. 21B. 25C. 26D. 36

In trapezoidal ABCD, AB is parallel to CD, AC and BD and intersects at point O. if AC = 5, BD = 12, the median line length is 6.5, the area of triangle AOB is S1, and the triangle cod is 1
If the area is S2, the root sign S1 + root sign S2 =?

Make be parallel AC AC AC AC DC extension line at e, we can see CE = AB, be = AC = 5, BD = 12, de = DC + AB = 6.5 * 2 = 13
5 ^ 2 + 12 ^ 2 = 13 ^ 2 is Pythagorean array, so BDE is right triangle area equal to trapezoid area = 12 * 5 / 2 = 30
Triangle AOB, COD, EBD similar, similar ratio = area ratio square
So root S1 / root (sbde) = AB / 13
Root sign S2 / root sign (sbde) = CD / 13
If the two formulas are added, the root sign S1 + root sign S2 = root sign (sbde) = root sign 30

Given that the trapezoid ABCD, AB / / CD, diagonal AC and BD intersect at point O, and AO: OC = 2:1, the area of triangle cod is 3, what is the trapezoid area

If the upper and bottom of a trapezoid is increased by 5 cm, it becomes a parallelogram. If the upper and bottom are reduced by 7 cm, it becomes a triangle, and the area is reduced by 21 square cm, Original area of trapezoid