![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
In the isosceles triangle ABCD, ab = AC, ∠ a = 36 ° and BD is the square of ∠ ABC, then calculate AD / AC
In the isosceles triangle ABC, the vertex angle a = 36 degrees, and BD is the bisector of the angle ABC
Angle ABC = angle c = 72 degree
Angle abd = angle CBD = 36 ° = angle a
So ad = BD
Angle BDC = angle a + angle abd = 36 ° + 36 ° = 72 ° = angle c
So BC = BD
So ad = BC
Triangle BCD similar triangle ABC
AC/BC=BC/CD
AD^2=BC^2=AC*CD=AC*(AC-AD)
(AC/AD)^2-AC/AD-1=0
AC / ad = (1 + 5 ^ (1 / 2)) / 2 or (1-5 ^ (1 / 2)) / 2 (rounding off)
So ad / AC = (5 ^ (1 / 2) - 1) / 2
The quadrilateral ABCD is an isosceles triangle, where ad = BC, if ad = 5, CD = 2, ab = 8, find the area of trapezoid ABCD?
Proof: link AC, BD
ABCD is a rectangle
∴AC=BD.
E.f.g.h is the midpoint of AB, BC, CD and Da respectively
∴ EF=GH=AC/2.EH=GF=DB/2
∴EF=FG=GH=HE
The efgh is a diamond
Must satisfy ad = BC, height is trapezoidal waist (actually Square)
In the isosceles triangle ABCD, ad is parallel to BC, ab = CD, and AC is vertical to BD, AF is the height of trapezoid, and the area of trapezoid ABCD is 49cm
The area of trapezoid ABCD is 49cm ^ 2, and the length of AF is calculated
The area is equal to ac * BD △ 2 = 49, and AC = BD, so AC = BD = 7 times root 2
After analysis, when AF is fixed, no matter how long AD and BC are, the sum of them is constant. Find a limit state, that is, when a and D coincide, the trapezoid becomes an isosceles right triangle, and you can find out the length of BC at the bottom. This value is the sum of ad at the upper bottom and BC at the lower bottom. Then you can find AF according to the area formula, and the result is AF = 7
As shown in the figure, in the isosceles triangle ABCD, ad is parallel to BC, BC-AD = 4cm, GH is the median line of the trapezoid, GH = 6cm, ab = CD = 4, calculate the trapezoid area
Sorry, it's isosceles trapezoid
For high AF, AF and GH intersect at e, because BC-AD = 4, so BF = 2, so Ge = 1, from GH = 6 to get ad = 4, we can get height = 2 times root number 3, trapezoid area = (4 + 8) * 2 times root number 3, and then divide by 2 = 12 times root number 3
RELATED INFORMATIONS
- 1. As shown in the figure, in the square ABCD, point P is a point on the diagonal AC, PE ⊥ AB, PF ⊥ BC, and the perpendicular feet are points E and f respectively. If the perimeter of the square ABCD is 8cm, then the perimeter of the quadrilateral EBFP is______ cm．
- 2. It is known that in trapezoidal ABCD, ad is parallel to BC, AB is parallel to de, AF is parallel to DC, e and F are on edge BC As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, ab ‖ De, AF ‖ DC, e and F are on the side BC, and the quadrilateral aefd is a parallelogram (1) What is the equivalent relationship between AD and BC (2) When AB = DC, it is proved that the quadrilateral aefd is a rectangle I want to prove that the process is not an equation
- 3. As shown in the figure, in trapezoidal ABCD, ad is parallel to BC, AF is vertical, BC and points F, m are the midpoint of CD, angle B is 45 degrees, AF is 4, EF is 7 Find ABCD area
- 4. In the trapezoidal ABCD, ab = CD, e is the midpoint of AD
- 5. As shown in the figure, in trapezoidal ABCD, ∠ d = 90 ° and M is the midpoint of ab. if cm = 6.5 and BC + CD + Da = 17, the area of trapezoidal ABCD is () A. 20B. 30C. 40D. 50
- 6. In trapezoidal ABCD, ad ∥ BC, ab = 4, BC = 9, CD = 5, Da = 6. (1) prove ab ⊥ BC; (2) find the area of trapezoidal ABCD
- 7. In trapezoidal ABCD, AD / / BC, ∠ ABC = 60 °, DCB = 30 °, ab = 4, BC-AD is calculated
- 8. There is a little E in the quadrilateral ABCD,
- 9. As shown in the figure, in ▱ ABCD, the diagonal AC = 21cm, be ⊥ AC, the perpendicular foot is e, and be = 5cm, ad = 7cm, then the distance between AD and BC is______ cm．
- 10. It is known that the perimeter of the parallelogram ABCD is 50cm, the diagonals intersect at point O, the perimeter of △ AOB is 5cm longer than that of △ BOC, and the lengths of AB and BC are calculated No picture
- 11. The quadrilateral ABCD is the inner boundary quadrilateral of the circle, the diagonal AC and BD intersect at e, the extended DA and CB intersect at F, the angle CAD = 60 degrees, DC = De, it is proved that a is the outer center of the triangle bef How does angle F + angle AEB equal to 60 degrees?
- 12. It is known that as shown in the figure, ad and BC intersect at point O, ∠ cab = ∠ DBA, AC = BD
- 13. As shown in Figure 1, in quadrilateral ABCD, ad ∥ BC, de bisects ∥ ADB, ∥ BDC = ∥ BCD, (1) (2) if the bisector of abd and the extension of CD intersect at f (Fig. 2), and F = 55 °, find ABC
- 14. As shown in the figure, in the quadrilateral ABCD, the diagonal AC and BD intersect at point E, and ∠ cab = ∠ CBD. Given AB = 4, AC = 6, BC = 5, de = 5.5, find the length of de. & nbsp; & nbsp; [please solve this problem from the Perspective of similar triangles,]
- 15. AB = AC = ad, ∠ DAC = 3 ∠ cab Angle BDC = 1 / 2 angle cab, angle DBC = 1 / 2 angle DAC How did you get this step?
- 16. Isosceles trapezoid ABCD, ab = CD = 5, ad = 4, BC = 10, e, f on AB, BC, EF bisects the circumference of trapezoid, BF = x, and uses X to express the area of △ bef AB DC is waist
- 17. As shown in the figure, in square ABCD, point m is a point on diagonal BD. through point m, make me parallel to CD, intersect BC at point E, and make MF parallel to BC at point F. prove am = EF As shown in the figure, in the square ABCD, point m is a point on the diagonal BD, passing through point m to make a point Me parallel CD intersects BC at point E, MF parallel BC at point F. prove am = EF
- 18. As shown in the figure, m and N are the midpoint of AD and DC on both sides of the square ABCD, cm and BN intersect at P
- 19. As shown in the figure, in the quadrilateral ABCD, ab = CD, points E and F are the midpoint of AD and BC respectively, GH ⊥ EF intersects point P
- 20. As shown in the figure, in ladder ABCD, ab ‖ DC and E are the midpoint of BC, and the extension lines of AE and DC intersect at point F to connect AC and BF. What quadrilateral is abfc? Please give reasons