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In trapezoidal ABCD, AB / / CD, AC is vertical to BD, angle BDC = 60 degrees, AC = 3 root sign 3, find the length of BD and the area of trapezoidal ABCD In the isosceles trapezoid ABCD, ab = 2CD, AC bisects DAB, ab = 4, ab ‖ CD, (1) Find the trapezoid angle; (2) find the trapezoid area
Because ab ‖ CD, angle cab = angle ACD
Because AC bisects ∠ DAB, angle DAC = angle cab
So angle ACD = angle DAC, that is ad = DC
Because AB = 4, ab = 2CD
So CD = 2, that is, ad = 2
Make de through point D and ab perpendicular to point E. make CF through point C and ab perpendicular to point F
Because AE + EF + FB = AB, EF = CD = 2, AE = FB
So AE = 1
In the right triangle AED, ad = 2, AE = 1 (1)
So angle DAB = 60 degrees, that is, angle B = 60 degrees, angle ADC = angle BCD = 120 degrees
From (1), de = √ 3
So trapezoid area = (AB + CD) * de / 2 = (4 + 2) * √ 3 / 2 = 3 √ 3
As shown in the figure, in the quadrilateral ABCD, ab = ad = 8, ∠ a = 60 ° and ∠ d = 150 °, the perimeter of the known quadrilateral ABCD is 32,
Finding the area of quadrilateral ABCD
Connecting BD ∵ AB = ad, ∵ a = 60 ∵ abd is equilateral triangle ∵ ADB = 60 ∵ ADC = 150 ∵ CDB = 90 ∵ equilateral triangle. The side length of abd is 8. The area of ∵ abd = 16 ∵ 3 ∵ ABCD perimeter = 32 ∵ BC + CD = 16. Let CD = X. then BC = 16-x according to the Pythagorean theorem x ^ 2 + 8 ^ 2 = (16-x) ^ 2, the solution of x = 6 ∵ C is obtained
If AB = 3cm, BC = 5cm, the area of triangle DEF is
According to Pythagorean theorem, ab = 4, so def area = 3 / 2 * 4 * 2 = 3
The rectangle paper ABCD is folded diagonally to make vertex B and point d coincide, and the crease is EF, ab = 3, BC = 5. Calculate the area of triangle def of overlapping part
I'm a student of grade two in junior high school. I haven't learned Pythagorean theorem. Can I solve it without Pythagorean theorem?
Because points B and D are symmetric to EF, BG = DG, and BD is perpendicular to ef. There is Pythagorean theorem BD = SQR (3 ^ 2 + 5 ^ 2) = SQR (34), so BG = DG = SQR (34) / 2. From the angle, the triangle Berg is equal to the triangle DGF. So eg = FG. From the angle, the triangle Berg is Acacia to the triangle BCD
If AB = 3cm, BC = 5cm, the area of the overlap △ DEF is ()
A. 7.5cm2B. 5.1cm2C. 5.2cm2D. 7.2cm2
If AE = x, then a ′ e = xcm, de = 5-x (CM), ∧ a ′ E2 + a ′ D2 = ED2, ∧ x2 + 9 = (5-x) 2, the solution is: x = 1.6, ∧ de = 5-1.6 = 3.4 (CM), ∧ def surface product
If AB = 3, ad = 3, the perimeter of a 'EFD is___ ．
In RT △ a ′ ed, let a ′ e = x, according to Pythagorean theorem x2 + 32 = (3-x) 2, the solution is x = 1, i.e. a ′ e = 1; similarly in RT △ DCF, let CF = x, according to Pythagorean theorem, we can get x = 1, i.e. CF = 1; for eg ⊥ BF, then eg = AB = 3, FG = ad-ae-cf = 3-1-1 = 1, EF = 32 + 12 = 2, the perimeter of quadrilateral a ′ EFD = 1 + 3 + 2 + 2 = 5 + 3
As shown in the figure, e f in square ABCD is the length of GC at the intersection of AE = 1 / 3 ad AF = FB EF and G AB = 8
Then AE = EH. It is easy to get that the triangle EHG is similar to the triangle fag
EH / AF = GH / Ag. And because eh = 8 / 3, Ag + GH = 8 √ 2 / 3. The rest is needless to say! The key of this problem is to make auxiliary lines
F is the midpoint on the side ab of the square ABCD, AE = 1 / 4 D, FG is perpendicular to EC, and the square of FG = eg * GC is proved
If the two sides of a triangle are proportional to the two sides of another triangle, and the included angles are equal, then the two triangles are similar. According to this point, we can get that the triangle AEF is similar to the triangle FBC, if the two triangles are similar, we can get that the angle AEF is equal to the angle BFC, we can get that the angle EFC is 90 degrees, then we can get that the angle FEG is equal to the angle GFC, and the angle EFG is equal to the angle FCG, And the angle FGE is equal to the angle FGC, so the triangle EGF is similar to the triangle FGC. If eg / FG = FG / GC, then FG ^ = eg * GC
As shown in the figure, in rectangular ABCD, AF, BH, CH and DF are bisectors of internal angles, AF and BH intersect at e, CH and DF intersect at G
It is known that the area of ladder ABCD is root sign 24, the height of ladder is root sign 3, and the ratio of upper bottom to lower bottom is 1:3
Let the upper bottom be x and the lower bottom be 3x
So s = 1 / 2 (x + 3x) * √ 3 = √ 24
We get x = √ 2
So the top of the trapezoid is √ 2, and the bottom is 3 √ 2
RELATED INFORMATIONS
- 1. In the parallelogram ABCD, be ⊥ ad, BF ⊥ CD, the perpendicular foot is e, F, the perpendicular center of △ bef is h, if DG ⊥ BC, the perpendicular foot is g It should be verification: BH = GF
- 2. In the parallelogram ABCD, M is the midpoint of BC, n is on AB, BN = 1 / 2An, NM intersects BD with G, and BG: GD is obtained
- 3. In the trapezoidal ABCD, point P starts from point a and moves along the edge of ad to point d at a speed of 1cm / s, and point Q starts from point C and moves along CB to point B at a speed of 2cm / s. if points P and Q start from two points at the same time, when one of them reaches the end point, the other stops moving. (1) what is the value of T when trapezoidal pbqd is a parallelogram? (2) What is the value of T when trapezoid pbqd is isosceles trapezoid?
- 4. The perimeter of parallelogram ABCD is 75cm, the height is 14cm with BC as the bottom (as shown in the figure); the height is 16cm with CD as the bottom. Calculate the area of parallelogram ABCD
- 5. As shown in the figure below, the area of rectangle ABCD is 12 square decimeters, so the area of circle is () square decimeters Well, there is no picture. Let's give a brief description There is a circle, no matter how big, half of it (that is, a semicircle). Draw a rectangle on the semicircle. The length of the rectangle is equal to the diameter of the circle. Then say that the area of the rectangle is 12 square decimeters. Tell the area of the whole circle. Thank you!
- 6. In the quadrilateral ABCD, BD bisects ∠ ABC, AC ⊥ BD at R, PQ intersects BC and ad at points Q and P respectively, and ∠ bad = ∠ BQP
- 7. As shown in the figure, ad ∥ BC in ladder ABCD, point E is the midpoint of edge ad, connecting be with AC at point F, the extension line of be with CD at point G, verify Ge / GB = AE / BC
- 8. As shown in the figure, in ladder ABCD, ab ‖ CD, e is the midpoint of DC, the line be intersects AC at F, and the extension line of ad intersects g; verification: EF · BG = BF · eg
- 9. Take a point G on the extension line of one side of the square ABCD and connect it with a, intersect BC at point F, connect diagonal BD and intersect AF at point E. AE = 5, AF = 3 are known, and calculate the length of FG chart
- 10. In quadrilateral ABCD, ad = BC, AE, CF are perpendicular to BD, be = DF. It is proved that quadrilateral ABCD is parallelogram I can't find the last condition in the verification. Please give me some advice. Thank you
- 11. In a circle with a radius of 12 decimeters, draw a square, draw a small circle in the square, and calculate the area of the small circle
- 12. In the diamond ABCD with side length of 2, ∠ bad = 60 °, e is the midpoint of AB, P is the upper moving point of AC, and the minimum value of Pb + PE is obtained
- 13. In rectangular ABCD, BC = 4, BG is perpendicular to diagonal AC and intersects AC, ad and ray CD at points e, F, G, ab = x respectively. (1) when point G coincides with point D, calculate the value of X; (2) when point F is the midpoint of AD, calculate the value of X and the sine value of ∠ ECF
- 14. As shown in the figure, points E and F are points on edge BC and CD of diamond ABCD respectively, and ∠ EAF = ∠ d = 60 ° and ∠ fad = 45 °, then ∠ CFE=______ Degree
- 15. As shown in the figure, in diamond ABCD, ∠ bad = 60 °, M is the midpoint of AB, P is a moving point on the diagonal AC, if the minimum value of PM + Pb is 3, then the length of AB is () A. 3B. 3C. 6D. 23
- 16. It is known that the four vertices of the quadrilateral ABCD are a (2,3), B (1. - 1), C (- 1. - 2) and D (- 2.2),
- 17. If the length of a diagonal line of diamond ABCD is 6 sides and the length of AB is a root of the square of equation x - 7x + 12 = 0, then the perimeter of diamond ABCD is In the two known equations x ^ 2 + ax + (A-1) = 0, the absolute value of negative root is greater than that of positive root, and the value range of a? It is known that the equation x ^ 2-mx + M-1 = 0 has two real roots x1, X2 and 1 / X1 + 1 / x2 = m, then M =? One diagonal of the diamond ABCD is 3
- 18. In diamond ABCD, if the angle ABC is equal to 120 degrees, then BD is equal to several times AC
- 19. If the central line and the high line on the hypotenuse of a right triangle are 6cm and 5cm respectively, then its area is
- 20. 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!