It is known that the area s of the triangle ABC satisfies that the root sign 3 is less than or equal to s less than or equal to 3, and the vector ab × vector BC = 6, and the angle between vector AB and vector BC is a,

It is known that the area s of the triangle ABC satisfies that the root sign 3 is less than or equal to s less than or equal to 3, and the vector ab × vector BC = 6, and the angle between vector AB and vector BC is a,

AB and BC are modules of vector AB and BC respectively
Vector AB * vector BC = AB * BC * cos α = 6
S=AB*BC*sin（π-α）/2=AB*BC*sinα/2
√3≤S≤3
√ 3 / 3 ≤ 2S / (vector AB * vector BC) ≤ 1
That is √ 3 / 3 ≤ Tan α ≤ 1
∴π/6≤α≤π/4

In the triangle ABC, vector AB * vector BC = 3, the area of triangle ABC belongs to [2 / 3,2 / 3] of the root sign, and the angle range between vector AB and vector BC? The area range is wrong [root of 2 / 3, 3 / 2] closed interval

(AB is the length of AB, AB is the vector AB, BC is the same) let the angle be θ, because vector AB * vector BC = 3 > 0, the angle between vector AB and vector BC is acute angle s = 1 / 2 * AB * BC * sin θ, because AB * BC * cos θ = 3, so 9 / (4S ^ 2 + 9) = cos θ * cos θ, and because the range of S is [root of 2,3 of 2]

The area of triangle ABC is s, vector ab · vector BC = 1, if 1 / 2 Math homework help users 2017-09-29 report Use this app to check the operation efficiently and accurately!

S = (1 / 2) | ab | ×| AC | sin α = the angle between vector AB and AC
＝（1/2）AB·ACtanα＝tanα/2.
1 ＜ Tan α ＜√ 3. 45 ＜ angle between vector AB and AC ＜ 60 °

In the triangle ABC, ab = 2, AC = √ 2BC, how do you come up with a circle to solve the maximum area of the triangle ABC? Can you tell me what you think?

Let BC = a, a ^ 2 + 2A ^ 2-4 = 2 root sign 2A ^ 2cosc, s = 0.5 root sign 2A ^ 2sinc;
(3a^2-4)^2/(8a^4)+4S^2/(2a^4）=1 ；
The result is: 16S ^ 2 = 8A ^ 4 - (3a ^ 2-4) ^ 2 = - A ^ 4 + 24a ^ 2-16 = - (a ^ 2-12) ^ 2 + 128

In the right triangle ABC, if the length of the hypotenuse AB is 2, then the maximum area of the triangle is obtained As the title

In the right triangle ABC, if the length of the hypotenuse AB is 2, then the maximum area of the triangle is obtained
X^2+Y^2=ab^2=4
Area of triangle = 0.5xy ≤ 0.25 (x ^ 2 + y ^ 2) = 1
The maximum area of a triangle is 1

As shown in the figure, △ ABC is known (1) Please take two points D, e (except the midpoint of BC) on the edge of BC, connect AD and AE, write the corresponding conditions that make only two pairs of triangles with equal area exist in this diagram, and show the triangles with equal area; (2) Please prove that ab + AC > AD + AE according to the corresponding conditions that make (1) hold

(1) As shown in Fig. 1, the corresponding condition should be BD = CE ≠ De, so that the areas of △ abd and △ AEC are equal. Since BD = CE, be = CD, then the areas of △ ADC and △ Abe are equal. (2) it is proved that as shown in Fig. 2, parallel lines of Ca and EA are made through points D and B respectively, the two lines intersect at point F, DF and ab intersect at point G

Known: triangle ABC is an equilateral triangle, extend AC to D, take BD as an equilateral triangle BDE, connect AE, prove: ad = AE + AC

Hehe, it's simple
prove:
∵∠ABC=∠EBD=60°
∠ABE=∠ABC-∠EBC
∠CBD=∠EBD-∠EBC
∴∠ABE=∠CBD
And ∵ AB = CB, be = BD
∴△ABE≌△CBD
∴AE=CD
∵AD=AC+CD
∴AD=AC+AE

As shown in the figure: in △ ABC, point D is the midpoint of edge BC, and point E is the point on line ad, and AE = 2ed is satisfied, then the area of △ ABC is the area of △ BDE______ Times

Because point D is the midpoint of edge BC, so s △ abd = s △ ACD = 12S △ ABC, because AE = 2ed, s △ BDE = 12S △ BEA, and because s △ BDE + s △ bea = s △ abd, that is, s △ BDE + 2S △ BDE = s △ ABC, so s △ BDE = 16S △ ABC

As shown in the following figure, in triangle ABC, D is the midpoint of BC, ad is perpendicular to de, AE = 4ce, ad = 8cm, de = 5cm. Calculate the area of triangle ABC

E point is on AC, right
∵ △ ade is a right triangle
∴ S△ADE=1/2*AD*DE=1/2*8*5=20
∵ AE=4CE,AC=AE+CE
∴ AE/AC=4/5
The height of △ ACD and △ ade is different
∴S△ADE/S△ACD=AE/AC=4/5
∴S△ACD=5/4*S△ADE=5/4*20=25
∵ D is the midpoint of BC
 s △ ABC = 2 * △ ACD = 2 * 25 = 50 (square centimeter)

In the known triangle ABC, the points D and E are on AB, AC and s triangle respectively ADE:S triangle BDE:S Triangle BEC = 4:2:3, find de ∥ BC

The idea of this problem is relatively simple, that is, to find the ratio through the relationship between the bottom and the high, that is, typing is more troublesome
In the triangle ade and the triangle BDE, the two triangles have the same height when AB is the bottom edge;
DF was perpendicular to AE, AE to F, BG to AE, AE to g;
Because the ADF of triangle is similar to AGB of triangle, DF: GB = 2:3;
DF is the height of the triangle ade with AE as the edge, BG as the height of the CBE with CE as the edge;
After that, we can calculate the ratio of dfae: 3:2:2 by substituting dfae: 3:3;
Combined with BD: ad = 1:2, we can know that de ∥ BC;
In drawing, typing and explaining so clearly, we should adopt and recommend it