There are three schools a, B, Let three schools a, B and C be located at the three vertices of an equilateral triangle. In the Internet age, to set up communication cables between the three schools, Xiao Zhang designed three connection schemes, as shown in the figure, scheme a AB + BC; scheme B AD + BC {D is the midpoint of BC}; scheme C Ao + Bo + CO (o is the intersection of the three heights of the triangle). Please help to calculate which of the following schemes has the shortest route

There are three schools a, B, Let three schools a, B and C be located at the three vertices of an equilateral triangle. In the Internet age, to set up communication cables between the three schools, Xiao Zhang designed three connection schemes, as shown in the figure, scheme a AB + BC; scheme B AD + BC {D is the midpoint of BC}; scheme C Ao + Bo + CO (o is the intersection of the three heights of the triangle). Please help to calculate which of the following schemes has the shortest route

The third connection mode should be selected
Let the side length of triangle be a, then
1. In the first scheme, the cable length is 2A
2. In the second scheme, the cable length is 1.866a
3. The cable length of the third scheme is 1.732a

A = a + 2, B = a ^ 2 + 5-19, where - 7 < a < 3
Find (1) factoring b-a;
(2) Compare the size of a and B
It is known that a = a + 2, B = a ^ 2 + 5a-19, where - 7 < a < 3

1、
B-A
=a²+5a-19-a-2
=a²+4a-21
=(a+7)(a-3)
2、
B-A
=a²+4a-21
=a²+4a+4-25
=(a+2)²-25
-7

If the ratio of the three sides of △ ABC is 3:4:5, and the circumference of △ DEF is 18, then s △ def=___ ．

According to the inverse theorem of Pythagorean theorem, △ def and △ ABC are right triangles. Let △ def be 3x, 4x and 5x respectively, then 3x + 4x + 5x = 18, x = 32, and the lengths of the three sides are 92 and 6152 respectively, so s △ def = 12 × 6 × 92 = 13.5

In △ ABC and △ def, ab = 2 DE.AC=2DF The perimeter of ∠ a = ∠ D. △ ABC is 16. The area is 12. Find the perimeter and area of △ def

Because in △ ABC and △ def, ab = 2DE, AC = 2DF, ∵ AB / de = AC / DF = 2, and ∵ a = ∵ D, ∵ ABC ∽ def, and the similarity ratio of △ ABC and △ DEF is 2, ∵ ABC's perimeter is 16, area is 12, ∵ def's perimeter is 16 △ 2 = 8, area is 12 △ 4 = 3

In △ ABC, De is parallel to BC. (1) if D is the midpoint of AB, find the perimeter ratio of △ def to △ ABC; (2) if de bisects the triangle area, find ad
In △ ABC, De is parallel to BC
(1) If D is the midpoint of AB, find the perimeter ratio of △ def and △ ABC;
(2) If de bisects the triangle area, find ad: dB

（1）
De is parallel to BC
Then triangle ade ∽ ABC
So the perimeter ratio of △ ade and △ ABC = similarity ratio = 1:2
（2）
If de bisects the area of a triangle
That is, the area ratio of △ ade to △ ABC = 1:2
De is parallel to BC
Then triangle ade ∽ ABC
So the area ratio of △ ade and △ ABC = the square of the similar ratio
So (AD: ab) square = 1:2
So. Ad: ab = 1: √ 2
AD：DB.=1:(√2-1)=(√2+1):1