A few junior 3 math problems, please ask a few junior 3 math problems

Solution: 1. Let the side length of the regular hexagon be a (a is not equal to 0), then the side length of the regular triangle is 2a.

It is easy to find the area of a regular triangle = 3a 2 (three times a square under the root).

A regular hexagon is actually made up of two isosceles trapezoids.

The upper bottom of the isosceles trapezoid is the side of the regular hexagon = a, and the lower bottom is the central axis of the regular hexagon = 2a, and the height = 3a 2.

The area can be found as 3 3a 2 2 (three times the root of three times multiplied by a squared divided by two).

Then the ratio of the area of a regular triangle to a regular hexagon = 2:3

2. Let their area be 3a 2. Then the regular triangle becomes longer = 2a.

According to the relationship between the regular hexagon and the upper and lower sides of the hexagon in the above question, the side length of the regular hexagon = 6a 3 (six times a divided by three under the root).

Then the ratio of the side length of a regular triangle to a regular hexagon = 6:1

3. Because each angle of a regular hexagon is 120°, and because AEF and ABC are isosceles triangles.

So, fae= bac=30°.

Because ace is a regular triangle, hag=60°. ah=hg=ga………

AF= AFB+ BFE, giving AFB=30°.

Then in AFH, haf= hfa=30°, then ah=hf.........

Similarly, in ABG, AG=BG.........

synthetic, BG=GH=HF.

1. Solution: Assuming that the side length of a regular triangle is a, its circumference is 3a

Then, the circumference of a regular hexagon is 3a, and the length of its side is 3a 6=a 2

This regular hexagon can be divided into 6 regular triangles of equal area along the center point, and each regular triangle has a side length of a 2

Find the area of the major positive triangle, the base of which is a, and its height is root number [a 2-(a 2) 2] = root number (3a 2 4) = [root number (3a 2)] 2

Therefore, the area is a*[root number (3a 2)] 4

In the same way, the area of the small triangle is found at a base, and its height is root number[(a 2) 2-(a 4) 2]=root number(3a 2 16)=[root number(3a 2)] 4, so the area is (a 2)*[root number(3a 2)] 8=a*[root number(3a 2)] 16

Therefore, the area of this regular hexagon is 6*[root(3a2)] 16=3*[root(3a2)] 8

Therefore, the area ratio is 2:3

All right! I got it! x (x y) 3 8, that's the relationship, of course you can make this formula look better.

x 10) (x y 10) 1 2, plus the above formula, just form a system of binary equations, and I don't need to do the work of solving the equation! Let's look at the next question: correct your description, "the probability of 2 people winning", it is impossible for 2 people to win at the same time, what should be the probability of each person winning.

There is no need to think about it, whether you win, you want to lose, you want to draw, only these 3 cases, the probability of winning is of course 1 3

The hour hand has 12 scales, and the degree of the code between each number is 30 degrees, and when the time is 3:30, the angle between 3 and 4 is 15°, so the angle between the hour hand and the minute hand is 75°.

2.(4,-four-thirds of the root number three) tan30° = h 4,h = four-thirds of the root number three, because the stool is clockwise rotation a2 ordinate is. (4,-four-thirds of the root number three).

I don't understand what it means.

The first, if it is not so deep and so exquisite, is the second one of 90 ° of the trembling ant bridge, as shown in the figure above, the angle AOB is equal to 30 degrees, bo is equal to 4, ab = sin30 ° times bo = (1 2) times 4 = 2

According to the Pythagorean theorem, AO=2 root number 3 can be obtained

So. The coordinates of A2 are (2 root number 3,2).

This kind of problem is typically solved using Veda's theorem, and any kind of quadratic equation that tells you that there are two unequal roots is definitely true.

Vedic theorem: If x1 and x2 are unequal to the equation ax + bx + c = 0, then x1 + x2 = -b a; x1*x2=c/a

Then look at this problem, find the range of values of k, where can there be a relationship between the size of k? B -4ac, of course. i.e. (2k-3) -4k >0, and the solution has k<3 4

Second question:( 3 -5=( +4 +3 -5=( +5.)

β=3-2k;αβ=k²

Then k +3-2k=6, i.e., k -2k-3=0, can be solved k=3 or -1Since k<3 4, round off k=3.

Then ( +5=(2k-3) -k -5=19

i.e. ( -3 -5 = 19

0, i.e. (2k-3) -4k 0

So k 4 3

From Veda's theorem: +=3-2k =k k=-1 or 3( -3 -5=( +5=(3-2k) -k -5=19 or -5