Find answers to 1 2 3 questions, classify math triangles in the first year of junior high school

1.When the circumference is 10, the longest side cannot exceed 4, and the three sides can only have 2 4 4 and 3 3 4.

2.The circumference is 11 and the waist length is an integer, and the possible scenarios are as follows:

This is because the sum of the two sides is greater than the third side to form a triangle.

Therefore, the first two cases are excluded.

That is: waist length is 3 or 4 or 5

3.Let the length of the waist of the isosceles corner be x, and the midline of the upper waist divides the perimeter of the triangle into two parts, then the circumference of one part is x 2+x, then the circumference of the other side is x 2+8, and the circumference of the two parts is different by 5, so x 2+x-(x 2+8)=5 or x 2+8-(x 2+x)=5, so the solution is x=13 or 3.

There is also by the triangular property: the sum of the two sides is greater than the third side and it is easy to know x=3 rounded. So the waist of this triangle is 13 to seek.

1.If the minimum side is 1, then the other two sides can only be ; 0 groups.

If the minimum side is 2, then the other two sides can only be 4,4;1 group.

If the smallest side is 3, then the other two sides are 4,3;1 group.

If the smallest side is 4, then the other two sides are 4,2 or 3,3;2 groups.

If the smallest side is 5, then the other two sides are 0;

Because 244 and 442 are a combination, there are 2, respectively or 2) 3, 3 3 5. 4 4 3。5 5 13)13 Let the waist length be x, so the two parts divided by the midline are x 2, x 2 because the midline is the common side of two triangles.

So x+x 2-(x 2+8)=5

x=13x/2+8-(x+x/2)=5

x=-7 (rounded).

So the waist length is 13

1., the degree sum of the outer angles of a, b and c is 360 degrees The degree ratio of the outer angles of a, b and c is 2:3:4, and the degree of the outer angle of a is 2 9 * 360 degrees = 80 degrees.

a = 180-80 = 100 degrees.

2.Angular bad= adc- b=40 degrees.

Angular cad = bad=40 degrees.

c = 180 - adc - angle cad = 30 degrees.

3.P = 180-angular PFE-angular PEF

180-1 2 (Angular AFE-Angular BEF).

180-1 2 (180-angular CFE + 180-angular CEF) = 180-180+1 2 (angular CFE + angular CEF) = 1 2 (angular CFE + angular CEF).

1 2 (180-angle c).

45 degrees. Therefore, the degree of p is only related to the degree of angle c.

It has nothing to do with the position of the two points f and e.

Solution: Because BM is parallel to AC

So angular c = angular cbm

Because the angle c = 45 degrees.

So the angle cbm = 45 degrees.

Because the angle EBC = angle CBM + angle MBE = 60 degrees.

So the angle mbe = 60-45 = 15 degrees.

Because bn parallel de

So angular e = angular ebn

Because the angle e = 30 degrees.

So the angle ebn = 30 degrees.

Because the angle EBN = angle MBE + angle MBN

So the angle mbn = 30-15 = 15 degrees.

So the degree of angular MBN is 15 degrees.

First, add a point F at the intersection of AE and Cd to prove: ab cd (known) 4= BAF (two straight lines are parallel, the isotopic angle is equal) 3= 4 (known).

3= BAF (Equal Substitution).

1 = 2 (known).

1+ CAF = 2+ CAF (property of equation) i.e. BAF = CAD

3= CAD (Equivalent Substitution).

ad be (equal internal misangle, two straight lines parallel).

You don't think you have a letter on it.

What does it mean to be in front of a straight line? Be clear.

Completely unaware of what is being said, can you insert **?

1) Angle A1 = 48 degrees, Angle A2 = 24 degrees, Angle A3 = 12 degrees.

2) Proof that because A1C bisects ACD and A1B bisects ABC, ACA1=1 2( A+ B).

a1bc=1/2∠a1bc

180 degrees - abc - acb) - 1 2 a1 2 a48 degrees.

In the same way, it can be proved that an=1 2 an-1

So an is a proportional series with 48 degrees as the initial value and 1 2 as the common ratio.

i.e. an=48*(1 2) (n-1).

Done! 3) From (2) an=48*(1 2) (n-1), so a5=48*(1 2) 4=3 degrees.

What do you think. Satisfied?..If you don't understand there, just ask me.