A few math problems in the second year of junior high Thank you

The first question of the landlord must have missed a condition, and the first sentence should say what the length of the ladder is.

The second question says that when the rope falls vertically, there is 1 meter more on the ground, which means that the length of the rope is 1 meter longer than the flagpole, and after pulling 5 meters, the rope is straight. Since the direction you pull away must be perpendicular to the direction of the flagpole, it is equivalent to a right triangle. The two right-angled sides are the length of the flagpole and the pull-away, which is 5 meters, and the hypotenuse is the rope.

After the mathematical model is established, the problem is very simple, let the length of the flagpole be x, then the length of the rope is x+1, according to the Pythagorean theorem, it is easy to solve x=12, and the answer is c

1. Select C to draw a triangle, the corner of the wall is the point C, the lower end of the ladder is B, BC=5M, the point of the ladder against the wall A, AB=5M, because the corners of the wall are all right angles, so the length of AC can be found to be 12M according to the Pythagorean theorem

2.Choose C According to the title, the rope is straight at the beginning to the ground, the length is 1m, the rope is oblique after pulling 5m, the flagpole is set to be x meters high, draw a triangle, the top of the flagpole is a, the bottom of the flagpole is b, and the lower end of the rope is c, then find the length of ab=x, the distance between the lower end of the rope and the bottom of the flagpole bc=5m, the flagpole is perpendicular to the ground, the rope length = flagpole length +1m = ab + 1 = x + 1, then according to the Pythagorean theorem ab + bc = ac, that is, x +5 = (x + 1), the solution x =12, so the flagpole is 12 meters long.

The first problem is c The ladder and the building and the ground form a triangle The Pythagorean theorem can find that the building is 12m high

The second problem is c is also the Pythagorean theorem, where the flagpole is x meters high and the rope is x + 1 meter.

x+1）²=x²+5²

x²+2x+1=x²+25

2x=24x=12

According to the Pythagorean theorem, 13 is the hypotenuse and 5 is the shortest right-angled side, then the height is the square root of 13 2-5 2 (you can know it according to the inscription).

Set the height of the flagpole xm, x 2+5 2=x 2+1+2x

1c 13 square minus 5 square.

2 I don't quite understand the meaning of the question.

Mainly use the Pythagorean theorem to study online.

c, is a right-angled triangle, one hypotenuse is 13, the right-angled side is 5, and the other is 12

I couldn't read it the next day

1. The two flat angles are 360°, and the outer angle is 245°

So the sum of a top angle + a bottom angle is 360°-245°=115°180°-115°=65° (the degree of a base angle), so the number of top angles is 180°-65°-65°=50°2, when 50° is the top angle, 180°-50°=130°, and the other two angles are 65° and 65°

When 50° is a bottom angle, 180-50*2=80° and the other two angles are 50° and 80°

3. When 13cm is the waist length, the circumference is 13 + 13 + 25 = 51cm and 25cm is the waist length.

The circumference is 25 + 25 + 13 = 63cm

4. When 11cm is the waist length + 1 2 waist length.

The waist length is equal to 22 3cm

So the bottom edge length is 15-22 3 = 23 3 cm

When 15cm is waist length + 1 2 waist length.

The waist length is equal to 5cm

So the bottom edge is 11-5=6cm long

5、be=ab,cd=ac

So bae= bea, dac= cdaIn ADE, ade+ aed+ dae=180° so DAC+ BAE+ DAE=180° so DAE+ EAC+ BAD+ DAE+ DAE=180°BAD+ DAE+ DAe=90°

So 2 dae = 90°

dae=45°

1.Be. Proof:

Triangle ABF and triangle CDE congruence.

So af=ce, and the angle afb is equal to the angle dec

So the angle AFE = the angle EFC

So AF is parallel and equal to CE

So the quadrilateral AFCE is a parallelogram.

Make AE vertical BC over BC to E

Because it is an isosceles trapezoid. So be=2

So the angle b = 60 degrees.

Extend DA to F

So the FDCB is a parallelogram at this point.

So fb=dc=4

So the triangle abf is an equilateral triangle.

So the angular fab is 60 degrees.

So the angle fbc is 60 degrees.

Make de vertical DC to e

Make AF vertical BC on F

DE intersects AF at G

Question 3 will come back later.

1) A in the third quadrant indicates that both coordinates are negative.

5 is obviously a negative number, so n-3<0 is required, i.e. n<3

b in the fourth quadrant shows that the abscissa is positive and the ordinate is negative, so -n+1

A coordinates are (-1, -5).

The b coordinate is (2,-1).

2 The law is to ask you why the left side is equal to the right side, and the following is the proof:

You already have the formula: n 2+n 2(n+1) 2+(n+1) 2 = n 2+n+1) 2

Now let's prove this equation and derive it from the left:

You might as well take the middle term on the left and get n2+n2(n 2+2n+1)+(n+1) 2

Again: n 4+2n 3+2n 2+(n+1) 2Put the 2n 2 in the middle of the above equation: n 4+2n 2(n+1)+(n+1) 2

Did the donor find that the above formula is perfectly squared?

That's right, it's equivalent to proving that the right side of the equation (n 2+n+1) 2 is complete.

So a(-1,-5) b(2,-1).

2.The truth is that the perfect square formula a 2+2ab+b 2=(a+b) 2

1 A 2 + B 2 = C 2, substituted m=-42 molecule 1 2[( 3 + 5 + 7) 2-5], denominator ( 3 + 5 + 7 + 5), simplified 3 + 7

3 (100（1+2r)-50)(1+r)=63 2r=4 n(n-1)/2=66 n=12

1.If a + b = c

then 5-m+2m+7=4-m

m = -4 if b + c = a

Then 2m+7+4-m=5-m

m = -3 if a + c = b

then 5-m+4-m=2m-7

m=2.This question is indeed a bit difficult, but it is simplified a little:

Not anymore ...

3.Set the first year with an annual interest rate of x

100+100x)-50]+(50+100x) x=The process is too complicated, simplified)

4.I'm sorry, I don't know, but I have summarized a rule:

If there are x people shaking hands, there will be a total handshake x+(x-3)+(x-4)+x-x+1) times.

When we know that a, b, c are the side lengths, we know that a, b, c are not zero, i.e. a= (5-m) is not 0; b= (2m+7) is not 0; c= (4-m) is not 0, and after finishing, we get:

Then let's discuss: when m is 0, there is a = 5, b = 7, c = 4. Obviously, a, b, and c are not right-angled edges, so it is not true that m is 0.

When 0 m, a is maximum, so there is the square of b + the square of c = the square of a a is solved to get m=-3, that is, when m = -3, a, b, c are the three sides of a right triangle and when 0 m 4, b is the largest, so there is a square of a + c squared = the square of b, and the solution is m=2 1, that is, when m=2 1, a, b, c are the three sides of a right triangle.