The application of the Pythagorean theorem, please give some mathematical examples? Thank you

The side length of the right triangle is 3, 4, 5 respectively, and what is his area.

A flagpole is broken 6 meters from the ground, and the top of the flagpole falls 8 meters from the bottom of the flagpole, how high is the flagpole before it breaks?

Because bc=6, ac=8, ab=10

So the triangle ABC is a right-angled triangle with an angle of c = 90 degrees.

So the s triangle abc = 6 * 8 2 = 24

Because o is the intersection of the three angular bisectors.

So the distance from the point o to the three sides is equal.

Let this distance be x

Because o is inside the triangle.

So the S triangle OAC + S triangle OAB + S triangle obc = S triangle abc = 24

So (ac*x+ab*x+bc*x) 2=24 because bc=6, ac=8, ab=10

So the solution is x=4

Then the distance from o to each variable is 4

Solution: connect AC, pass d as DE perpendicular AC to E, then ABC, AED, CDE are all right triangles, because AB=90, BC=120, therefore, from the Pythagorean law AC=150M and because, AD=140, CD=130 Let CE=X, 140 squared - (150-x) squared = 130 squared - x squared (solve the equation yourself) x = ?

Whereas, in CDE, the length of DE is determined by the Pythagorean law.

Then find the area of the ACD.

The area of the quadrilateral ABCD = ACD+ ABC

(1) According to the Pythagorean theorem; ad=2 root number 5

dc = root number 10

bc = root number 17

ab = root number 37

Circumference = AD+DC

BC+AB=Area=36

2) Not at right angles.

Triangle AOD

It is not similar to the triangle DMC, and the corresponding angles are not equal or redundant.

Solution: (1) Because the area of each small square is 1

Therefore, the side length of each small square is 1Therefore: AD

20, the area of the hypotenuse with ad as 1 2 2 4 4 so: ad 2 5

Similarly: cd 10, the area with cd as the hypotenuse 1 2 1 3 3 2bc = 17, the area with bc as the hypotenuse 1 2 1 4 2

ab = 37, with ab as the hypotenuse area 1 2 1 6 3

Therefore: the perimeter of the quadrilateral ab bc cd ad 2 5 10 17 37

2) Connect AC, so: AC 3 5 34 again: AD

20，cd²＝10

Because AC ≠AD

CD Therefore: ADC is not a right angle.

Solution: From the question of the sparrow, abc is a right triangle.

ab²=ac²+bc²

ab = (ab-1) +5

The solution is ab = 13 meters of open eyes.

A: The length of the rope is 13 meters.

Connect AB, extend B to C directly below A, from the problem we know AC=2+4=6 km, BC=7+1=8km, in the right-angled triangle ABC, the square of AB = the square of AC + the square of BC = 36 + 64 = 100=10 square so AB = 10 km.

Draw a dotted line from ac to divide the figure into two triangles.

The triangle abc is equal to 20x15 2 = 150 square meters by the Pythagorean theorem:

20x20+15x15=625 625 square = 25 That is to say, AC=25M

And then it's interesting that although we don't know how many degrees d is, (24x24+7x7) is also 25 degrees, so d is also 90 degrees. (I don't know if you understand here.) The language is a bit confusing. ）

Therefore, the area of the triangle CDA = 24x7 2 = 84 square meters, and then add the two triangles together to be the total area: 84 + 150 = 234 square meters, if you don't understand or still want to know what, you can ask, oh, I hope to adopt thank you

1. Cosine theorem:

c 2 = a 2 + b 2-2abcosc (c is one side, c is the angle of c side pairs, cos is the cosine of c).

So, in the Pythagorean theorem, because when the triangle is at right angles, c = 90 degrees, cos 90 degrees = 0, so we get:

2. The Pythagorean theorem (special case of the cosine theorem).

c 2 = a 2 + b 2 (c is the hypotenuse).

3. Applying this theorem, you can set unknowns, and you can apply it directly. For example, to draw the root number 20, you first need to write 20 as the sum of square values, for 20, because 20 = 16 + 4 = 4 2 + 2 2, so that the two sides of the right angle are equal to 4 and 2 respectively, then the hypotenuse is equal to the root number 20 (according to the Pythagorean theorem). However, for example, for root number 3, because 3 cannot be written as the sum of two square values, it is necessary to draw the root number 2 first, then take the root number 2 as a right-angled side, 1 as another right-angled side, and the hypotenuse is the root number 3, and for the root number 2, draw both right-angled sides are 1, and the hypotenuse side is the root number 2 (according to the Pythagorean theorem).

Therefore, for some other root numbers, if they can be directly written as the sum of two square values, then they will be bounded by the root number 20, and if they cannot be written as the sum of two square values, they will be decomposed into two or more steps according to the root number 3. For example, for the root number x, you can draw the root number x-1 first, so that the root number x-1 is a right-angled side, 1 is the other side, and the hypotenuse is the root number x, and for the root number x-1, you can draw the root number x-2 first, and so on, until the root number 2, then the two right-angled sides are 1 respectively, and the hypotenuse is the root number 2In this way, any root number can be solved using this circular method.