Pythagorean theorem proof question in 8th grade

Let a=2x, then b=3x, according to the Pythagorean theorem, there is (2x) 2+(3x) 2=25 2, that is, 13x 2=625, and the solution is x=25 root number 13. So a=50 root size 13cm, b=75 root number 13cm. Then the area of abc s=ab 2=(50 root number 13*75 root number 13) 2=1875 13cm 2

This is based on the k-value assumption. Let a and b be 2k and 3k

Because c=90°, it is obtained by the Pythagorean theorem:

The square of 2k plus the square of 3k equals the square of 25. It can be calculated.

I'm a junior high school student, and I hope we become friends, so I don't know how to ask me in the future.

Hee-hee.

The Pythagorean theorem... 2*2+3*3=13, then 25 is divided into 13 root numbers, that is, how much is each part, a is 2*25 root number 13; b By the same token, the area can be obtained.

Let the length of the straight angle side be 3a, then the length of the other right angle side is 4a, and (3a) 2+(4a) 2=20 2

9a^2+16a^2=400

25a^2=400

a^2=400/25

a=20/5

4 So the length of the straight corner edge is 3a=3*4=12, then the length of the other right-angled edge is 4a=4*4=16

Let the length of the two right-angled edges be 3x and 4x, respectively

Then 20*20=(3x) 2+(4x) 2

x = 4, 12 and 16 on both sides

1), 30 minutes later, Xiao Ming walked 39 30 = 1170 meters, Xiao Hua walked 52 30 = 1560 meters.

The distance between the two is the signal radius = (1170 + 1560) 1950 meters.

Then the radius r: 52 30 = 5:4

r=52/4×5×30=1950

The distance of the car at point C is 30m, and the distance of AB is 50m

According to the Pythagorean theorem, BC = (50 -30 ) = 40m, and the speed of the car is (40 1000) (2 3600) = 72km h>70km h

So overspeeding.

2) Through calculation, can you find the connection between the data in the question and the Pythagorean number 3, 4, 5 Try to find a simple algorithm to solve the problem. (1), after 30 minutes, Xiao Ming walked 39 30 = 1170 meters, and Xiao Hua walked 52 30 = 1560

Let bc be point e, let ae=x, then ae=ce, ae+be=be+ce=bc=4

and abe is at right angles, so ae 2 = ab 2 + be 2 i.e. x 2 = (4-x) 2 + 3 2

x^2=16-8x+x^2+9

x=25/8，be=4-25/8=7/8

In addition, EF 2 = 3 2 + (BC-2*BE) 2, that is, EF 2 = 9 + (4-2 * 7 8) 2 = 9 + 81 16 EF = 15 4

If you look at it yourself, you can see their relationship.

Two congruent right-angled rectangles are obtained.

ae=cf de=bf bf+ae=4 ab=3 folded, the right triangle abf, ab 2+bf 2=cf 2 set bf=x, x 2+3 2=(4-x) 2 x=7 8bf=7 8 ae=25 8 ef 2=(ae-bf) +ab 2=225 162 ef=15 4

Classmate, this question is actually very simple, because it is a rectangle, so there is a 90° warp fold in the RT triangle ABC (BC is the EF side), BC EF 3, root number 2

Just now I saw the median line, and I should set the waist to x.

x^2+x^2=16 x^2=8

Half of the waist length is under the root number 2

So the length of the midline on the waist is 8+2=10, and then the 10 root number is opened! It's the middle line for the root number under 10, I hope you can understand!

Let a waist be x2=8

Let the midline be y2=4 y=2

In the first question, let the length of the bamboo pole be xm, then it is known by the meaning of the question, x 2=(x-1) 2+3 2; x=5m can be found

The second problem, from the meaning of the question, we know that the triangle BCD is a right triangle, because ab=ac, let ad=x, then ab=ac=x+12, from the Pythagorean theorem, x 2+16 2=(x+12) 2, the solution is x

1 Set the door height x, the bamboo bar is x+1 3 squared plus x squared equals x+1 squared so x=4

2. The angle CDB is a right angle, and the AC=X ad=x-12 cd=16 satisfies the Pythagorean condition.

1.Let the bamboo pole be x long.

x×x=3×3+(x-1)(x-1)

X is the Pythagorean number, so the triangle BCD is a right triangle.

Let ac be x, and get x x=(x-12)(x-12)+16 16x=85 6

1. The north straight side is 7km, the west straight side is 24km, and the distance between the two places is 25km.

2. The bottom end slide is also 2 meters. (6m to 8m).