2 math problems of the Pythagorean theorem for junior 2 and 2 math problems of the Pythagorean theor

a^2+b^2=c^2

ab=cha^2*b^2=(a^2+b^2)h^2

1/a^2+1/b^2=1/h^2

Rule: (n 2-1) 2+(2n) 2=(n 2+1) 2Example: 35 2+12 2=37 2,48 2+14 2=50 2,63 2+16 2=65 2,80 2+18 2=82 2,99 2+20 2=101 2

2) The proof is simple, and the formula is disassembled and calculated to prove that the left and right are equal.

The latter ones are counted as ha.

If the hypotenuse is c, then the triangle area = ab 2 = ch 2

ab=cha²+b²=c²

1/a²+1/b²=(a²+b²)/(ab)²=(a²+b²)/(c²*h²)=1/h²

Pythagorean theorem proof, please do it yourself.

ab²=39999²+400²=(40000-1)²+160000=(40000+1)²

ab=40001

1,be^2=ae^2+ab^2

ef^2=ed^2+df^2

bf^2=fc^2+bc^2

Substitution data. bf^2=ef^2+eb^2

After proof 2, the Pythagorean theorem is used to prove fe 2 + de 2 = fd 2, and the proof method is the same as that of question 1.

1.Because it's a square.

Therefore ab=ad=bc=cd=4, then.

ed=2, cf=3, there is the Pythagorean theorem known be 2=20;ef^2=5;bf^2=bc^2+cf^2=25;So.

bf^2=be^2+ef^2;The problem is proven.

2。Let the square become as.

4 and the first question.

It is also possible to set position variables.

x, it shouldn't be difficult to follow the first question.

Question 1: 13 squared + 84 squared = 85 squared (the first number is an ordinary odd number: 3, 5, 7, 9.......)The second is 4 times 1 3 6 10 ......1,3,6,10 This can be seen as 1+2=3 3+3=6 6+4=10, that is, add 1 more each time. The last number is the second number plus one).

Question 2 below. in doing.

AB edge height to obtain the length of the height is 15, area = 1 2 * 20 * 15 = 150;

1. The first question of the first question is 50cm for the (30 2 + 40 2) prescription: the second question is 60cm for the (100 2-80 2) Zhao Xiao prescription is 60cm

30 40 = 1200 80 100 = 8000 Selected calendars are noisy, Zheng A Ken bends and cleverly sets the ode to the limbs.