Junior 2 math problems Pythagorean theorem 2nd year math problems about the Pythagorean theorem

The sum of the areas of the two small triangles is equal to the area of the large triangle.

Let the side length of the regular triangle be 2a

Rule. There can be an angular relationship to prove that it is high at any bottom edge.

Root No. 2) A area s = 1 2

2a root number 2) a

Root number 2) a*a

i.e. the area is.

The side length is multiplied by.

Root number 2) set the right-angle side lengths are a

b Hypotenuse length c

You can't see the picture, so correspond to it yourself).

Yes. a*a+b*b=c*c

Root number 2) a*a

1 2 * (root number 2) b*b =

1 2 * (root number 2) c*c

Some symbols won't hit.

It's up to you to understand.

Also, check it out for yourself.

It's been a long time since I've done a junior high school question.

According to the Pythagorean theorem: AC2+BC2=AB2, so: BC= (13 2-5 2)= (169-25)= 144=12

Area of the triangle = ab*cd 2 = ac * bc 2

So: cd=ac*bc ab=5*12 13=60 13

Right triangle ABC, according to the Pythagorean law, 169 25 12cm under the root number BC

The key utilization area of this problem is unchanged, that is, bc*ac=ab*cd, 12*5=13*cd, so cd=60 13 is correct.

This problem uses a trigonometric function, ab=13 ac=5 from the Pythagorean theorem bc=12 in the triangle abc sina=12 13 in the triangle acd sina = cd 5, so 12 13=cd 5 cd=60 13

Solution: In RT ABC.

ab=13 ac=5

So bc=12

Due to the equal area.

So ac*bc=ab*cd

Bring in 12*5=13*cd

So cd=60 13

It's actually that simple. It is known that BC=12cm Since AB*CD=AC*BC (equal area), CD=60 13

BC2 (squared) = 13 * 13 - 5 * 5 = 144 BC = 12 by equal area method:

ab*cd=ac*bc

13*cd=5*12

Solution: cd = 60/13

Pythagorean theorem bc=12cm

s△abc=12*5/2=30cm^2

Area method cd=30*2 13=60 13cmA: The length of cd is 60 13cm

2. It's best to have a picture.

3，170cm

Can you give a picture for question 2?

The answer to the third question is 80 squared plus 150 squared, and then you can open it.

After the rectangle is folded, it is easy to know af=fc, using the right triangle bfc, using the pythagorean theorem to find the cf length, that is, af length, s afc = 1 2af bc

Answer: Solution: Let af=x, according to the meaning of the question, the rectangle is folded in half along the diagonal ac, there is d = b=90°, afd = cfb, bc=ad ad f cbf

cf=af=x∴bf=8-x

In RT BCF there is BC2+BF2=FC2

i.e. 4 2 + (8-x) 2 = x 2

Solution x=5

s afc= 1 2af bc= 1 2 5 4=10 Comments: For the more complex calculations in folding, it is necessary to find the corresponding right triangle after folding, and use the Pythagorean theorem to solve the required line segment

AC = 8 * 8 + 4 * 4 = 80 square meters.

The midpoint of ac is E

ec=ac 2=20 squares.

s ecf=(20 square dc) 2* s acd=20 64*16=5

safc=2*5=10

Hello! Take the midpoint of AC as E and connect it to EF.

It can be proved that AFC is isosceles. Therefore EF is a high for AFC because AFE is similar to ACB

AC = 80 squares, AE = 20 squares.

ef/bc=ae/ab

So ef = 5 squares.

afc=1 2ac*ef=10 thanks.

Due to the folding, apparently ACD'≌△cab

acd'＝∠cab

If fc fa is set to fb x, then fc fa is 8 x

Applying the Pythagorean theorem in BFC, we get:

8－x)^2＝4^2＋x^2

Solution: x 3

Thereby af 5

s△acf＝1/2·af·bc＝1/2×5×4＝10