Mathematics in the second day of junior high school, about Pythagorean theorem, please draw a picture and explain, choose C.
If BD length is x, then AD is 2x. Because CD is perpendicular to AB, and CD is on the same side, so bring AC 2-(2x) 2 = BC 2-X 2 into 16-4 * X 2 = 4-X 2 to calculate x=2.
Pythagorean Theorem of Senior Two Mathematics Pythagorean Theorem of Senior Two Mathematics It is known that the area of right triangle ABC is 30㎝? , the sum of two right angles is 17㎝, find the height on the hypotenuse.
Pythagorean Theorem 3 of Mathematics in Senior Two? +4? =5? ,8? +6? = 10? , 15? +8? = 17? ,24? + 10? =26?
Answer? +B？ =C？
A: 3 815 24 ... = 3 (3+5) (3+5+7) (3+5+7+9). ......
B: 46810 ... = 2+2 *1(2+2 * 2) (2+2 * 3) .....
35*35+ 12* 12=37*37
3999*3999+400*400=(4000- 1)*(4000- 1)+400*400
= 16000000-8000+ 1+ 160000= 16 160000-8000+ 1= 16 152000+ 1
Let the hypotenuse of the first triangle be x, then the hypotenuse of the second triangle is (root number 2)x, the third one is (root number 2)*√ (root number 2)x= square x of (root number 2), and the ninth one is eight times of (root number 2). The eighth power of (root number 2) = 16. So x=√3. Then the hypotenuse of the first isosceles right triangle is √3.
1) isosceles right triangle and right trapezoid
2)
s=(AB+DE)(BC+CD)/2=(a+b)^2/2
s=a*b/2*2+c*c/2=ab+c^2/2
3)
(a+b)^2/2=ab+c^2/2
a^2+b^2=c^2
Draw the conclusion of Pythagorean theorem
1. The question means: AC 2+BC 2 = 100.
AC^2=AD^2+CD^2
BC^2=BD^2+CD^2
∴AD^2+CD^2+BD^2+CD^2= 100
AD = 8，BD=2
∴ 8 2+2 2+2cd 2 = 100 When AD 2+CD 2+BD 2+CD 2 is substituted, that is, 64+4+2cd 2 = 100.
Get CD=4
I can't do it. I'll have my brother take a look at it for you when I have time.
The three sides of a triangle are respectively (assuming that one side is 3m, the second side is 4m and the third side is 5m, then the angle formed by the first side and the second side must be a right angle).
Let OE=x, then BE=4-x,
And OA=3, then AD=3, then BD=AB-AD,
According to Pythagorean Theorem, AB= radical sign (4*4+3*3)=5.
So BD=2,
In the RT triangle, DE +BD =OE +BD =BE.
X squared +4=(4-x) squared.
Solution, OE=3/2.
e(0.3/2)
Alas ~ ~ finally figured it out. Well, the first question is actually quite simple, and the second question is a bit complicated. Let's start with the first question. Sorry, I drew a picture, but I believe you can draw a picture.
The first question: this question does not need to consider the height of the floor. Suppose there are two BC points on the road, and the distance to A's residence is 41m. Then, make a vertical line perpendicular to the road through House A, which is the distance from the house to the road. According to the meaning of the question, AD=9m, and then use Pythagorean theorem to calculate BD=CD=40M, then BC=80M, you will definitely do this. Then,
Question 2: Suppose that the distance from point M to the road is a diagonal. Suppose the distance from point O to point M on the road is exactly 4 1M, that is, the car can hear the sound at point O, and there are two such points, just like BC in the first question. Suppose it's O and O2, and then M lives on the 9th floor. Attention! Don't think that the ninth floor is 27M. In fact, the ninth floor is only the height of the eighth floor, 24M. So we can calculate the distance from O to A, which is the root number 1 105. Don't be frightened by the numbers. In fact, the latter is very easy to calculate. Then we can put it on the ground now. We have calculated OA=O2A= root number 1 105. AD=9m, then we can calculate OD=O2D=32M, that is, 002=64M, that is, within this distance, the sound can be heard on the 9th floor, so we can get 25.6S by dividing 64 by 2.5m per second. A: In 25.6 seconds, M can hear the sound.
After reading it patiently, you will understand that the pictures of the first question and the second question can be common ~ the main points are the same.