The process of mathematical Pythagorean theorem application problems !! Urgent

Question 1. The depth of the basin is x feet, and the length of this reed is x + 1 foot x 2 + 5 2 = (x + 1) 2

x^2+25=x^2+2x+1

2x=24x=12 pools of depth and the length of this reed are 12 feet and 13 feet respectively.

Question 2: The height of the breaking place from the ground is x feet.

Then the broken part is 10-x feet.

The part that is obviously broken is the hypotenuse.

So (10-x) 2=x 2+3 2

x^2-20x+100=x^2+9

20x=91

x=so it's a ruler.

Solution 1: If the water depth is H, then a right triangle is formed by the water depth H, half of the side length of 5, and the length of the reed H+1.

h+1）^2=h^2+5^2

Solution: h=12

h+1=13

Solution 2: Set the depth of the pool to be x, and the length of the reed is y

From the Pythagorean theorem, it can be seen that x 2 + 5 2 = y 2

And the reed is one foot above the water: y-x=1 ; i.e. y=x+1x 2+25=(x+1) 2

Solution: x=12, y=13

A: The depth of this water is 12 feet, and the length of this reed is 13 feet.

How many feet is a zhang?

I'm still young, you tell me!

I didn't even learn the Pythagorean theorem, but I just looked at the computer and didn't know much

If the water depth is x meters, the reed height is x + 1 meter.

Because the surface of the water is a square with a side length of 10 feet, and the reeds are in the middle of the pool.

So the square of 5 + the square of x = the square of (x + 1).

One zhang moral education is 10 feet, and after being broken, it is three feet from the bottom, that is, 2x+3 10 x

Therefore, the height of the break from the ground is 7 feet.

The height of the tree is 4 meters.

a Shadow length Tree height = tree height b Shadow length.

So: Tree height 2 = shadow length when a * shadow length when b.

Tree height 2=2*8

Tree height 2=16

So the height of the tree is 4 meters.

Solution: Let the intersection of two red rays at the top of the tree be c, the letter at the root of the tree is d, the lower left side is e, and the lower right side is f

ecd+∠dcf=90°;∠ecd+∠e=90°.

Then: e= dcf; cde= cdf=90°

So, edc cdf, cd fd=ed cd, cd 2=fd*ed=8*2, cd=4 (m).- The height of the tree.

Let it be x, according to the projective theorem.

x²=2*8=16

x = 4 (m).

The most straightforward way is to set the tree height to x, the hypotenuse of a is y, and the edge of b is z. A system of ternary quadratic equations is obtained.

y square — x square = 4

z square - x square = 64

y square + z square = 100

This system of equations is not difficult to solve... Solve x, i.e. the tree height is more than 4 meters.

1.Rectangle ADBC, AD is long, AB is wide, fold one side of the rectangle AD, so that the point D falls at the point F on the edge of BC, if AB=8cm, BC=10cm, find the length of EC.

- Solution: The quadrilateral ABCD is rectangular.

ad=bc ab=cd

ad=bc bc=10cm

ad=10cm

AEF is similar to AED with respect to straight line AE symmetry.

af=ad ef=ed

ad=10cm af=ad

af=10cm

The quadrilateral ABCD is rectangular.

fba=90° ∠eda=90°

fba=90°

abf is a right triangle.

ABF is a right triangle AF=10cm AB=8cm

bf=6cm

bf=6cm bf+cf=bc bc=10cm

cf=4cm

ab=8cm ab=cd

cd=8cm

Let ce=x ce+de=cd cd=8cm

de=8-x

ef=ed de=8-x

ef=8-x

eda=90°

ADE is a right-angled triangle.

AEF is symmetrical with AE with respect to straight lines and ADE is a right triangle.

AEF is a right triangle.

ce=x ef=8-x cf=4cm

CE = 3cm (evaluated using the Pythagorean theorem for right triangles).

2.There is a red lotus on the calm lake, 1m above the water surface, a gust of wind blows, the red lotus is blown to the other side, the flowers reach the water surface, it is known that the horizontal distance of the red lotus movement is 2m, then the water depth?

Let the water depth be h, a right triangle, the two sides are h and 2 meters, and the hypotenuse is h+1 meter, according to the Pythagorean theorem, it is easy to know that h=meter.

3.Xiao De and Xiao Zhi came home from school, Xiao De walked due east at a speed and arrived home in 10 minutes, Xiao Zhi rode at a speed of 26 m min in the south direction, and arrived home in 15 minutes, how many meters apart were the homes of these two classmates?

The Pythagorean theorem, the square of the distance between two classmates' homes = 125 2 + 390 2 = 25 (625 + 6084).

Home distance of two classmates = 5 6709

4.There are two monkeys at a height of 10m from a tree, one of them climbs down the tree to the pond 20m from the tree, while the other climbs to the top of the tree and goes straight to the pond, if the two monkeys pass the same distance, how tall is the tree?

Let the height of the tree be (10+x) meters.

10+20）^2 =（10＋x）^2+20^2 x=5

The height of the tree is 15 meters.

Set the eagle to catch it at a distance of x meters from the snake hole.

The distance the snake travels when caught is (9-x).

Since the two velocities are the same, the distance that the eagle flies is also (9-x), which is solved by the Pythagorean theorem equation 3 2+(9-x) 2=x 2 to get x=4

The eagle caught it 4 meters from the snake hole.

Let the distance be x, the velocity is the same, and the time is the same, so hypotenuse = 9-x, and x can be solved according to the Pythagorean theorem.

Solution: Grab x meters from the hole.

Because the speed of the eagle and the snake are the same, the distance is also equal.

3^2+x^2=(9-x)^2

9+x^2=81-18x+x^2x=4

1. Pass point A as AD, parallel to CB, and connect DC to form a rectangular ABCD

2 In ADC, ADC=90°, cut ACD=60° DAC=30° CD=1 2AC

3 If CD is X, AC is 2X

4 In ABC, AB2+CB2=AC2, 2x, 2-500, 2=BC2, 1

5 AC2-CD2=AD2 2x 2-x 2=AD2 in ACD

6 ad=cb Bring 1 into 2 to find the root number 3 with x as 500 times Narrator (if the distance traveled by ** is cb, it will be 500 times the root number 3, I don't know which line is traveling).

This diagram illustrates the problem clearly, ab=500m, abc=90°, where 60° north-east gives acb=30°. Then AC=AB sin30°=1000m, according to the Pythagorean theorem, BC= AC2-AB2

3/2 （km）

It is known that ab=500 and angle a=60°

So bc=sqrt(3)*ab=500sqrt(3).

Note: sqrt(3) indicates the root number 3

Because c=30° and ab=500, ac=1000So cos30° = bc 1000, that is, the root number of 2 parts of the root number 3 = bc 1000, so bc = 500 times the root 3, because the root 3 is approximately equal, so bc = 866

Solution: CE AB, ECB=90°

A= ECA=60°,BC=AB Tan60°=500 3=5003M A: The distance traveled by this ** is 5003m

Solution: ab=500m, angle acb=30 degrees, angle b=90 degrees, so ac=1000m is obtained by the Pythagorean theorem; bc = 500 times the root number 3