According to your experience in physical education, the speed of Olympic 100 meter runners should be () A. About 2m / s B. 6m / s C. 10m / s D. 20m / S

According to your experience in physical education, the speed of Olympic 100 meter runners should be () A. About 2m / s B. 6m / s C. 10m / s D. 20m / S

(1) 100m junior high school students run 14s, junior high school students 100m speed: v = st = 100m14s ≈ 7.14m/s. Olympic 100 meter race speed is slightly greater than 7.14m/s, so choose 10m / S. (2) 20m / S = 72km / h, this speed is car speed, Olympic athletes 100 meter speed is less than car speed. So choose C

3 4 10 - 6 24 o'clock

Hello!
3 × (4 - 6 + 10)=24
Hope to help you!

It is known that in the triangle ABC, the angle ACB is 90 degrees, CD is perpendicular to AB, D, the bisector of angle a intersects CD to F, BC to F, EH is made through point E, AB is perpendicular to h, CE = CF = EH is proved

EH is perpendicular to AB, angle ACB is 90 degrees, bisector of angle a is AE
It is easy to know that ace of triangle is equal to ahe
Then CE = eh angle CEF = FEH 1)
EH is perpendicular to AB, CD is perpendicular to ab
So the angle FEH = CFE 2)
From 1) 2) to know the angle CEF = CFE
So CF = CE
So CE = CF = EH
The title should be "hand in BC to e"“

If the bisector of CD ⊥ AB, ∠ cab intersects CD and F, it is called BC and E. if eh ⊥ AB passes through point E and h, then CE = CF = EH. Why?
In △ ABC, ∠ ACB = 90 ° CD ⊥ AB, the bisector of ∠ cab intersects CD and F, which is called BC and E. if eh ⊥ AB passes through point E, then CE = CF = EH. Why?

Because AE is ∠ cab, it is called bisector ∠ ace and ∠ ahe = 90 degree
So he is a triangle
Connect FH
Because AE is an angular bisector, CF = HF
Because eh ‖ CD
Because triangle ACF ≌ triangle AHF, so ∠ AHF = ∠ ACF
Because triangle ace ≌ triangle ahe, so ∠ ahe = ∠ ace
So, fhe FCE
Because ∠ ECD = ∠ beh, ∠ fhe = ∠ Heb
So FH ‖ CB
So the quadrilateral FHEC is a parallelogram. So FH = CE
And because CF = FH, CF = CE
So CE = CF = EH

As shown in the figure, in △ ABC, BD ⊥ AC, CE ⊥ AB, vertical feet are D and e respectively, BD and CE intersect at h, ∠ a = 60 °, DH = 2, eh = 1
1) Find the length of BD and CE
2) If ∠ ACB = 45 °, calculate the area of △ ABC

The solution is: 1) ∠ DCH = 90 ° - 60 ° = 30 ° so HC = 2hd = 4 (the side opposite the 30 ° angle in a right triangle is half of the hypotenuse), so EC = HC + eh = 4 + 1 = 5. Similarly, BD = BH + HD = 2eh + HD = 2 + 2 = 4.2) ∠ ACB = 45 °△ DBC is a right angle equal to the triangle DC = BD = 4, because ∠ a = 60 ° is in the right triangle abd

As shown in the figure, in △ ABC, BD is vertical to AC and CE, AB is vertical, D is perpendicular, e is perpendicular, a is equal to 60 degrees, DH = 2, eh = 11. Find the length of BD and CE

Idea: BD is perpendicular to AC, CE is perpendicular to AB, angle a = 60 degrees. So ∠ DCH = 90-60 = 30 degrees, the side opposite 30 degrees in a right triangle is half of the hypotenuse. HC = 2hd = 4, then EC = HC + eh = 4 + 1 = 5. Similarly, BD = BH + HD = 2eh + HD = 2 + 2 = 4

Known: as shown in the figure, in the triangle ABC, ab = AC, points D and E are on AB and AC respectively, and BD = CE, DG is perpendicular to BC, EH is perpendicular to BC, and the perpendicular feet are g and h respectively
To prove that a quadrilateral DGHE is a rectangle

Results: ab = AC, BD = CE
∴AD=AE
‖ de ‖ BC (parallel line bisection)
∵DG⊥BC,EH⊥BC
∴DG∥EH
A quadrilateral DGHE is a parallelogram
∵∠DGH＝90°
The parallelogram DGHE is a rectangle

The perimeter of △ ABC is 20, and the bisectors of ∠ A and ∠ B intersect at p. if the distance from P to edge AB is 4, then the area of △ ABC is______ ．

The bisectors of ∵ - A and ∵ - B intersect at P, the distance from P to edge AB is 4, the distance from point P to AB and BC is 4, the perimeter of ∵ △ ABC is 20, and the area of ∵ △ ABC = 12 × 20 × 4 = 40

In △ ABC, point P is the intersection of triangle bisectors, the distance between point P and ab is 0.8, and the perimeter of triangle is 9, then the area of △ ABC is ()

Because point P is the intersection of the bisectors of the three interior angles of a triangle
So the distance from P to the three sides is equal to 0.8
So s △ ABC = (1 / 2) AB * 0.8 + (1 / 2) BC * 0.8 + (1 / 2) * ac * 0.8
=(/2)(AB+CB+AC)*0.8
=(1/2)*9*0.8
=3.6

The perimeter of △ ABC is 60, and the bisectors of angle A and angle B intersect at point P. if the distance between point P and edge AB is 10, the area of △ ABC is----------
In the triangle ABC, the vertical bisectors de on the sides of AB = a, AC = B, BC intersect BC and BA at points D and e respectively, then the perimeter of △ AEC is equal to

Question 1: from the meaning of the question, P is the heart of △ ABC, that is, the distance from P to the three sides is equal to 10,
S△ABC=1/2(AB+BC+AC)*h=300
The second question: from the meaning, be = EC,
So the perimeter of △ AEC = AE + EC + AC = AE + be + AC = AB + AC = a + B