Math Problem 1 Junior 3 I .

1.When t=3 Yes, bpq is a right triangle.

When t=3, bp=3 is the midpoint of ba, q coincides with point c, and because abc is an equilateral triangle, abc is an equilateral triangle with a side length of 6cm.

2. It can be seen that bp=6-t, bq=2t, and because abc is an equilateral triangle with a side length of 6cm.

So pbq=60°

If the BC perpendicular line crosses the BC at the P point and crosses the BC at D, then PD BQ, PD=sin60° PB= 3 2 (6-t).

Then s=pd, bq= 3, 2(6-t), 2t= 3(6-t)t

3, qr ba and abc is an equilateral triangle, then ar=bq=2t, and ap=t, and because abc is an equilateral triangle, we can get par=60°, ar=2ap, we can know that arp is a right triangle arp=30°

There is also qr ba, qrc=60°, so prq is always unchanged at 90°

When t=3, bpq is a right triangle.

s=√3(6-t)t

qr ba and abc is an equilateral triangle, then ar=bq=2t and ap=t, and because abc is an equilateral triangle, we can get par=60° and ar=2ap, we can see that arp is a right triangle arp=30°

There is also qr ba, qrc=60°, so prq is always unchanged to 90° approval.

tan bac=bc ac, because there is that angular bisector and perpendicular easy to know that dac is similar to abc, so bac= adc, so tan bac=ac cd=ab ad, and then do your own algebra calculations.

Do a perpendicular line over o, vertical cd, even od, and you get it. In a right triangle, cd=r, cf=r, r2-r2=1 is. ∏r2-∏r2=∏（r2-r2）=∏。

Amount! The shadow area is the large circle minus the minor circle, r2- r2= x1=

[Solution]:

The path traversed by the point o is a straight line parallel to the horizontal ground. There are 2 ways to get it.

Method 1]: The fan-shaped OAB is a part of the circle, and the fan-shaped is completed into a circle.

Because, when the circle rolls on the horizontal ground, the height of the center of the circle is constant.

Therefore, as part of a circle, when the sector OAB rolls on the horizontal ground, the height of the point O is also constant, and the trajectory of the point O is a straight line parallel to the horizontal ground.

Or, more popularly, look at a part of the wheel of the fan, when the wheel rolls on the horizontal ground, the axis of rotation of the wheel is always parallel to the horizontal ground, then the trajectory of the axis of the wheel is also a straight line parallel to the horizontal ground.

Therefore, the forward trajectory of the circle O of the sector OAB is a straight line parallel to the horizontal ground.

Method 2]: When the fan-shaped OAB moves forward on the horizontal ground, it has been tangent to the ground, that is, the ground is a fan-shaped tangent, and the line between the tangent point and the center of the circle O is perpendicular to the ground.

The trajectory of the tangent point is the straight line formed when the edge of the fan moves forward on the ground, because the line between the tangent point and the center o is perpendicular to the straight line formed by the trajectory of the tangent point, and the distance between the center o and the tangent point at each moment is always equal to the radius.

Therefore, the trajectory of the center of the circle is also a straight line, and the straight line is parallel to the straight line formed by the fan-shaped edge advancing on the ground.

Because the straight line formed by the advance of the center o is parallel to the straight line formed by the tangent point on the ground, the line between the center o and the corresponding tangent point remains perpendicular to the ground at all times, and the distance between the center o and the tangent point is always equal to the fan radius.

Then, at any two moments, the two radii of the line connecting the center of the circle o and the tangent point, the line segment formed by the center of the circle o, and the line segment formed by the front of the tangent point, together form a rectangle.

Therefore, the distance between the center o and the tangent is equal.

Let the central angle of the sector OAB aob=x The arc length l of the sector OAB and the circumference of the circle c

Then, s sector s circle = x 360°

l Sector c Circle = x 360

So, s sector l sector = s circle c circle.

Because, s circle c circle = r 2

So, s sector l sector = r 2

s sector = 30

Then, l sector = 2s sector r

The distance at which the tangent advances is the distance formed by the arc of the sector on the ground.

Because, the distance at which the center of the circle is advanced is equal to the distance at which the tangent is advanced.

Therefore, the distance traveled by the center of the circle is 10 cm

During the fan rolling, the path traveled by the dot o is 10 cm

PS: Change the point b to the point O, and the path of the point b is a cycloid (triangular curve), which is very complicated to calculate.

Make an auxiliary line to connect OD, and O is the midpoint of AB to get OD=1 2BC=3, and the triangle ODF is an isosceles triangle. db=3

1. Because the three-family ABC is an isosceles right triangle, and AC cuts a circle o in D. So the OD is perpendicular to the AC. So od bc

2. By OD BC, the angle DFO = angle bfg (to the apex angle), the angle dob = angle FBG (the wrong angle), the angle ODF = the angle BGF (the wrong angle 0

3. From the result of 2, it can be obtained that the triangle ODF is similar to the triangle bfg, so FBG is also an isosceles triangle, bf=bg, bf=ob-of=1 2AB-of=6 times the root number 2-3

So bg = 6 times the root number 2-3

Math symbols are really hard to beat. Let's just give some points. Hehe.

x 2+1 x 2+x+1 x=0, x 2+2+1 x 2+x+1 x-2=0(x+1 x) 2+(x+1 x)-2=0, let x+1 x=t, the original equation is:

t^2+t-2=0

t+2)(t-1)=0

t1 = when t = -2, x + 1 x = -2, x 2 + 2x + 1 = 0, (x + 1) 2 = 0, x = -1

When t = 1, x + 1 x = 1, x 2-x + 1 = 0, discriminant = 1-4 = -3<0, no solution, from above, x = -1

Solution: Set up rent X trucks of type A and rent trucks (10-x) of type B; then x type A trucks can transport 4x tons of type A goods and x tons of type B goods, and (10-x) type B trucks can transport 2 (10-x) tons of type A goods and 2 (10-x) tons of type B goods; Depending on the title, groups of inequalities can be listed:

4x+2(10-x)≥30 (1)

x+2(10-x)≥13 (2)

Solve inequality (1) to get x 5

Solve inequality (2) to get x 7

The solution for the group of inequalities is 5 x 7

When x=5, 10-x=10-5=5

When x = 6, 10-x = 10-6 = 4

When x=7, 10-x=10-7=3

There are three options for renting a car:

Plan 1: Rent 5 cars of type A and 5 cars of type B.

Plan 2: Rent 6 cars of type A and 4 cars of type B.

Plan 3: Rent 7 cars of type A and 3 cars of type B.

Set rent A x B y.

x+y=10 （1）

x（4a+1b）+y（2a+2b）≥30a+13b （2）

Thus there is 1, x+y=10

2， 4x+2y≥30

3， x+2y≥13

So x 5 so x=5 y=5 at this time a 5*4+5*2=30 b 5*1+5*2 13

x=6 y=4 yes.

x=7 y=3 yes.

x=8 y=2 yes.

x=9 y=1 yes.

x=10 y=0 yes.

It can be seen that there are a total of 6 ways to rent a car.

The watchtower lord adopts!

Let A be x car, and find its range is [5,7], so there are 3 schemes, A5 B5, A6 B4, A7 B3

The process teacher will speak, typing is too troublesome.

[Solution]:Set up car A to rent car A Rent car B (10 - a) 4x + 2 (10 - x) greater than or equal to 30

x +2 (10 - x) greater than or equal to 13

Solution: 5 is less than or equal to x less than or equal to 7

To sum up, there are three car rental options.

a 5 6 7

b 5 4 3

Since AC tangentiates the arc ab at the point A, A=90 degrees, OAC is a right triangle.

Find the area of the right triangle: s=ac*ao 2=4*6 2=12 square centimeters find the entire circumference of the circle composed of the fan: l=2 *r=2*find the angle of the fan:

Because the circumference of the whole circle is, and the length of the fan arc ab is equal to 3cm, it can be seen that the fan occupies 3 of the whole circle, and the area of the circle is calculated s= *r 2=, so the area of the fan is s1 = square centimeter.

Therefore, the area of the shaded part of the ACB is: 12-9 = 3 square centimeters.