A math problem about similarity is worth 30 points

Let the triangle formed by Master Li and the mirror be DFC (d is Master Li's eyes) DFC ABC ( DCF ACB, DFC ABC) df CF ab CB i.e.

CB (again D'f'c'∽△abc'(Same reason as above)d'f'∶c'f'＝ab∶c'b i.e.

ab 10 A: Write it yourself, it's hard to type two words!

The question mark is 8 stupid ways. Count from both ends. Divide the previous one by 7 and get to the question mark 6? , in their sixties.

The mantissa at the back is 52 at the question mark

Therefore, only 63 and 52 can be combined, so the number is 3, 5, 8 divided by 7 for 158730159 746031746

Let's assume that the height of the tree is y, and the distance from b to c is x, and we can get c from the graph I made'f'/(cc'+bc)=cf bc, and from these two equations, y= can be solved, so the height of the tree is.

Please see ** (click for clarity).

m is the midpoint of bc, bm=mc, and bme= cmf(to apex), bem= cfm(inner wrong angle), bme cmf, cf=be=3 4ab

and ane = cnf (to the vertex angle), aen = cfn (inner wrong angle), ane cmf, and ae = ab + be = 7 4ab

s△cnf：s△ane=cf²/ae²=3²/7²=9/49

9/16

The ratio of the area of a similar triangle is equal to the square of the similarity ratio.

s△cnf:s△ane=3:14

Let the parallelogram ab be the base height of x

By EBM FMC

Get: eb=cf

and ab=4 3eb (do your own math).

s cnf:s ane=1 3x:7 3*2 3x=3:14 thank you.

The shadow length of the 1m high benchmark indicates that the shadow length is half of the benchmark.

Therefore, if the shadow of the tree on the ground is 3 meters, it means that the corresponding tree is 6 meters, plus 2 meters parallel to the wall. The height of this large tree is 6+2 8 meters.

Columnable proportion: Let the part of the tree cast on the ground be x meters.

x:3=1：

The length of a large tree corresponding to a shadow length of 3m on the ground is 6m, and the length of a large tree corresponding to a shadow length of 2m on the wall is 2m

So the height of the big tree is 6+2=8m

This has to do with the position of the point p.

For the convenience of illustration, let the three sides of abc bc=a, ac=b, ab=c. According to the title, b is the obtuse angle and b is the largest side.

The triangle is obviously similar to the original triangle by passing the point P as the parallel line of AB and the parallel line of BC, which is an ordinary situation, that is, there are at least 2 L that meet the meaning of the title, which may be recorded as L1 and L2, as shown in Figure 1.

In the obtuse angle ABC, b is the largest side, so the same can be proved for C 2, A 2 B On the other hand, in the obtuse angle ABC, B is the obtuse angle, so A 2 + C 2C 2.

So, ap2=ac-cp2=b-a2 b=(b2-a2) b>c 2 b=ap1. Therefore, the position of points p1, p2 is shown in Figure 2.

Joining bp1, by ap1=c 2 b, i.e., ap1=ab 2 ac, we get ap1 ab=ab ac, and bap1= cab, so bap1 cab.

Linking BP2 can also be proved to BCP2 ACB.

The sub-scenarios are discussed as follows:

If the point p=p1, then the straight line bp1 is the straight line l3 that matches the meaning of the question;

If the point p=p2, then the straight line bp2 is the straight line l3 that conforms to the topic;

Solution: Only 2 lines can be made, which are parallel lines that cross the P point as AB and BC edges.

It can be seen that the triangle BCD is similar to the triangle ABC, which is proportional to the corresponding sides as:

bc/cd=ac/bc

i.e. BC2-AC*cd=0

And because bc=ad, cd=ac-ad

So get :

ad^2-ac*(ac-ad)=0

Solving the equation about ad yields:

ad=ac*(√5-1)/2

So ad ac=( 5-1) 2

Not similar.

The aspect ratio of the lawn rectangle is not 1, and the ratio of length to width increase is 1, and the two are unequal, so they are not similar.

Dissimilar: The proportions are different before and after.