Nine Ren Xia Teaching Edition Mathematics

The first question:

Answer: ∫ quadrilateral EFGH is similar to quadrilateral KNML.

∴∠E=∠K=67，∠G=∠M= 107，∠N=∠F=43，

X/35= 10/z= 6/y =4/ 10

X= 14 y= 15 z=25。

The second question:

Solution: Let the other two sides of △DEF be x and y respectively.

∫△ABC∽△DEF, the minimum side length of△ def is 15.

∴5/ 15= 12/x= 13/y

X = 36 y = 39。

So the perimeter of △DEF is 15+36+39=90.

The third question:

Solution (1) △ fgh ∽△ jik x = 4 y =10.

(2) △GFH∽△KJH x= 124 y=33

The fourth question:

Solution: According to the meaning of the question, when the side length is expanded to three times and the area is expanded to nine times, the advertising fee is 180×9= 1620 (yuan).

Question 7:

Solution: ∫AB∶CD = OA∶OC = N, ∠ COD = ∠ AOB.

∴△ABO∽△COD,CD∥AB，

CD = b

∴AB=nb

∴nb+2x=a

∴x=a-nb/2

Question 8:

Proof: as shown in the figure, link AC, BC.

∵AB is the diameter ⊙ O, ∴∠ ACB = 90.

∵CD⊥AB,∴∠A=∠BCP

∴△ACP∽△CBP

∴AP/PC=PC/BP

∴CP？ =PA PB

Question 9:

Solution: let the ball touch the wall x meters above the ground.

According to the meaning of the question, it is1.8: 2 = x: 6.

The solution is X=5.4.

Set the ball to contact the wall 5.4 meters above the ground.

Question 10:

Solution: when the focal length is 50 mm, the width of the scene that can be photographed is X.

From the meaning of the title, it is 35: 50 = x: 5.

The solution is x=3.5.

When the focal length is 70mm, the width of the photographed scene is y meters.

Judging from the meaning of the question, it is 35 ∶ 70 = y ∶ 5.

The answer is y=2.5.

So when the focal length is 50mm, the width of the shooting scene is 3.5 meters, and when the focal length is 70mm, the width of the shooting scene is 2.5 meters.

Question 1 1:

Evidence: ∵AC∥BD,∴△AOC∽△BOD.

∴AO∶BO=OC∶OD

∴OA OD=OB OC

Question 13:

Solution: Let the side length of a square part be x mm.

∫ef∨ BC ∴△AEF∽△ABC,△AKF∽△ADC.

∴EF/BC=AF/AC,AK/AD=AF/AC，

∴EF/BC=AK/AD=AD-EG/AD

∫ BC =120mm AD = 80mm.

∴x/ 120=80-x/80

x=48