Toán 9 bài 1 hình học chương 1 – Một số hệ thức về cạnh và đường cao trong tam giác vuông

Bạn đang xem: Toán 9 bài 1 hình học chương 1 – Một số hệ thức về cạnh và đường cao trong tam giác vuông tại thptnguyenquannho.edu.vn

Math 9 lesson 1 geometry chapter 1 students will have to solve a lot of exercises related to formulas and relationships in the math part. To master those contents, in addition to memorizing the available formulas, students also need to practice doing a lot of math problems with different types of problems.

I. Maths knowledge of grade 9 geometry lesson 1 chapter 1

Right in the first lesson of chapter 1 geometry, you will have access to the basic relation related to triangles. Before going into the instructions to solve the detailed exercises, Kienguru will help you systematize the 9th grade math theory lesson 1.

1. The relation between the side of a right angle and its projection on the hypotenuse

Consider the following triangle:

Theorem 1: In a right triangle, the square of each right angled side is equal to the product of the hypotenuse of the projection of that right angle on the hypotenuse.

Formula:

\

2. Some relations related to highways

– Theorem 2: In a right triangle, the square of the altitude corresponding to the hypotenuse is equal to the product of the projections of the two sides of the right angle on the hypotenuse.

Formula:

– Theorem 3: In a right triangle, the product of two sides of a right angle is equal to the product of the hypotenuse and the altitude respectively.

Formula:

– Theorem 4: In a right triangle, the inverse of the square of the altitude corresponding to the hypotenuse is equal to the sum of the inverses of the squares of the two sides.

Formula:

Example exercise: Given triangle ABC, right angled at A, altitude AH. Know AB:AC = 3:4 and AB + AC = 21cm.

a) Calculate the sides of triangle ABC.

b) Calculate the lengths of the segments AH, BH, CH.

Prize:

a) According to the assumption: AB:AC = 3:4, so

Hence AB = 3.3 = 9 (cm); AC = 3.4 = 12 (cm).

Triangle ABC is right-angled at A, according to Py – ta – go theorem we have:

So BC = 15cm

b) Triangle ABC is right-angled at A, we have AH.BC = AB.AC, deduce

.

Set BH = x (0 < x < 9) then HC = 15 – x, we have:

(7,2)2 = x(15 – x) x2 – 15x + 51.84 = 0

⇔ x(x – 5.4) = 9.6(x – 5.4) = 0

(x – 5.4)(x – 9.6) = 0

⇔ x = 5.4 or x = 9.6 (type)

So BH = 5.4cm. Then HC = BC – BH = 9.6 (cm).

Note: BH can be calculated as follows:

AB2 = BH.BC inferred

II. Support to solve 9 problems 1 geometry textbook

To help you understand the relationship in the above theoretical content, right now, Kienguru will help you solve the 9th grade math exercise in geometry lesson 1. The solution guide is fully compiled and follows the textbook. faculty, so that you can refer to it to practice doing related exercises.

1. Lesson 1 page 68

Calculate x and y in each of the following figures (h.4a, b):

Prize:

1. Applying Pythagorean theorem we have:

⇒ x + y = 10

Applying theorem (1) we have:

1. Applying theorem (1) has:

=> y = 20 – x = 20 -7.2 = 12.8.

2. Lesson 2 page 68

Calculate x and y in each of the following figures (h.5):

Prize:

Applying theorem 1 we have:

3. Lesson 3 page 69

Calculate x and y in each of the following figures (h.6):

Prize:

Applying Pythagorean theorem we have:

Applying theorem (3) we have:

4. Lesson 4 page 69

Calculate x and y in each of the following figures (h.7):

Prize:

Applying theorem (2) we have:

Applying theorem (1) we have:

III. Guide to solving sbt

Kienguru has just supported you to solve the exercises in the math textbook 9 lessons 1 geometry. To practice and firmly grasp the theorems as well as the content of the lesson, let’s solve some more problems in the workbook!

1. Lesson 1 page 102

Calculate x and y in the following figures:

The answer:

a. Picture A

According to the Pythagorean theorem, we have:

According to the relation between the side of the right angle and its projection, we have:

b. Figure b

According to the relation between the right angle side and the projection, we have:

142 = y.16

x + y = 15 x = 16 – y = 16 – 12.25 = 3.75

2. Lesson 2 page 102

Calculate x and y in the following figures:

Prize:

a. Picture A:

According to the relation between the right angle side and the projection, we have:

x2 = 2.(2 + 6) = 2.8 = 16 x = 4

y2 = 6.(2 + 6) = 6.8 = 48 y = 48 = 4√3

b. Figure b:

According to the relationship between the altitude and the projection of the two right angles, we have:

x2 = 2.8 = 16 x = 4

3. Lesson 3 page 103

a. Picture A

According to the Pythagorean theorem, we have:

According to the relation between the altitude and the side in a right triangle, we have:

b. Figure b

According to the relation between elevation and projection, we have:

According to the relation between the right angle side and the projection, we have:

4. Lesson 5 page 103

Let ABC be a right triangle at A, altitude AH. Solve the problem in each of the following cases:

a. Let AH = 16, BH = 25. Calculate AB, AC, BC, ONLY

b. Let AB = 12, BH = 6. Calculate AH, AC, BC, ONLY

Prize:

a. According to the relation between elevation and projection, we have: AH2 = BH.CH

BC = BH + CH = 25 + 10.24 = 35.24

According to the relation between the right angle side and the projection, we have:

b. According to the relation between the right angle side and the projection, we have:

CH = BC – BH = 24 – 6 = 18

According to the relation between the sides of the right angle and the projection, we have:

According to the relation between the altitude and the projection of the right angle, we have:

5. Lesson 7 page 103

The altitude of a right triangle divides the hypotenuse into two lines of lengths 3 and 4. Calculate the right angles of this triangle.

Prize:

Suppose triangle ABC has angle

AH BC, BH = 3, CH = 4

According to the relation between the right angle side and the projection, we have:

= 3.(3 + 4) = 3.7 = 21 AB = √21

= 4.(3 + 4) = 4.7 = 28 ⇒ AC = √28 = 2√7

IV. Conclusion

Thus, Kien Guru helped you review the theoretical focus of math 9 lesson 1 geometry chapter 1, as well as solve some related exercises in textbooks and workbooks. Hopefully, through the above article, you can master the content of the lesson as well as practice doing the exercises easily and quickly.

To read more related articles or refer to the content of other subjects, please visit Kien Guru’s website.

Wish you always study well and get high scores

Bạn thấy bài viết Toán 9 bài 1 hình học chương 1 – Một số hệ thức về cạnh và đường cao trong tam giác vuông có khắc phục đươc vấn đề bạn tìm hiểu ko?, nếu ko hãy comment góp ý thêm về Toán 9 bài 1 hình học chương 1 – Một số hệ thức về cạnh và đường cao trong tam giác vuông bên dưới để Trường THPT Nguyễn Quán Nho có thể thay đổi & cải thiện nội dung tốt hơn cho các bạn nhé! Cám ơn bạn đã ghé thăm Website: thptnguyenquannho.edu.vn của Trường THPT Nguyễn Quán Nho

Nhớ để nguồn bài viết này: Toán 9 bài 1 hình học chương 1 – Một số hệ thức về cạnh và đường cao trong tam giác vuông của website thptnguyenquannho.edu.vn

Chuyên mục: Giáo dục