## Math 9 Chapter 1 Lesson 1: Some relationships about sides and altitudes in a right triangle

## 1. Summary of theory

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

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

Specifically, in Figure 1 we have:** **Triangle ABC is right-angled at A, so:

\(b^2=a.b’\) , \(c^2=a.c’\)

### 1.2. 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 two projections of the two sides of the right angle on the hypotenuse.

Specifically, in Figure 1 we have:** **Triangle ABC is right-angled at A, so:

\(h^2=b’.c’\)

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

Specifically, in Figure 1 we have:** **Triangle ABC is right-angled at A, so:

\(bc=ah\)

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

Specifically, in Figure 1 we have:** **Triangle ABC is right-angled at A, so:

\(\frac{1}{h^2}=\frac{1}{b^2}+\frac{1}{c^2}\) or \(h=\frac{bc}{\sqrt{ b^2+c^2}}\)

**Attention: **In the examples and numerical exercises of this chapter, the length measurements in each lesson, if not written in units, are conventionally the same.

## 2. Illustrated exercise

### 2.1. Type 1. Problem using Theorem 1

**Question 1: **Calculate: \(x, y\) in the picture below

**Solution guide**

Applying theorem 1 we have: \(x^2=3,6.(3,6+6.4)=3,6.10=36\Rightarrow x=6\)

Similarly: \(y^2=6.4.(3,6+6.4)=6.4.10=64\Rightarrow y=8\)

### 2.2. Type 2. Problem using Theorem 2

**Verse 2: **Calculate: \(x,y\)

**Solution guide**

Applying theorem number 2, we have: \(4^2=2.y\Rightarrow y=8\).

Applying theorem 1, we have: \(x^2=2.(2+8)=2.10=20\Rightarrow x=2\sqrt{5}\)

### 2.3. Type 3. Problem using Theorem 3

**Question 3: **Let ABC be a right triangle at A, altitude AH. Know AB:AC = 3:4 and AH=12. Calculate the perimeter of triangle ABC.

**Solution guide**

Set: \(AB=3k, AC=4k\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{9k^2+16k^2}=5k\)

Applying theorem 3, we have: \(AB.AC=BC.AH\Leftrightarrow 3k.4k=5k.12\Rightarrow k=5\)

\(\Rightarrow AB=15; AC=20; BC=25\) and \(P=60\)

### 2.4. Type 4. Problem using Theorem 4

**Question 4: **Calculate: \(x,y\)

**Solution guide**

Applying theorem 4, we have: \(\frac{1}{x^2}=\frac{1}{b^2}+\frac{1}{c^2}\Rightarrow x=\frac{ bc}{\sqrt{b^2+c^2}}=\frac{3.4}{\sqrt{3^2+4^2}}=\frac{12}{5}\)

Applying theorem 3, we have: \(xy=3.4\Rightarrow y=\frac{3.4}{x}=\frac{12}{\frac{12}{5}}=5\)

(can calculate \(y\) first using the Pythagorean theorem and then \(x\))

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Calculate x and y in each of the following figures

**Verse 2:** Let ABC be a right triangle at A, AB : AC = 7 : 24, BC = 625 cm. Calculate the length of the projection of the two sides of the right angle on the hypotenuse.

**Question 3:** Let ABC be a right triangle at A, altitude AH. Know AC = 20 cm, BH = 9 cm. Calculate the lengths BC and AH

**Question 4: **Let ABC be a right triangle at A, altitude AH. Knowing AB/AC = 20/21 and AH = 420. Calculate the perimeter of triangle ABC

**Question 5:** Let ABC be a right triangle at A, altitude AH

Given that AC/AB = √2; HC – HB = 2cm.Calculate:

a) Ratio HC : HB

b) The sides of triangle ABC

### 3.2. Multiple choice exercises

**Question 1. **Triangle ABC is right-angled at A, altitude AH. AB = 9cm, AC = 12cm. Find the false statement

A. BC = 15cm

B. AH=6.2cm

C. BH=5.4cm

D. CH=9.6cm

**Verse 2. **Let OEF triangle right at O with altitude OI. Yes IE = 3, IF = 12. Calculate OE,OF

A. \(OE=3\sqrt5;OF=6\sqrt5\)

B. \(OE=5\sqrt3;OF=3\sqrt2\)

C. \(OE=4\sqrt2;OF=6\sqrt3\)

D. Another answer

**Verse 3.** Triangle ABC is right-angled at A, altitude AI has AB = 13, AI = 12. The area of triangle ABC is:

A. 90.8

B. 189.5

C. 202.8

D. 220

**Verse 4. **Let ABC be a right triangle at A with altitude AH=12, know BH-CH=7. What is the length of side BC?

A. BC=23

B. BC=24

C. BC=25

D. BC=26

**Question 5. **Right triangle ABC has: AB:AC is proportional to 3:4. Knowing AH=6. What is the length of side BC?

A. BC=11.5

B. BC=12

C. BC=12.5

D. BC=13

## 4. Conclusion

Through this lesson, you should know the following:

- Master the relations between the side of a right angle and its projection on the hypotenuse, some of which are related to the altitude.
- Do related math problems.

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