## Math 8 Chapter 2 Lesson 2: Area of a rectangle

## 1. Theoretical Summary

### 1.1. The concept of polygon area

The measure of a portion of the plane bounded by a polygon is called the area of the polygon.

Each polygon has a defined area. Polygonal area is a positive number.

The area of a polygon has the following properties:

– Two similar triangles have the same area.

– If a polygon is divided into polygons that have no interior points in common, its area is equal to the sum of the areas of those polygons.

### 1.2. Formula to calculate the area of a rectangle

Area of a rectangle is the product of its two dimensions \( S = ab\)

(\(S\) is the area, \(a\) is the length, \(b\) is the width of the rectangle).

### 1.3. Formula to calculate area of square, right triangle

The area of a square is equal to the square of its side: \(S = {a^2}\)

The area of a right triangle is half the product of the two sides of the right angle: \(S = \dfrac{1}{2}ab\)

## 2. Illustrated exercise

### 2.1. Exercise 1

How will the area of the rectangle change if :

a) The length is doubled, the width remains the same

b) Length and width increase 3 times.

c) Length increased 4 times, width decreased 4 times.

**Solution guide**

Let the length and width of a rectangle be a,b . respectively

The area of the rectangle is Shcn = ab

a) If the length is increased by 2 times and the width remains the same, then the new length and width are 2a and b.

The area of the new rectangle is Sm = 2a.b = 2S.

⇒ The area of the rectangle is increased by 2 times.

b) If the length and width are increased by 3 times, the new length and width will be 3a,3b

The area of the new rectangle is Sm = 3a.3b = 9S.

⇒ The area of the rectangle is increased 9 times.

c) If the length is increased by 4 times and the width is decreased by 4 times, the new length and width are 4a, 1/4b.

The area of the new rectangle is Sm = 4a. 1/4b = ab = S.

⇒ The area of the rectangle remains the same.

### 2.2. Exercise 2

Find the lengths of the sides of a rectangle knowing that the square of the length of one side is 16cm and the area of the rectangle is 28cm.^{2}

**Solution guide**

Let the two dimensions of the rectangle be a,b ( a > 0, b > 0 ). Then the area of the rectangle is Shcn = ab

According to the problem, we have: xy = 28 ( 1 ) and x2 = 16 = 42 ⇔ x = 4 (because x > 0 ), the case y2 = 16 is similar.

Substituting x = 4 into equality ( 1 ) we have: 4y = 28 ⇔ y = 7.

With x = 4, y = 7 satisfying the condition requirement.

So the two dimensions of the rectangle are 4cm, 7cm

### 2.3 Exercise 3

A room with a rectangular background with dimensions \(4,2\,m\) and \(5,4\,m\) has a rectangular window with dimensions \(1\,m\) ) and \(1,6\,m\) and a rectangular door whose dimensions are \(1,2\,m\) and \(2\,m.\)

We consider a room to meet the standard of light if the area of the doors is equal to \(20\%\) the area of the floor. Ask if the room above meets the standard level of light or not?

**Solution guide**

The area of the floor is: \(S = 4,2.5,4 = 22.68\) (\({m^2}\))

The window area is: \({S_1}= 1. 1.6 = 1.6\) (\({m^2}\))

The area of the door is: \({S_2}=1,2.2 = 2,4\) (\({m^2}\))

The area of the doors is: \(S’ = {S_1} + {S_2}= 1.6 + 2.4 = 4\) (\({m^2}\))

We have \(\dfrac{S^{‘}}{S} = \dfrac{4}{22.68}.100\%≈ 17.64\% < 20\%.\)

So the room does not meet the standard level of light.

### 3.2. Multiple choice exercises

**Question 1:** Rectangle is 5 times shorter in length and 5 times wider in width, then the area of the rectangle

A. No change

B. Increase 5 times

C. Reduce 5 times

D. Increase 3 times

**Verse 2: **A rectangle has an area of 24 cm2 and a length of 8 cm. The perimeter of that rectangle is:

A. 11 cm

B. 20 cm

C. 22 cm

D. 16 cm

**Question 3:** Select the correct answers

A. The area of an equilateral triangle is half the product of the two sides of the right angle

B. The area of a rectangle is half the product of its two dimensions

C. The area of a square with side a is 2a

D. All of the above answers are correct

**Question 4: **Given triangle ABC, the area of triangle is 16cm2 and side BC = 8cm. The altitude corresponding to side BC is:

A. 5 cm

B. 8 cm

C. 6 cm

D. 4 cm

**Question 5: **Given triangle ABC, take M on BC such that BM = 3CM. Choose the wrong sentence:

A. \({S_{ABM}} = \frac{3}{4}{S_{ABC}}\)

B. SABM = 3SAMC

C. \({S_{AMC}} = \frac{1}{3}{S_{ABC}}\)

D. SABC = 4SAMC

## 4. Conclusion

Through this lesson, you will learn some of the main topics as follows:

- Mastering the formula for calculating the area of a rectangle, square, and right triangle.
- Understand that in order to prove the above formulas for area calculation, it is necessary to use the properties of polygon area.

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