## Math 8 Review Chapter 2: Polygons. Area of Polygon

## 1. Theoretical Summary

### 1.1. Polygon – Convex Polygon

**Define:**

A convex polygon is a polygon that always lies in a half-plane whose border is a line containing any side of that polygon.

A regular polygon is a polygon with all sides equal and all angles equal.

**Attention:**

– Polygons \(n\) top \(\left( {n \ge 3} \right)\) called figure \(n\) sense or shape \(n\)-edge.

– Sum of the angles of the polygon \(n\) equal side \(\left( {n – 2} \right).180^\circ \) .

– Each angle of a regular polygon \(n\) equal side \(\dfrac{{\left( {n – 2} \right).180^\circ }}{n}\).

– Number of diagonals of a convex polygon \(n\) equal side \(\dfrac{{n\left( {n – 3} \right)}}{2}\) .

### 1.2. Formula to calculate area of polygon

**Rectangular area:**

The area of a rectangle is the product of its two dimensions: \(S = ab\) .

**Area square**

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

**Area of a right triangle**

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

**Area of a regular triangle**

The area of a triangle is half the product of one side and its height: \(S = \dfrac{1}{2}ah\).

**Area of trapezoid**

The area of a trapezoid is half the product of the sum of the two bases and the height: \(S = \dfrac{{\left( {a + b} \right)h}}{2}\)

**Area of parallelogram**

The area of a parallelogram is equal to the product of one side and its height: \(S = ah\) .

**Area of a quadrilateral with two perpendicular diagonals**

The area of a quadrilateral with two perpendicular diagonals is half the product of the two diagonals

**Area of rhombus**

The area of a rhombus is half the product of the diagonals: \(S = \dfrac{1}{2}{d_1}. {d_2}\)

## 2. Illustrated exercise

### 2.1. Exercise 1

Given rectangle ABCD. Let H, I, E, K be the midpoints of BC, HC, DC, EC, respectively (h.159). Calculate

a) Area of triangle DBE

b) Area of quadrilateral EHIK

**Solution guide**

a) We have: \(DE = \dfrac{1}{2}DC = \dfrac{1}{2}.12 = 6\left( {cm} \right)\) (midpoint property)

\({S_{DBE}} = \dfrac{1}{2}.DE.BC = \dfrac{1}{2}.6.6,8\)\(\, = 20.4\) \(\left ( {c{m^2}} \right)\)

b) We have : \(HC = \dfrac{1}{2}BC = \dfrac{1}{2}.6,8 = 3.4\,\left( {cm} \right)\) (calculation) midpoint)

\(HI = \dfrac{1}{2}HC = \dfrac{1}{2}.3.4 = 1.7\left( {cm} \right)\) (midpoint property)

\(EC = DE = 6cm\) (midpoint property)

\(EK = KC = \dfrac{1}{2}EC = \dfrac{1}{2}.6 = 3\,\left( {cm} \right)\) (midpoint property)

Therefore

\({S_{EHIK}} = {S_{EHK}} + {S_{HKI}} \)

\( = \dfrac{1}{2}EK.HC + \dfrac{1}{2}HI.KC\)

\( = \dfrac{1}{2}EK.HC + \dfrac{1}{2}EK.HI \)

\(= \dfrac{1}{2}EK\left( {HC + HI} \right)\)

\({S_{EHIK}} = \dfrac{1}{2}.3.\left( {3.4 + 1.7} \right) \)\(\,= \dfrac{1}{2} .3.5,1 = 7.65\,(c{m^2})\)

### 2.2. Exercise 2

On figure 160 (AC // BF), find a triangle whose area is equal to the area of quadrilateral ABCD.

**Solution guide**

Let \(O\) be the intersection of \(AF\) and \(BC\).

We have \(\Delta A{\rm{D}}F\) whose area is equal to the area of the quadrilateral \(ABCD\).

Indeed, since \( AC// BF\) should \({S_{ABC}} = {S_{AFC}}\) because the same base \(AC\) and the same height is the distance between the two lines parallel \(AC, BF.\)

We have: \({S_{ABC}} = {S_{AFC}}\) (proof above)

\( \Rightarrow {S_{ABO}} + {S_{ACO}} = {S_{CF{\rm{O}}}} + {S_{AC{\rm{O}}}}\)

\(\Rightarrow {S_{ABO}} = {S_{CFO}}\).

So \({S_{ADF}} = {S_{AOCD}} + {S_{CFO}} \)\(= {S_{AOCD}} + {S_{ABO}}= {S_{ABCD}}\ )

So \({S_{ADF}} = {S_{ABCD}}\)

### 2.3. Exercise 3

Given a triangle \(ABC.\) Let \(M, N\) be the respective midpoints of \(AC, BC.\) Prove that the area of a trapezoid \(ABNM\) is equal to \(\dfrac {3}{4}\) area of triangle \(ABC.\)

**Solution guide**

Drawing two medians \(AN, BM\) of \(∆ABC.\) We have:

\({S_{MNA}} =\dfrac{1}{2}{S_{ACN}}\)

(same height from top \(N\), bottom \(AM = \dfrac{1}{2}AC)\)

\({S_{ACN}} =\dfrac{1}{2}{S_{ABC}}\)

(same height from top \(A\), bottom \(CN = \dfrac{1}{2}BC)\)

\({S_{ABN}} =\dfrac{1}{2}{S_{ABC}}\)

(same height from top \(A\), bottom \(BN = \dfrac{1}{2}BC)\)

So \({S_{AMN}}= \dfrac{1}{2}{S_{ACN}} =\dfrac{1}{2}.\dfrac{1}{2}{S_{ABC}}\ )\(=\dfrac{1}{4}{S_{ABC}}\)

So \({S_{ABN}} + {S_{AMN}} = \dfrac{1}{2}{S_{ABC}} +\dfrac{1}{4}{S_{ABC}} \)\( =\dfrac{3}{4}S_{ABC}\)

That is, \({S_{ABNM}} = \dfrac{3}{4}{S_{ABC}}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Let \(ABC\) triangle with three altitudes \(AA’,\, BB’,\ CC’.\) Let \(H\) be the orthocenter of that triangle.

Prove that: \(\eqalign{{HA’}

**Verse 2: **Given triangle \(ABC\)

a) Calculate the ratio of the altitudes \(BB’\) and \(CC’\) from the vertices \(B\) and \(C\)

b) Why if \(AB < AC\) then \(BB' < CC' ?\)

**Question 3:** Through the center \(O\) of the square \(ABCD\) side \(a,\) draw a line \(l\) that intersects \(AB\) and \(CD\) at \(M\, respectively) ) and \(N.\) Knowing \(MN = b.\) Sum the distances from the vertices of the square to the line \(l\) by \(a\) and \(b\) ( \(a\) and \(b\) have the same unit of measure)

**Question 4:** The triangle \(ABC\) has two medians \(AM\) and \(BN\) perpendicular to each other. Calculate the area of the triangle in terms of \(AM\) and \(BN\)

**Question 5: **Given a parallelogram \(ABCD.\) Let \(K\) and \(L\) be two points on the side \(BC\) such that \(BK = KL = LC.\) Calculate the ratio of the areas of :

a) The triangles \(DAC\) and \(DCK\)

b) Triangle \(DAC\) and quadrilateral \(ADLB\)

c) The quadrilaterals \(ABKD\) and \(ABLD\)

### 3.2. Multiple choice exercises

**Question 1: **A 10-sided convex polygon has the number of diagonals:

A. 35

B. 30

C. 70

D. 27

**Verse 2: **Given an sided polygon, the number of diagonals of that polygon is:

A. 36

B. 27

C. 20

D. 18

**Question 3: **Let ABCD be a rectangle with AD = 8cm, AB = 9cm. The points M, N on the diagonal BD such that BM = MN = ND. Calculate the area of triangle CMN.

A. 12 \(cm^{2}\)

B. 24 \(cm^{2}\)

C. 36 \(cm^{2}\)

D. 6 \(cm^{2}\)

**Question 4: **A trapezoid has a small base of 9 cm , a height of 4 cm , and an area of 50 \(cm^{2}\). The big bottom is:

A. 25 cm

B. 18 cm

C. 16 cm

D. 15 cm

**Question 5:** A rhombus has two diagonals of length 15 cm and 20 cm. Calculate the length of the altitude of the rhombus.

A. 12 cm

B. 7, 5 cm

C. 15 cm

D. 24 cm

## 4. Conclusion

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

- Master the formula for calculating the area of simple polygons, especially the ways to calculate the area of triangles and trapezoids
- Know how to properly divide the polygon to find the area into simple polygons that can calculate the area
- Know how to make the necessary drawings and measurements.

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