## Math 9 Review chapter 3: Angle with a circle

## 1. Summary of theory

### 1.1. Angle in the center

An angle whose vertex coincides with the center of the circle is called a central angle.

The two sides of the central angle intersect the circle at two points, thus dividing the circle into two arcs.

For angles α ( 0 < α < 180°), the arc inside the angle is called the minor arc.

The arc outside the angle is called the major arc.

### 1.2. Inscribed angle

**Define**

An inscribed angle is an angle whose vertex lies on a circle and whose sides contain two chords of that circle.

The arc inside the angle is called the intercepted arc.

**Theorem**

In a circle, the measure of the inscribed angle is half the measure of the intercepted arc.

**Consequences**

In a circle:

Congruent inscribed angles intercept congruent arcs

Inscribed angles that intercept the same arc or intercept equal arcs are congruent

Inscribed angle (less than or equal to 900) whose measure is half the measure of the central angle intercepting the same arc

An inscribed angle that intercepts a semicircle is a right angle

### 1.3. Angle formed by tangent and chord

**Concept**

An angle formed by a tangent and a chord is an angle whose vertex lies on the circle, two sides of the angle include a ray that is tangent to the circle, the other ray containing the chord.

Angle \(\widehat{BAx}\) (or \(\widehat{BAy}\)) is the angle formed by the tangent and the chord.

**Theorem**

The measure of the angle formed by the tangent and the chord is half the measure of the intercepted arc.

### 1.4. Angle whose vertex is inside the circle

**THEOREM:** The measure of the angle whose vertex inside the circle is half the sum of the measures of the intercepted arcs.

Angle \(\widehat{BEC}\) is an angle whose vertex \(E\) lies inside the circle, so \(\widehat{BEC}=\frac{1}{2}\)(sđ\(\stackrel\) frown{BnC}\)+sd\(\stackrel\frown{AmD}\))

### 1.5. Angle whose vertex is outside the circle

**THEOREM: **The measure of the angle whose vertex outside the circle is half the difference of the intercepted arcs.

Angle \(\widehat{AED}\) has a vertex \(E\) outside the circle so \(\widehat{AED}=\frac{1}{2}\)(sđ\(\stackrel\frown{BnC) }-\)sđ\(\stackrel\frown{AmD}\))

### 1.6. The locus problem “Arc contains angle”

Given the given line segment \(AB\) and angle \(\alpha(0^0<\alpha<180^0)\) then the locus of points \(M\) satisfies \(\widehat{AMB}= \alpha\) are two arcs containing the angle \(\alpha\) built on the segment \(AB\)

**Attention:**

– The two arcs containing the angle \(\alpha\) above are two arcs that are symmetrical to each other through \(AB\)

– Two points \(A,B\) are considered to be in the locus

– In case \(\alpha=90^0\), the locus above is two semicircles of diameter \(AB\)

### 1.7. Inscribed quadrilateral

**Define:** A quadrilateral with four vertices lying on the same circle is called a cyclic quadrilateral (or inscribed quadrilateral).

For example, the quadrilateral \(ABCD\) has four vertices \(A,B,C,D\) lying on the same circle, so \(ABCD\) is called an inscribed quadrilateral.

**Theorem: **In an inscribed quadrilateral, the sum of the measures of two opposite angles is 180 .^{0}

\(ABCD\) is an inscribed quadrilateral, so we have \(\widehat{A}+\widehat{C}=\widehat{B}+\widehat{D}=180^0\)

**Island definition: **If a quadrilateral has the sum of the measures of two opposite angles equal to 1800, then the quadrilateral is inscribed in a circle

Specifically in the image above, if there is \(\widehat{A}+\widehat{C}=180^0\) or \(\widehat{B}+\widehat{D}=180^0\) then the quadrilateral is \(ABCD\) is inscribed in a circle.

### 1.8. Formula for calculating the length of a circle

The “circle length” denoted by C, aka the circumference of a circle is calculated by the formula \(C=2\pi R\) where \(R\) is the radius of the circle

### 1.9. Formula for calculating arc length

On a circle of radius \(R\), the length of \(l\) an entire arc \(n^0\) is calculated by the formula \(l=\frac{\pi Rn}{180}\)

### 1.10. Area of a circular fan

\(S=\frac{\pi R^2n}{360}\) or \(S=\frac{lR}{2}\)

## 2. Illustrated exercise

### 2.1. Exercise 1

Let a circle (O) with diameter AB equal to 12cm. A line through A intersects the circle (O) at M and intersects the tangent to the circle at B at N. Let I be the midpoint of MN. Knowing that AI=13cm, the length of line segment AM is:

**Solution guide**

Set \(AM=x, MI=NI=y\)

Then according to the problem we have \(x+y=13\) (1) (AI=13cm)

On the other hand, applying the trigonometric system for right triangle ABN with altitude BM we have \(AB^2=AM.AN\) or \(12^2=x(x+2y)\) (2)

From (1) we have \(y=13-x\) and substitute (2) we get: \(x(x+2(13-x))=12^2\Leftrightarrow -x^2+26x-144 =0\)

Easily solve the above equation with the solution formula, we get \(x=18 ,x=8\)

Combined with the condition, we deduce AM=8cm

### 2.2. Exercise 2

Given the simulated ceramic tile as shown, calculate the area to be colored, knowing the square brick has side 40cm

**Solution guide**

We have the area of the square brick \(S_{hv}=40.40=1600(cm^2)\)

The four uncolored corners are the area of a circle with a radius of 20cm.

So, the area of the uncolored part is: \(S_{ktm}=\pi r^2=20.20.\pi=400\pi(cm^2)\)

The area of the shaded part is: \(S=1600-400\pi\approx 344(cm^2)\)

### 2.3. Exercise 3

The graph above shows the student distribution of a school in a rural area, in which, blue shows primary school students, yellow shows middle school students, and red shows high school students.

know that the angle value \(\alpha=30^{\circ}\) and the sum of middle and high school students is only equal to \(\frac{1}{4}\) elementary school students. There are 720 children in school. Calculate the number of students in each level.

**Solution guide**

We see that the number of students in middle school and high school has a sum of \(\frac{1}{4}\) so the number of students of these two levels is \(\frac{720}{4}=180\) you.

The number of primary school students in this school is \(720-180=540\) children

Since the angle \(\alpha =30^{\circ}\Rightarrow\) the number of high school students is equal to \(\frac{30}{90}=\frac{1}{3}\) the number of middle school students and 3.

The number of high school students is: \(\frac{180}{3}=60\) children.

The number of primary school students is \(180-60=120\) children

### 2.4. Exercise 4

Given a circle (O). Draw two chords AC and BD equal and perpendicular to each other at point I (B is in minor arc AC). Prove that ABCD is an isosceles trapezoid.

**Solution guide**

Let the intersection of AC and BD be H

We have two wires AC and BD equal and perpendicular to each other, so:

sdAD=sdBC.

Infer that the two triangles HCD and HAB are right-angled at H

\(\widehat{BDC}=\widehat{ABD}\)

ABCD is a trapezoid.

**Note:** A trapezoid inscribed in a circle is always an isosceles trapezoid

### 2.5. Exercise 5

An equilateral triangle ABC is inscribed in circle (O). M is midpoint of minor arc BC, AM intersects BC at E. prove \(AB.BM=AM.BE\)

**Solution guide**

We have, M is the midpoint of arc BC so \(\widehat{BAM}=\widehat{CAM}\)

Otherwise, ABC is an equilateral triangle, so AM is the perpendicular bisector of BC.

And AM is the diameter of the circle (O).

\(\Rightarrow \widehat{MBA}=90^o\)

Easily prove \(\Delta ABM\sim \Delta BEM(gg)\)

So \(\frac{AB}{BE}=\frac{AM}{BM}\Leftrightarrow AB.BM=AM.BE\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Given a circle of diameter \(AB.\) Through \(A\) and \(B\) draw two tangents to that circle. Let \(M\) be a point on the circle. The lines \(AM\) and \(BM\) intersect the above tangents at \(B’\) and \(A’.\) respectively.

\(a)\) Prove that \({\rm{AA}}’.BB’ = A{B^2}\)

\(b)\) Prove that \(A'{A^2} = A’M.A’B\)

**Verse 2: **Given a hexagon \(ABCDEF.\) Prove that the diagonal \(BF\) divides \(AD\) into two lines by the ratio \(1: 3.\)

**Question 3:** Let \(ABC\) triangle with three acute angles. Construct a point \(M\) inside the triangle \(ABC\) such that \(\widehat {AMB} = \widehat {BMC} = \widehat {CMA}\)

**Question 4:** Two pulleys have center \(O, O’\) and radius \(R = 4a,\) \(R’ = a.\) Two common tangents \(MN\) and \(PQ\) intersect each other. at \(A\) at angle \(60^\circ.\) Find the length of the belt that passes through the two pulleys.

### 3.2. Multiple choice exercises

**Question 1: **Given a circle with diameter AB and a chord AC such that the measure of the arc AC is 60 degrees. The measure of angle OCB is:

A. \(\widehat{OCB}=15^o\)

B. \(\widehat{OCB}=20^o\)

C. \(\widehat{OCB}=25^o\)

D. \(\widehat{OCB}=30^o\)

**Verse 2: **Given a circle (O;R). Draw the string AB such that the measure of minor arc AB is equal to \(\frac{1}{2}\) the measure of major arc AB. The area of triangle AOB is:

A. \(\frac{R^2\sqrt{3}}{4}\)

B. \(\frac{R^2\sqrt{3}}{2}\)

C. \(R^2\sqrt{3}\)

D. \(2R^2\sqrt{3}\)

**Question 3: **Given the simulated ceramic tile shown in the figure, what is the shaded area? (I know the square brick has side 80cm)

A. \(\approx 1373.45(cm^2)\)

B. \(\approx 1375.55(cm^2)\)

C. \(\approx 1385.55(cm^2)\)

D. \(\approx 1345.65(cm^2)\)

**Question 4: **Given that the triangle OBC is equilateral, how many degrees are the two line segments OB and CD to each other \((\leq 90^{\circ})\)

A. \(65^o\)

B. \(75^o\)

C. \(105^o\)

D. \(115^o\)

## 4. Conclusion

Through this lesson, you will learn some key topics as follows:

- Know how to apply the knowledge to solve problems on calculating quantities related to circles and circles.
- State concepts, read, draw pictures, …. Apply to solve problems

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