Luyện tập môn toán 9 đường kính và dây của đường tròn – Kiến thức và bài tập

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Diameter and cord of a circle is an extremely important part of the 9th grade math program and the 10th exam. Here, we will introduce you to the knowledge and practice exercises for 9 diameter and 10th grade math. The wire of the circle with detailed and clear explanations according to the textbook program framework.

Please refer!

Practice math 9 problems of diameters and strings of a circle

1. Compare the length of the diameter and wire

Of the strings of a circle, the largest wire is the diameter.

Example: Let AB be any chord of the circle (O; R). Prove that AB 2R

+ Case 1: AB is the diameter

⇒ AB = 2R

math 9 diameters and chords of a circle

+ Case 2: AB is not a diameter

Considering triangle AOB, applying the triangle inequality we have:

AB < AO + OB = R + R = 2R

So we always have AB 2R

2. Perpendicular relationship between diameter and wire

In a circle, a diameter perpendicular to a string passes through its midpoint.

In a circle, the diameter passing through the midpoint of a string but not passing through the center is perpendicular to the chord.

Example: Given the following figure, calculate the length of string AB when OA = 13cm; AM = MB; OM = 5cm.

Instruct:

Applying the theorem: “In a circle, the diameter through the midpoint of a string that does not pass through the center is perpendicular to the string.”

Then we have: OM ⊥ AB.

Applying the Pythagorean theorem, we have:

⇒ AB = 2.AM = 2.12 = 24 (cm)

Solution for 9 problems of diameter and string of a circle

1. Lesson 10 page 104

Given triangle ABC, altitudes BD and CE. Prove that:

a) Four points B, E, D, C lie on the same circle.

b) DE < BC.

The answer:

a) Let M be the mid point of BC.

=> ME = MB = MC = MD

Therefore, four points B, E, D, C belong to the circle with center M. (dpcm)

b) In the circle with center M above, we have DE is the string, BC is the diameter, so DE < BC.

2. Lesson 11 page 104

Given a circle (O) of diameter AB, a chord CD that does not intersect diameter AB, Let H and K be the feet of the perpendiculars drawn from A and B to CD, respectively. Prove that CH = DK.

Hint: Draw OM perpendicular to CD.

The answer:

Draw OM ⊥ CD.

Since AH // BK (the same perpendicular to HK), quadrilateral AHKB is a trapezoid.

Trapezoid AHKB has:

AO = OB (radius).

OM // AH // BK (same perpendicular HK)

=> OM is the median of the trapezoid.

=> MH = MK (1)

Since OM ⊥ CD, MC = MD (2)

From (1) and (2) it follows that CH = DK. (dpcm)

Hints to answer sbt

1. Lesson 15 page 158

Given triangle ABC, altitudes BH and CK. Prove:

a. Four points B, C, H, K are on the same circle

b. HK < BC

The answer:

a. Let M be the mid point of BC.

Triangle BCH is right-angled at H and HM is the median, so:

HM = (1/2).BC (property of right triangle)

Triangle BCK is right-angled at K and KM is the median, so:

KM = (1/2).BC (property of right triangle)

Infer: MB = MC = MH = MK

So four points B, C, H, K lie on a circle with center M and radius equal to (1/2).BC.

b. In a circle with center M we have KH is the chord that does not pass through the center, BC is the diameter, so: KH < BC

2. Lesson 16 page 159

Quadrilateral ABCD has

a. Prove that the four points A, B, C, D are on the same circle

b. Compare the lengths AC and BD. If AC = BD then what is the quadrilateral ABCD?

The answer:

a. Let M be the midpoint of AC .

In triangle ABC, right angled at B, BM is the median, so:

BM = (1/2).AC (property of right triangle)

Triangle ACD is right-angled at D and DM is the median, so:

DM = (1/2).AC (property of right triangle)

Infer: MA = MB = MC = MD

So four points A, B, C, D lie on a circle with center M and radius equal to (1/2).AC.

b. In a circle with center M, BD is the chord that does not pass through the center, and AC is the diameter, so: BD < AC

AC = BD if and only if BD is the diameter. Then quadrilateral ABCD is a rectangle.

3. Lesson 17 page 159

Given a semicircle with center O, diameter AB, and a string EF that does not intersect the diameter. Let I and K be the feet of the perpendiculars drawn from A and B to EF, respectively. Prove that IE = KF.

The answer:

We have: AI EF (gt)

BK EF (gt)

Inferred: AI // BK

So quadrilateral ABKI is a trapezoid

OH EF

Infer: OH // AI // BK

We have: OA = OB (= R)

Derived: HI = HK

Or: HE + EI = HF + FK (1)

Again: HE = HF (diameter of chord) (2)

From (1) and (2) infer: IE = KF

4 – Lesson 18 page 159

Given a circle (O) with radius OA = 3cm. The chord BC of the circle is perpendicular to OA at the midpoint of OA. Calculate the length BC.

The answer:

Let I be the midpoint of OA

Infer: IO = IA = (1/2).OA = 3/2

We have: BC OA (gt)

Infer: angle (OIB) = 90o

Applying the Pythagorean theorem to right triangle OBI we have: OB2 = BI2 + IO2

Infer: BI2 = OB2 – IO2

We have: BI = CI (diameter of chord)

5. Lesson 19 page 159

Lesson 19 page 159 Math Workbook 9 Volume 1: Given a circle (O), diameter AD = 2R. Draw an arc D with radius R, which intersects the circle (O) at B and C.

a. What is the quadrilateral OBDC? Why?

b. Calculate the measure of angles CBD, CBO, OBA

c. Prove that triangle ABC is an equilateral triangle.

The answer:

a. We have:

OB = OC = R (because B, C lie on (O; R))

DB = DC = R (because B, C are on (D; R))

Infer: OB = OC = DB = DC

So quadrilateral OBDC is a rhombus

b. We have: OB = OC = BD = R

6 – Lesson 20 page 159

a. Given a semicircle with center O, diameter AB, chord CD. The lines perpendicular to CD at C and D intersect AB at M and N respectively. Prove that AM = BN

b. Given a semicircle with center O and diameter AB. On AB take the points M, N such that AM = BN. Through M and N draw parallel lines, which intersect the semicircle at C and D respectively. Prove that MC and ND are perpendicular to CD.

The answer:

a. We have: CM CD

DN CD

Infer: CM // DN

The OI CD

Infer: OI // CM // DN

We have: IC = ID (diameter of chord)

Infer: OM = ON (1)

Which: AM + OM = ON + BN (= R) (2)

From (1) and (2) infer: AM = BN

b. We have: MC // ND (gt)

The quadrilateral MCDN is a trapezoid

Yes again: OM + AM = ON + BN (= R)

Where AM = BN (gt)

Infer: OM = ON

Draw OI ⊥ CD (3)

Derived: IC = ID (chord diameter)

Then OI is the median of trapezoid MCDN

Infer: OI // MC // ND (4)

From (3) and (4) infer: MC ⊥ CD, ND ⊥ CD.

7. Lesson 21 page 159

Given a circle with center O and diameter AB. The string CD cuts the diameter AB at I. Let H and K, respectively, be the feet of the perpendiculars drawn from A and B to CD. Prove that CH = DK

The answer:

Draw OM CD cut AD at N

We have: MC = MD (diameter of chord)

Or MH + CH = MK + KD (1)

We have: OM // BK (same perpendicular to CD)

Hay: MN // BK

Which: OA = OB (= R)

Infer: NA = NK (property of the median of the triangle)

Again: OM // AH (same perpendicular to CD)

Hay: MN // AH

Which: NA = NK (proved above)

Infer: MH = MK (property of the median of the triangle) (2)

From (1) and (2) infer: CH = DK

8. Lesson 22 page 159

Let the circle (O; R) and point M lie inside the circle.

a. State how to construct a string AB with M as its midpoint

b. Calculate the length of AB in question a knowing that R = 5cm, OM = 1.4cm

The answer:

a. * How to build

– Construct the OM . segment

– Through M construct a line perpendicular to OM intersecting O at A and B.

Connecting A and B we get the wire to build

*Prove

We have: OM ⊥ AB MA = MB

b. Applying Pythagorean theorem to right triangle OMB we have:

OB2 = OM2 + MB2

Inference: MB2 = OB2 – OM2 = 52 – 1.42 = 25 – 1.96 = 23.04

MB = 4.8 (cm)

So AB = 2.MB = 2.4.8 = 9.6 (cm)

9. Lesson 23 page 159

Given a circle (O), point A is inside the circle, and point B is outside the circle so that the midpoint I of AB is inside the circle. Draw cord CD perpendicular to OI at I. What is ACBD? Why?

The answer:

We have: OI ⊥ CD (gt)

Derived: IC = ID (chord diameter)

Which: IA = IB (gt)

Quadrilateral ACBD has two diagonals that intersect at the midpoint of each line, so it is a parallelogram.

10. Lesson 2.1, page 159

The length of the side of an equilateral triangle inscribed in the circle (O; R) is

A. R/2; B. (R√3)/2;

C. R√3; D. Another answer.

Please select the correct option.

The answer:

Choose the answer C

11. Lesson 2.2 page 159

Given a circle (O; 2cm). Draw two strings AB and CD perpendicular to each other. Find the maximum area of ​​quadrilateral ABCD.

The answer:

We have AB ≤ 4cm, CD ≤ 4cm. Since AB ⊥ CD, S.ACBD = ½.AB.CD ≤ 1/2.4.4 = 8 (cm^2)

The maximum value of S.ACBD is 8cm2 when AB and CD are both diameters of the circle.

12. Lesson 2.3 page 159

Given a circle (O;R), string AB has a different diameter. Draw on both sides of AB the strings AC, AD. Let H and K, respectively, be the feet of the perpendiculars drawn from B to AC and AD. Prove that:

a) Four points A, H, B, K belong to the same circle;

b) HK < 2R.

The answer:

a) Four points A, H, B, K are on the same circle with diameter AB.

b) We have HK ≤ AB ≤ 2R.

Above is the math practice of 9 problems of diameter and string of a circle for students’ reference. In it, we introduced the comparison of lengths and the perpendicular relationship of diameters and strings, and gave answers to pages 158-159 SBT. Hopefully, these knowledge will help students review and master the basic knowledge of Math and support students in exams in the 9th grade year.

Please follow other articles of Kien Guru to update the lessons of other subjects too!

Good luck with your studies!

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