Saturday, April 20, 2024

grade 9 Circles Review Sheet Geometry Honors

Circles Review Sheet Geometry Honors     

For the following questions, solve using your knowledge of circles. 







































1)1) In the diagram below of circle o  , chords AE and DC  intersect at point B, such that mAC = 36 and mDE =20. Find m angle ABC and m angle ABD

1.) m<ABC = 28, m<ABD = 152

2) In the accompanying diagram, chords AB and CD intersect at E. If mAD = 70 and mBC = 40, find m < CEA.

2.) m<CEA = 125

3) In the following diagram of circle O, AB and BC are chords and m < AOC = 94. What is m < ABC?

In the given diagram, angle AOC is an inscribed angle subtended by arc AC. By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of its intercepted arc.

So, m∠AOC = 94° implies that arc AC measures 94°.

Now, chord AB and chord BC intersect at point B. According to the Intersecting Chords Theorem, the measure of an angle formed by two intersecting chords is equal to half the sum of the measures of the arcs intercepted by the angle. Since angle ABC is formed by chords AB and BC, it intercepts arc AC.

Therefore, m∠ABC = (1/2) * m(arc AC) = (1/2) * 94° = 47°.

So, the measure of angle ABC is 47°.


4) In the accompanying diagram, BA is a diameter and mBC = 50. Find m < CВА.

5) In the accompanying diagram of circle 0, MABC = 260. What is m < ABC?



7) In the accompanying diagram of circle 0, mAB = 64 and m < AEB = 52. What is the mCD?

Answer: The measure of CD is 40.. 

Explanation: m∠ AEB= 1/2 (m AB+mCD)

52= 1/2 (64+m CD)

m CD=40

The degree of the inner angle of I a circle is equal to half of the sum of the degrees of the arc 

subtended by this angle and its opposite vertex angle


8.

8) In the accompanying diagram of circle O, the measure of RS = 64. What is m < RTS?

9) If m DE = 121 and m BC = 83, find m ∠A.


 Answer: m∠ A=19°. 

Explanation: ∵ m PE=121 m BC=83

∴ m∠ DBE= 1/2 m DE= 121/2 

m∠ BEC= 1/2 m BC= 83/2 

∴ m∠ A=m∠ DBE-m∠ BEC= 121/2 - 83/2 =19

Tie circumscribed Angle of an arc is equal to half the central Angle

10) In the diagram below, PA is tangent to circle O, and AB is a chord. If mACB = 300, find the measure of ∠BAP.


13) In circle O, BC is a chord and CD is a tangent.<BCD = 75° Find the value of x.

In circle O, if BC is a chord and CD is a tangent, then angle BCD is an inscribed angle and angle BCD = 75°. 

Now, let's use the property that the measure of an inscribed angle is equal to half the measure of its intercepted arc.

So, angle BCD = (1/2) * arc BC.

Given that CD is a tangent to the circle at point C, angle BCD also equals angle OCD (angle formed by the tangent and the chord).

Therefore, angle OCD = 75°.

Now, in a circle, the angle formed by a tangent and a chord is equal to half the measure of the intercepted arc that's farther from the tangent.

So, angle OCD = (1/2) * arc BC.

Since arc BC is the full circle except for the arc subtended by the chord CD, which is x, we have:

arc BC = 360° - x.

(1/2) * (360° - x) = 75°

180° - x/2 = 75°

-x/2 = 75° - 180°

-x/2 = -105°

x = 210°

14) In the accompanying diagram, tangent PA and secant PBC are drawn to circle O from point P. If mAC = 110 and m < P = 20, find mAB.


15) In the accompanying diagram, ∆ABC is inscribed in circle O and AB is a diameter.

What is the number of degrees in m∠C?

1if a triangle is inscribed in a circle and one of its sides is a diameter of the circle, then the angle opposite to that side is a right angle 2Since AB is a diameter, mC is the angle opposite to AB3Therefore∠=90∘ mC=90.So, the number of degrees in mC is 90∘

16) Find the value of x, y, and z.



17) A and Care points on a circle with center 0.

(a) Draw a point B on the circle so that AB is a diameter. Then, draw the angle ABC.

(b) What angle in your diagram is an inscribed angle?

(c) What angle in your diagram is a central angle?

(d) What is the intercepted arc of < ABC?

(e) What is the intercepted arc of < AOC?

Angle ACB is an inscribed angle

Angle AOC is a central angle

The intercepted arc of angle ABC is arc AC

The intercepted arc of angle AOC is arc ACB


18) Toby says that ABEA is a right triangle because < BEA = 90°. Is he correct? Justify your answer.

D

X=60

21) In the accompanying diagram, tangent PB and secant PCA are drawn to circle O and mBC: MAC: MAB = 3:4:5

Find:

P

(a) mBC=

(b) m < PAB=

(c) m < APB=

 the total degrees in a circle: 360°

mBC: 3/12 * 360° = 90°

 m < PAB: 1/2 * 90° = 45°

 m < APB: 180° - 90° - 45° = 45°




23) In the accompanying diagram of circle O, EA is a tangent, EBC is a secant, Dis a midpoint of AC, mAD = 94°, and mAB = 86°. Find:

A

E

(a) mBC=

(b) m < E=

(c) m < ABD=

(d) m < AFB=

0

D

F

B

(e) m < EBA=_

C


Remember: There will be review questions! This topic is VERY IMPORTANT. 

You need to make sure you study for this exam!

Answer Key: 

Circles Review Sheet: 

1.) m<ABC = 28, m<ABD = 152 2.) m<CEA = 125 3.) m<ABC = 47 4.) m<CBA = 65 5.) m<ABC = 50 6.)  

7.) 8.) m<RTS = 32 9.) m<A = 19 10.) m<BAP = 30 11.) m<DBC = 72 12.) (4) 13.) x = 210 14.) 15.) m<C = 90 16.) y = 84, x = 168, z = 192 17.) SKIP 

18.) No he is not correct because m<BEA = 95. 19.) x = 60 20.) x = 60 21.) (a) 90 (b) 45 (c) 30 22.) (a) 100 (b) 40 (c) 50 (d) 40 (e) 50 23.) (a) 86 (b) 51 (c) 47 (d) 90 (e) 43 Review Chapters  

1.) y + 1 = -2(x – 3) 2.) x = 5; m<FHB = 105; m<BHE = 75 3.) correct construction 4.) right scalene 5.) (0,5), (1,6) 6.) (5,3) 7.) TS = 3; SU = 2.7; p = 9.7 8.) correct proof 9.) (a) x = 12 (b) x = 4 10.) r = 6


Wednesday, April 17, 2024

formula

☑️ Very Important formula

1. (α+в)²= α²+2αв+в²
2. (α+в)²= (α-в)²+4αв
3. (α-в)²= α²-2αв+в²        
4. (α-в)²= (α+в)²-4αв
5. α² + в²= (α+в)² - 2αв.
6. α² + в²= (α-в)² + 2αв.
7. α²-в² =(α + в)(α - в)
8. 2(α² + в²) = (α+ в)² + (α - в)²
9. 4αв = (α + в)² -(α-в)²
10. αв ={(α+в)/2}²-{(α-в)/2}²
11. (α + в + ¢)² = α² + в² + ¢² + 2(αв + в¢ + ¢α)
12. (α + в)³ = α³ + 3α²в + 3αв² + в³
13. (α + в)³ = α³ + в³ + 3αв(α + в)
14. (α-в)³=α³-3α²в+3αв²-в³
15. α³ + в³ = (α + в) (α² -αв + в²)
16. α³ + в³ = (α+ в)³ -3αв(α+ в)
17. α³ -в³ = (α -в) (α² + αв + в²)
18. α³ -в³ = (α-в)³ + 3αв(α-в)

ѕιη0° =0
ѕιη30° = 1/2
ѕιη45° = 1/√2
ѕιη60° = √3/2
ѕιη90° = 1

➡️ ¢σѕ ιѕ σρρσѕιтє σƒ ѕιη

тαη0° = 0
тαη30° = 1/√3
тαη45° = 1
тαη60° = √3
тαη90° = ∞

➡️ ¢σт ιѕ σρρσѕιтє σƒ тαη

ѕє¢0° = 1
ѕє¢30° = 2/√3
ѕє¢45° = √2
ѕє¢60° = 2
ѕє¢90° = ∞

➡️ ¢σѕє¢ ιѕ σρρσѕιтє σƒ ѕє¢

1. ѕιη2α = 2ѕιηα¢σѕα
2. ¢σѕ2α = ¢σѕ²α − ѕιη²α
3. ¢σѕ2α = 2¢σѕ²α − 1
4. ¢σѕ2α = 1 − ѕιη²α
5. 2ѕιη²α = 1 − ¢σѕ2α
6. 1 + ѕιη2α = (ѕιηα + ¢σѕα)²
7. 1 − ѕιη2α = (ѕιηα − ¢σѕα)²
8. тαη2α = 2тαηα / (1 − тαη²α)
9. ѕιη2α = 2тαηα / (1 + тαη²α)
10. ¢σѕ2α = (1 − тαη²α) / (1 + тαη²α)
11. 4ѕιη³α = 3ѕιηα − ѕιη3α
12. 4¢σѕ³α = 3¢σѕα + ¢σѕ3α

» ѕιη²Θ+¢σѕ²Θ=1
» ѕє¢²Θ-тαη²Θ=1
» ¢σѕє¢²Θ-¢σт²Θ=1
» ѕιηΘ=1/¢σѕє¢Θ
» ¢σѕє¢Θ=1/ѕιηΘ
» ¢σѕΘ=1/ѕє¢Θ
» ѕє¢Θ=1/¢σѕΘ
» тαηΘ=1/¢σтΘ
» ¢σтΘ=1/тαηΘ
» тαηΘ=ѕιηΘ/¢σѕΘ

CH2 Lines and Angles WORKSHEET CLASS 6

Subject: Maths                        WORKSHEET   - CH-2 LINES AND ANGLES                  Class-VI             Q1. How many lines ...