contestada

Part A
Construct a circle of any radius, and draw a chord on it. Then construct the radius of the circle that bisects the chord. Measure the angle between the chord and the radius. What can you conclude about the intersection of a chord and the radius that bisects it? Take a screenshot of your construction, save it, and insert the image below your answer.

Part B
Write a paragraph proof of your conclusion in part A. To begin your proof, draw radii OA and OC.

Part C
In this part of the activity, you will investigate the converse of the theorem stated in part A. To get started, reopen GeoGebra.

Draw a circle of any radius, and draw a chord on it. Construct the radius of the circle that is perpendicular to the chord. Measure the line segments into which the radius divides the chords. How are the line segments related? What can you conclude about the intersection of a chord and a radius that is perpendicular to it? Take a screenshot of your construction, save it, and insert the image below your answer.

Part D
Write a paragraph proof of your conclusion in part C. To begin your proof, draw radii OA and OC.

Respuesta :

Answer:

All answers are bold and links are real photos  

BTW THE PHOTOS ARE IN ORDER OF THE ANSWERS

Brainlyist??

Step-by-step explanation:

Task 1

Part A

Tangent Line

Part B

PHOTO WITH LARGE CIRCLE AND LINE OUT SIDE BUT CONNECTED

Part C

2034.7

PHOTO WITH EQUATION

We calculate the point from the center of the earth to the satellite and because we have the radius we can subtract it getting the distance from the satellite to the earth.

Task 2

Gears

Part A

Approximately 10.47 inches

Part B

300°

Part C

Approximately 125.66 inches

Part D

The distance traveled increases by approximately 62.83 inches

Task 3

Circle Theorems

Part A

PHOTO WITH LINE THROUGH CIRCLE AND ONE 90 DEGREE ANGLE

Part B

Prove OC is perpendicular to AB

We know that ΔAEC≅AED by SSS criteria (SSS criteria is when all three sides of a triangle is equal making them congruent)

To prove this that they apply to the SSS criteria: m∠ADO=m∠BDO m∠ADO=90°=m∠BDO

Part C

A radius is perpendicular to a chord when the chord is equally divided in half by the radius.

PHOTO OF LINE THROUGH CIRCLE AND LINE DISTANCES

Part D

Prove OC is perpendicular to AB

AO=BO because AO and BO are both radii

OD≅OD by reflexive theorem

ΔADO≅ΔBDO because they are both right triangle with the hypotenuse and a side

AB≅BC means that OC is perpendicular to AB

Ver imagen blakeantarbowman
Ver imagen blakeantarbowman
Ver imagen blakeantarbowman
Ver imagen blakeantarbowman
anuksha0456

Disclaimer: The images to part A and part C is the same.

Two theorems we are proving are:

  1. Line drawn from the center of the circle to bisect a chord is perpendicular to it.
  2. Perpendicular drawn from the center of the circle to a chord, bisects it.

How do we solve the given question?

Part A: The image is attached

Part B:

Let there be a circle with the center at O. Let AC be a chord. Let OE be a radius of the chord such that OE bisects AC at B, that is, AB = BC.

To prove: A line drawn from the center of the circle to bisect a chord is perpendicular to it.

Construction: We join OA and OB.

Proof:

In ΔOAB and ΔOCB,

OA = OC (Radius of the same circle)

AB = BC (Given, at B, the chord is bisected)

OB = OB (common side)

∴ ΔOAB ≅ ΔOCB [SSS axiom]

∴ ∠OBA = ∠OBC [Common parts of Congruent triangles]

Let ∠OBA = ∠OBC = x.

Since AC is a straight line,

∠OBA + ∠OBC = 180° [Linear pair]

or, x + x = 180°

or, 2x = 180°

or, x = 90°.

∴ ∠OBA = ∠OBC = 90°.

So, we can say that OB ⊥ AC, hence the theorem is proved.

Part C: The image is attached

Part D:

Let there be a circle with the center at O. Let AC be a chord. Let OE be a radius of the chord such that OE is perpendicular to AC at B, that is, AB ⊥ BC.

To prove: A line drawn from the center of the circle perpendicular to the chord bisects it.

Construction: We join OA and OB.

Proof:

In ΔOAB and ΔOCB,

OA = OC (Radius of the same circle)

∠OBA = ∠OBC = 90° (∵ OB ⊥ AC)

OB = OB (common side)

∴ ΔOAB ≅ ΔOCB [RHS axiom]

∴ AB = AC [Common parts of Congruent triangles]

So, we can say that OE perpendicular to the chord AC bisects it, hence the theorem is proved.

Learn more about chord properties at

https://brainly.com/question/13950364

#SPJ2

Ver imagen anuksha0456