1 / 20

# 9 th Grade Geometry - PowerPoint PPT Presentation

9 th Grade Geometry. Lesson 10-5: Tangents. Main Idea . Use properties of tangents! Solve problems involving circumscribed polygons. New Vocabulary. Tangent Any line that touches a curve in exactly one place Point of Tangency The point where the curve and the line meet. Theorem 10.9.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Lesson 10-5: Tangents

Main Idea
• Use properties of tangents!
• Solve problems involving circumscribed polygons

New Vocabulary

• Tangent
• Any line that touches a curve in exactly one place
• Point of Tangency
• The point where the curve and the line meet
Theorem 10.9
• If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency.
• Example: If RT is a tangent, OR RT

T

R

O

Example: Find Lengths

ALGEBRARS is tangent to Q at point R. Find y.

S

20

16

Q

P

R

y

Because the radius is perpendicular to the tangent at the point of tangency, QRSR. This makes SRQ a right angle and SRQ a right triangle. Use the Pythagorean Theorem to find QR, which is one-half the length y.

Example: Find Lengths

(SR)2 + (QR)2= (SQ)2Pythagorean Theorem

162 + (QR)2 = 202 SR = 16, SQ = 20

256 + (QR)2 = 400 Simplify

(QR)2 = 144 Subtract 256 from each side

QR = +12 Take the square root of each side

Because y is the length of the diameter, ignore the negative result. Thus, y is twice QR or y = 2(12) = 24

Example

CD is a tangent to B at point D. Find a.

• 15
• 20
• 10
• 5

C

a

B

A

D

40

25

Theorem 10.10
• If a line is perpendicular to a radius of a circle at its endpoint on the circle, then the line is tangent to the circle.
• Example: If OR RT, RT is a tangent.

R

T

O

Example: Identify Tangents

Determine whether BC is tangent to A

C

7

9

7

A

B

7

First determine whether ABC is a right triangle by using the converse of the Pythagorean Theorem

Example: Identify Tangents

(AB)2 + (BC)2 = (AC)2 Converse of the Pythagorean Theorem

72 + 92 = 142AB = 7, BC = 9, AC = 14

130 ≠ 196 Simplify

Because the converse of the Pythagorean Theorem did not prove true in this case, ABC is not a right triangle

Answer:So, BC is not a tangent to A.

?

?

Example: Identify Tangents

Determine whether WE is tangent to D.

E

16

24

10

D

W

10

First Determine whether EWD is a right triangle by using the converse of the Pythagorean Theorem

Example: Identify Tangents

(DW)2 + (EW)2 = (DE)2 Converse of the Pythagorean Theorem

102 +242 = 262DW = 10, EW = 24, DE = 26

676 = 676 Simplify.

Because the converse of the Pythagorean Theorem is true, EWD is a right triangle and EWD is a right angle.

Answer:Thus, DW WE, making WE a tangent to D.

?

?

Quick Review

Determine whether ED is a tangent to Q.

A. Yes

B. No

C. Cannot be

determined

D

√549

18

Q

E

15

Quick Review

Determine whether XW is a tangent to V.

A. Yes

B. No

C. Cannot be

determined

W

10

17

10

V

X

10

Theorem 10.11
• If two segments from the same exterior point are tangent to a circle, then they are congruent
• Example: AB ≈ AC

B

C

A

Example: Congruent Tangents

ALGEBRA Find x. Assume that segments that appear tangent to circles are tangent.

ED and FD are drawn from the same exterior point and are tangent to S, so ED ≈ FD. DG and DH are drawn from the same exterior point and are tangent to T, so DG ≈ DH

H

x + 4

F

y

D

G

y - 5

E

10

Example: Congruent Tangents

ED = FD Definition of congruent segments

10 = y Substitution

Use the value of y to find x.

DG = DH Definition of congruent segments

10 + (y - 5) = y + (x + 4) Substitution

10 + (10 - 5) = 10 + (x + 4) y = 10

15 = 14 + x Simplify.

1 = x Subtract 14 from each side

Quick Review

Find a. Assume that segments that appear tangent to circles are tangent.

• 6
• 4
• 30
• -6

30

N

b

6 – 4a

R

A

Example: Triangles Circumscribed About a Circle

Triangle HJK is circumscribed about G. Find the perimeter of HJK if NK = JL +29

H

N

18

K

L

M

16

J

Example: Triangles Circumscribed About a Circle

Use Theorem 10.11 to determine the equal measures:

JM = JL = 16, JH = HN = 18, and NK = MK

We are given that NK = JL + 29, so NK = 16 + 29 or 45

Then MK = 45

P = JM + MK + HN + NK + JL + LH Definition of

perimeter

= 16 + 45 + 18 + 45 + 16 + 18 or 158 Substitution

Answer:The perimeter of HJK is 158 units.

Quick Review

Triangle NOT is circumscribed about M. Find the Perimeter of NOT if CT = NC – 28.

• 86
• 180
• 172
• 162

N

52

C

T

A

B

10

O