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

• 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

• 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

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.

(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

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

• 15

• 20

• 10

• 5

C

a

B

A

D

40

25

• 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

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

(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.

?

?

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

(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.

?

?

Determine whether ED is a tangent to Q.

A. Yes

B. No

C. Cannot be

determined

D

√549

18

Q

E

15

Determine whether XW is a tangent to V.

A. Yes

B. No

C. Cannot be

determined

W

10

17

10

V

X

10

• If two segments from the same exterior point are tangent to a circle, then they are congruent

• Example: AB ≈ AC

B

C

A

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

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

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

• 6

• 4

• 30

• -6

30

N

b

6 – 4a

R

A

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

H

N

18

K

L

M

16

J

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.

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