Geometry Sections 10.1-10.3

Circle P, H is a tangent point.

PH = 22, HT = 20

Solve for length of PT.

• 26.8

• 31.3

• 29.7

• 34.1

Tags

CCSS.HSG.C.A.2

Circle H, Tangent point T

BT = 61, BH = 72

Solve for PH

• 35.7

• 38.2

• 32.6

• 31.5

Tags

CCSS.HSG.C.A.2

P and H are tangent points

PT = (x² + 18)

TH = 112

Solve for x.

• 9.7

• 11.4

• 94

• 130

Tags

CCSS.HSG.C.A.2

Circle P

mArc TM = 47°

Solve for m<RPT

• 143°

• 145°

• 133°

• 123°

Tags

CCSS.HSG.C.A.2

Circle P

m<QPR = 111°

m<TPM = 88°

mArc RQT = ?°

• 268°

• 92°

• 198°

• 312°

Tags

CCSS.HSG.C.B.5

Circle P

Arc WR = Arc MQ

m<RPQ = 127°

Solve for mArc WQM

• 233°

• 243°

• 183°

• 213°

Tags

CCSS.HSG.C.A.2

Circle P

mArc AC = 164°

Solve for mArc BAC

• 312°

• 196°

• 98°

• 262°

Tags

CCSS.HSG.C.A.2

Circle P

mArc AWD = 310°

Solve for mArc AB

• 25°

• 50°

• 335°

• 75°

Tags

CCSS.HSG.C.B.5

Circle P

Chord AD = Chord NW

mArc AB = 3x°, mArc NW = (5x + 12)°

Solve for x

• 6

• 12

• -6

• -12

Tags

CCSS.HSG.C.B.5

Circle P

PK = PT, AD = 44 cm, m< W = 36°

Solve for PW

• 27.2

• 29.1

• 31.4

• 22

Tags

CCSS.HSG.C.A.2