Sec 3 E-Math Circle properties worksheet with answers

A free Sec 3 E-Math worksheet on circle properties: 9 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 E-Math worksheet: Circle properties

Total: 30 marksSuggested time: 45 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

In the diagram,

A

B

and

C

lie on the circle with centre

O

. Find

x

Answer°[1]

In the diagram,

A

B

C

and

D

lie on the circle. Find

x

Answer°[1]

In the diagram,

PA

and

PB

are tangents to the circle with centre

O

, and

OP

is joined. Find

x

Answer°[2]

In the diagram,

AB

is a chord of the circle with centre

O

, and

OM

is perpendicular to

AB

. Find

x

Answer°[2]

In the diagram,

AT

is a tangent to the circle with centre

O

A

, and

AB

is a chord. Find

x

Answer°[3]

In the diagram,

AB

is a diameter of the circle with centre

O

, and

C

lies on the circle. Find

x

, correct to 3 significant figures.

Answercm[3]

Section B

In the diagram,

PA

and

PB

are tangents to the circle with centre

O

A

and

B

C

lies on the major arc

AB

and angle

APB = 56^\circ

(a)Find angle

AOB

Answer (a)°[2]

(b)Find angle

ACB

Answer (b)°[1]

(c)Find angle

OAB

Answer (c)°[2]

In the diagram,

A

B

C

and

D

lie on the circle with centre

O

, and

AC

is a diameter. Angle

BAC = 28^\circ

and angle

CAD = 43^\circ

(a)Find angle

BCA

Answer (a)°[2]

(b)Find angle

BCD

Answer (b)°[2]

(c)Find angle

BDC

Answer (c)°[1]

In the diagram,

AB

and

CD

are parallel chords of the circle with centre

O

, on opposite sides of

O

AB = 24

cm and

CD = 18

cm. The line through

O

perpendicular to the chords meets

AB

M

and

CD

N

MN = 21

cm and

OM = x

cm.

(a)Write down the length of

AM

, giving a reason.

Answer (a)cm[1]

(b)Show that

x = 9

[3]

(c)Find the radius of the circle.

Answer (c)cm[2]

(d)Find angle

AOB

, correct to 1 decimal place.

Answer (d)°[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

x = 26

°[1]

\mathstrut \angle ACB = \dfrac{52^\circ}{2}

(∠ at centre = 2 ∠ at circumference)

x = 26

Question 2

Answer:

x = 67

°[1]

\angle ADB = \angle ACB

(∠s in same segment)

x = 67

Question 3

Answer:

x = 56

°[2]

\angle OPA = \angle OPB

(tangents from ext. pt.)

\angle OPA = 34^\circ

\angle OAP = 90^\circ

(tan. ⊥ rad.)

x = 180^\circ - 90^\circ - 34^\circ

(∠ sum of △)

x = 56

Question 4

Answer:

x = 136

°[2]

\angle OBA = 22^\circ

(base ∠s of isos. △)

\mathstrut x = 180^\circ - 2 \times 22^\circ

(∠ sum of △)

x = 136

Question 5

Answer:

x = 86

°[3]

\angle OAT = 90^\circ

(tan. ⊥ rad.)

\mathstrut \angle OAB = 90^\circ - 43^\circ

\angle OAB = 47^\circ

\angle OBA = 47^\circ

(base ∠s of isos. △)

\mathstrut x = 180^\circ - 2 \times 47^\circ

(∠ sum of △)

x = 86

Question 6

Answer:

\phantom{x} = 16.6

cm (3 s.f.)[3]

\angle ACB = 90^\circ

(∠ in semicircle)

AB^2 = AC^2 + BC^2

\mathstrut x^2 = 9^2 + 14^2

x^2 = 277

x = \sqrt{277}

x = 16.6433\ldots

\phantom{x} = 16.6

cm (3 s.f.)

Question 7

(a)Answer:

124^\circ

°[2]

\angle OAP = 90^\circ

(tangent ⊥ radius)

\angle OBP = 90^\circ

(tangent ⊥ radius)

\angle AOB = 360^\circ - 90^\circ - 90^\circ - 56^\circ

\phantom{\angle AOB} = 124^\circ

(∠ sum of quad.)

(b)Answer:

62^\circ

°[1]

\angle ACB = \dfrac{124^\circ}{2}

\phantom{\angle ACB} = 62^\circ

(∠ at centre = 2 ∠ at circumference)

(c)Answer:

28^\circ

°[2]

OA = OB

(radii)

\angle OAB = \dfrac{180^\circ - 124^\circ}{2}

\phantom{\angle OAB} = 28^\circ

(base ∠s of isos. △)

Question 8

(a)Answer:

62^\circ

°[2]

\angle ABC = 90^\circ

(rt. ∠ in semicircle)

\angle BCA = 180^\circ - 90^\circ - 28^\circ

\phantom{\angle BCA} = 62^\circ

(∠ sum of △)

(b)Answer:

109^\circ

°[2]

\angle BAD = 28^\circ + 43^\circ

\phantom{\angle BAD} = 71^\circ

\angle BCD = 180^\circ - 71^\circ

\phantom{\angle BCD} = 109^\circ

(∠s in opp. segments)

(c)Answer:

28^\circ

°[1]

\angle BDC = \angle BAC

\phantom{\angle BDC} = 28^\circ

(∠s in same segment)

Question 9

(a)Answer:

12

cm[1]

AM = \dfrac{24}{2}

\phantom{AM} = 12

cm (⊥ from centre bisects chord)

(b)[3]

CN = 9

(⊥ from centre bisects chord)

OA^2 = AM^2 + OM^2

\phantom{OA^2} = 12^2 + x^2

(Pythagoras' theorem)

ON = 21 - x

OC^2 = CN^2 + ON^2

\phantom{OC^2} = 9^2 + (21 - x)^2

OA = OC

(radii)

12^2 + x^2 = 9^2 + (21 - x)^2

144 + x^2 = 81 + 441 - 42x + x^2

42x = 378

x = 9

(shown)

(c)Answer:

15

cm[2]

OA^2 = 12^2 + 9^2

\phantom{OA^2} = 225

OA = \sqrt{225}

\phantom{OA} = 15

(d)Answer:

106.3^\circ

°[2]

\tan \angle AOM = \dfrac{AM}{OM}

\phantom{\tan \angle AOM} = \dfrac{12}{9}

\angle AOM = 53.1301\ldots^\circ

\angle AOB = 2 \times 53.1301\ldots^\circ

\phantom{\angle AOB} = 106.2602\ldots^\circ

\angle AOB = 106.3^\circ

Practice questions by skill

Sec 3 E-Math Circle properties worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Circle properties notes. Or practise it online with step-by-step help when you get one wrong.

Answers and worked solutions

Question 1

Answer:

x = 26

°[1]

\mathstrut \angle ACB = \dfrac{52^\circ}{2}

(∠ at centre = 2 ∠ at circumference)

x = 26

Question 2

Answer:

x = 67

°[1]

\angle ADB = \angle ACB

(∠s in same segment)

x = 67

Question 3

Answer:

x = 56

°[2]

\angle OPA = \angle OPB

(tangents from ext. pt.)

\angle OPA = 34^\circ

\angle OAP = 90^\circ

(tan. ⊥ rad.)

x = 180^\circ - 90^\circ - 34^\circ

(∠ sum of △)

x = 56

Question 4

Answer:

x = 136

°[2]

\angle OBA = 22^\circ

(base ∠s of isos. △)

\mathstrut x = 180^\circ - 2 \times 22^\circ

(∠ sum of △)

x = 136

Question 5

Answer:

x = 86

°[3]

\angle OAT = 90^\circ

(tan. ⊥ rad.)

\mathstrut \angle OAB = 90^\circ - 43^\circ

\angle OAB = 47^\circ

\angle OBA = 47^\circ

(base ∠s of isos. △)

\mathstrut x = 180^\circ - 2 \times 47^\circ

(∠ sum of △)

x = 86

Question 6

Answer:

\phantom{x} = 16.6

cm (3 s.f.)[3]

\angle ACB = 90^\circ

(∠ in semicircle)

AB^2 = AC^2 + BC^2

\mathstrut x^2 = 9^2 + 14^2

x^2 = 277

x = \sqrt{277}

x = 16.6433\ldots

\phantom{x} = 16.6

cm (3 s.f.)

Question 7

(a)Answer:

124^\circ

°[2]

\angle OAP = 90^\circ

(tangent ⊥ radius)

\angle OBP = 90^\circ

(tangent ⊥ radius)

\angle AOB = 360^\circ - 90^\circ - 90^\circ - 56^\circ

\phantom{\angle AOB} = 124^\circ

(∠ sum of quad.)

(b)Answer:

62^\circ

°[1]

\angle ACB = \dfrac{124^\circ}{2}

\phantom{\angle ACB} = 62^\circ

(∠ at centre = 2 ∠ at circumference)

(c)Answer:

28^\circ

°[2]

OA = OB

(radii)

\angle OAB = \dfrac{180^\circ - 124^\circ}{2}

\phantom{\angle OAB} = 28^\circ

(base ∠s of isos. △)

Question 8

(a)Answer:

62^\circ

°[2]

\angle ABC = 90^\circ

(rt. ∠ in semicircle)

\angle BCA = 180^\circ - 90^\circ - 28^\circ

\phantom{\angle BCA} = 62^\circ

(∠ sum of △)

(b)Answer:

109^\circ

°[2]

\angle BAD = 28^\circ + 43^\circ

\phantom{\angle BAD} = 71^\circ

\angle BCD = 180^\circ - 71^\circ

\phantom{\angle BCD} = 109^\circ

(∠s in opp. segments)

(c)Answer:

28^\circ

°[1]

\angle BDC = \angle BAC

\phantom{\angle BDC} = 28^\circ

(∠s in same segment)

Question 9

(a)Answer:

12

cm[1]

AM = \dfrac{24}{2}

\phantom{AM} = 12

cm (⊥ from centre bisects chord)

(b)[3]

CN = 9

(⊥ from centre bisects chord)

OA^2 = AM^2 + OM^2

\phantom{OA^2} = 12^2 + x^2

(Pythagoras' theorem)

ON = 21 - x

OC^2 = CN^2 + ON^2

\phantom{OC^2} = 9^2 + (21 - x)^2

OA = OC

(radii)

12^2 + x^2 = 9^2 + (21 - x)^2

144 + x^2 = 81 + 441 - 42x + x^2

42x = 378

x = 9

(shown)

(c)Answer:

15

cm[2]

OA^2 = 12^2 + 9^2

\phantom{OA^2} = 225

OA = \sqrt{225}

\phantom{OA} = 15

(d)Answer:

106.3^\circ

°[2]

\tan \angle AOM = \dfrac{AM}{OM}

\phantom{\tan \angle AOM} = \dfrac{12}{9}

\angle AOM = 53.1301\ldots^\circ

\angle AOB = 2 \times 53.1301\ldots^\circ

\phantom{\angle AOB} = 106.2602\ldots^\circ

\angle AOB = 106.3^\circ