Sec 4 A-Math Proofs worksheet with answers

A free Sec 4 A-Math worksheet on proofs: 8 questions, 40 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 4 A-Math worksheet: Proofs

Total: 40 marksSuggested time: 1 hour

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 triangle

PQR

X

is the mid-point of

PQ

and

Y

is the mid-point of

PR

\angle PQR = 48^\circ

. Show that

XY

is parallel to

QR

and

XY = \dfrac{1}{2}QR

. Hence find

\angle PXY

Answer°[2]

The lines

PR

and

QS

meet at

M

, and

PQ

is parallel to

SR

MP = 4

cm,

MR = 5

cm and

PQ = 4

cm. Prove that

\triangle MPQ

is similar to

\triangle MRS

. Hence find

SR

Answercm[3]

XY

is the tangent to the circle at

Y

, and the straight line

XHK

cuts the circle at

H

and

K

XH = 4

cm and

XY = 6

cm. Prove that

\triangle XYH

is similar to

\triangle XKY

, and hence that

XY^{2} = XH \times XK

. Find

HK

Answercm[3]

In triangle

PQR

S

lies on

PQ

and

T

lies on

PR

, with

ST

parallel to

QR

PS = 3

cm,

SQ = 6

cm and

QR = 9

cm. Prove that

\triangle PST

is similar to

\triangle PQR

. Hence find

ST

Answercm[3]

(a)Prove that

\sin^{4} x - \cos^{4} x = \sin^{2} x - \cos^{2} x

Answer (a)[2]

(b)Prove that

\dfrac{2\tan x}{1 + \tan^{2} x} = \sin 2x

Answer (b)[3]

Section B

(a)Prove that

\cot x - \tan x = 2\cot 2x

[3]

(b)Hence solve the equation

\cot x - \tan x = 4

for

0^\circ < x < 360^\circ

Answer (b)°[3]

In the diagram,

A

B

and

C

lie on a circle. The tangent to the circle at

C

meets

AB

produced at

T

. The bisector of angle

ATC

meets

BC

Q

and

AC

P

(a)Explain why

\angle TCB = \angle CAB

[1]

(b)Prove that

CP = CQ

[4]

(c)Given that

\angle CAB = 35^\circ

and

\angle ATC = 28^\circ

, find

\angle CQP

Answer (c)°[2]

(d)Find

\angle ACB

Answer (d)°[2]

In the diagram,

O

is the centre of a circle. The point

T

lies outside the circle, and

TA

and

TB

are tangents to the circle at

A

and

B

(a)Prove that triangles

OAT

and

OBT

are congruent.

[3]

(b)Hence show that

TA = TB

and that

OT

bisects angle

AOB

[2]

(c)Given that

OA = 6

cm and

OT = 10

cm, find the length of

TA

Answer (c)cm[1]

(d)Find angle

AOB

, correct to 1 decimal place.

Answer (d)°[2]

(e)Find the area of the quadrilateral

OATB

Answer (e)cm²[1]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

\angle PXY = 48^\circ

°[2]

PX = XQ

and

PY = YR

(given)

XY \parallel QR

and

XY = \dfrac{1}{2}QR

(midpoint theorem)

\angle PXY = \angle PQR

(corr. ∠s (XY ∥ QR))

\angle PXY = 48^\circ

Question 2

Answer:

SR = 5

cm[3]

\angle MPQ = \angle MRS

(alt. ∠s (PQ ∥ SR))

\angle PMQ = \angle RMS

(vert. opp. ∠s)

\triangle MPQ

is similar to

\triangle MRS

(AA).

\dfrac{SR}{PQ} = \dfrac{MR}{MP}

\dfrac{SR}{4} = \dfrac{5}{4}

SR = 5

Question 3

Answer:

HK = 5

cm[3]

\angle YXH = \angle KXY

(common ∠)

\angle XYH = \angle XKY

(alt. segment theorem)

\triangle XYH

is similar to

\triangle XKY

(AA).

\dfrac{XY}{XK} = \dfrac{XH}{XY}

\dfrac{6}{XK} = \dfrac{4}{6}

XK = 9

\mathstrut HK = 9 - 4

HK = 5

Question 4

Answer:

ST = 3

cm[3]

\angle PST = \angle PQR

(corr. ∠s (ST ∥ QR))

\angle SPT = \angle QPR

(common ∠)

\triangle PST

is similar to

\triangle PQR

(AA).

\mathstrut PQ = 3 + 6

PQ = 9

\dfrac{ST}{QR} = \dfrac{PS}{PQ}

\dfrac{ST}{9} = \dfrac{3}{9}

ST = 3

Question 5

(a)Answer:

= \text{RHS}

[2]

\text{LHS} = \sin^{4} x - \cos^{4} x

Difference of two squares

= (\sin^{2} x - \cos^{2} x)(\sin^{2} x + \cos^{2} x)

Use sin²x + cos²x = 1

= \sin^{2} x - \cos^{2} x

= \text{RHS}

(b)Answer:

= \text{RHS}

[3]

\text{LHS} = \dfrac{2\tan x}{1 + \tan^{2} x}

Use 1 + tan²x = sec²x below

= \dfrac{2\tan x}{\sec^{2} x}

Dividing by sec²x is multiplying by cos²x; tan x = sin x ÷ cos x

= 2 \cdot \dfrac{\sin x}{\cos x} \cdot \cos^{2} x

Cancel cos x

= 2\sin x\cos x

Use 2 sin x cos x = sin 2x

= \sin 2x

= \text{RHS}

Question 6

(a)[3]

LHS

= \dfrac{\cos x}{\sin x} - \dfrac{\sin x}{\cos x}

= \dfrac{\cos^{2} x - \sin^{2} x}{\sin x\cos x}

= \dfrac{\cos 2x}{\dfrac{1}{2}\sin 2x}

= 2\cot 2x

= RHS (proved)

(b)Answer:

x = 13.3^\circ, 103.3^\circ, 193.3^\circ, 283.3^\circ

°[3]

By (a), the equation is

2\cot 2x = 4

2\cot 2x = 4

\tan 2x = \dfrac{1}{2}

0^\circ < 2x < 720^\circ

x = 13.3^\circ, 103.3^\circ, 193.3^\circ, 283.3^\circ

Question 7

(a)[1]

TC

is a tangent and

CB

is a chord, so by the alternate segment theorem

\angle TCB = \angle CAB

(b)[4]

\angle CQP = \angle QTC + \angle QCT

(ext. ∠ of

\triangle TQC

)

\angle CPQ = \angle PTA + \angle PAT

(ext. ∠ of

\triangle TPA

)

\angle QTC = \angle PTA

(

TP

bisects

\angle ATC

)

\angle QCT = \angle TCB

\phantom{\angle QCT} = \angle CAB

\phantom{\angle QCT} = \angle PAT

(part (a))So

\angle CQP = \angle CPQ

.Triangle

CPQ

is isosceles:

CP = CQ

(sides opp. equal ∠s).

(c)Answer:

49^\circ

°[2]

\angle QTC = \dfrac{1}{2}(28^\circ)

\phantom{\angle QTC} = 14^\circ

\angle QCT = \angle CAB

\phantom{\angle QCT} = 35^\circ

\angle CQP = 14^\circ + 35^\circ

= 49^\circ

(d)Answer:

82^\circ

°[2]

\angle ABC = \angle BTC + \angle BCT

(ext. ∠ of

\triangle TBC

)

= 28^\circ + 35^\circ

\phantom{} = 63^\circ

\angle ACB = 180^\circ - 35^\circ - 63^\circ

(∠ sum of

\triangle ABC

)

= 82^\circ

Question 8

(a)[3]

\angle OAT = \angle OBT

\phantom{\angle OAT} = 90^\circ

(tangent ⊥ radius)

OT

is common (the hypotenuse of both triangles)

OA = OB

(radii)So

\triangle OAT \equiv \triangle OBT

(RHS).

(b)[2]

From part (a),

\triangle OAT \equiv \triangle OBT

.So

TA = TB

(corresponding sides).And

\angle AOT = \angle BOT

(corresponding angles), so

OT

bisects

\angle AOB

(c)Answer:

8

cm[1]

TA^{2} = 10^{2} - 6^{2}

(Pythagoras' theorem)

= 64

TA = 8

(d)Answer:

106.3^\circ

°[2]

\cos \angle AOT = \dfrac{OA}{OT}

\phantom{\cos \angle AOT} = \dfrac{6}{10}

\angle AOT = 53.13^\circ

\angle AOB = 2\angle AOT

\phantom{\angle AOB} = 106.3^\circ

(e)Answer:

48

cm²[1]

Area

= 2 \times

area of

\triangle OAT

= 2 \times \dfrac{1}{2} \times 6 \times 8

= 48

cm²

Practice questions by skill

Sec 4 A-Math Proofs worksheet with answers

A free Sec 4 A-Math worksheet on proofs: 8 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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

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

Answers and worked solutions

Question 1

Answer:

\angle PXY = 48^\circ

°[2]

PX = XQ

and

PY = YR

(given)

XY \parallel QR

and

XY = \dfrac{1}{2}QR

(midpoint theorem)

\angle PXY = \angle PQR

(corr. ∠s (XY ∥ QR))

\angle PXY = 48^\circ

Question 2

Answer:

SR = 5

cm[3]

\angle MPQ = \angle MRS

(alt. ∠s (PQ ∥ SR))

\angle PMQ = \angle RMS

(vert. opp. ∠s)

\triangle MPQ

is similar to

\triangle MRS

(AA).

\dfrac{SR}{PQ} = \dfrac{MR}{MP}

\dfrac{SR}{4} = \dfrac{5}{4}

SR = 5

Question 3

Answer:

HK = 5

cm[3]

\angle YXH = \angle KXY

(common ∠)

\angle XYH = \angle XKY

(alt. segment theorem)

\triangle XYH

is similar to

\triangle XKY

(AA).

\dfrac{XY}{XK} = \dfrac{XH}{XY}

\dfrac{6}{XK} = \dfrac{4}{6}

XK = 9

\mathstrut HK = 9 - 4

HK = 5

Question 4

Answer:

ST = 3

cm[3]

\angle PST = \angle PQR

(corr. ∠s (ST ∥ QR))

\angle SPT = \angle QPR

(common ∠)

\triangle PST

is similar to

\triangle PQR

(AA).

\mathstrut PQ = 3 + 6

PQ = 9

\dfrac{ST}{QR} = \dfrac{PS}{PQ}

\dfrac{ST}{9} = \dfrac{3}{9}

ST = 3

Question 5

(a)Answer:

= \text{RHS}

[2]

\text{LHS} = \sin^{4} x - \cos^{4} x

Difference of two squares

= (\sin^{2} x - \cos^{2} x)(\sin^{2} x + \cos^{2} x)

Use sin²x + cos²x = 1

= \sin^{2} x - \cos^{2} x

= \text{RHS}

(b)Answer:

= \text{RHS}

[3]

\text{LHS} = \dfrac{2\tan x}{1 + \tan^{2} x}

Use 1 + tan²x = sec²x below

= \dfrac{2\tan x}{\sec^{2} x}

Dividing by sec²x is multiplying by cos²x; tan x = sin x ÷ cos x

= 2 \cdot \dfrac{\sin x}{\cos x} \cdot \cos^{2} x

Cancel cos x

= 2\sin x\cos x

Use 2 sin x cos x = sin 2x

= \sin 2x

= \text{RHS}

Question 6

(a)[3]

LHS

= \dfrac{\cos x}{\sin x} - \dfrac{\sin x}{\cos x}

= \dfrac{\cos^{2} x - \sin^{2} x}{\sin x\cos x}

= \dfrac{\cos 2x}{\dfrac{1}{2}\sin 2x}

= 2\cot 2x

= RHS (proved)

(b)Answer:

x = 13.3^\circ, 103.3^\circ, 193.3^\circ, 283.3^\circ

°[3]

By (a), the equation is

2\cot 2x = 4

2\cot 2x = 4

\tan 2x = \dfrac{1}{2}

0^\circ < 2x < 720^\circ

x = 13.3^\circ, 103.3^\circ, 193.3^\circ, 283.3^\circ

Question 7

(a)[1]

TC

is a tangent and

CB

is a chord, so by the alternate segment theorem

\angle TCB = \angle CAB

(b)[4]

\angle CQP = \angle QTC + \angle QCT

(ext. ∠ of

\triangle TQC

)

\angle CPQ = \angle PTA + \angle PAT

(ext. ∠ of

\triangle TPA

)

\angle QTC = \angle PTA

(

TP

bisects

\angle ATC

)

\angle QCT = \angle TCB

\phantom{\angle QCT} = \angle CAB

\phantom{\angle QCT} = \angle PAT

(part (a))So

\angle CQP = \angle CPQ

.Triangle

CPQ

is isosceles:

CP = CQ

(sides opp. equal ∠s).

(c)Answer:

49^\circ

°[2]

\angle QTC = \dfrac{1}{2}(28^\circ)

\phantom{\angle QTC} = 14^\circ

\angle QCT = \angle CAB

\phantom{\angle QCT} = 35^\circ

\angle CQP = 14^\circ + 35^\circ

= 49^\circ

(d)Answer:

82^\circ

°[2]

\angle ABC = \angle BTC + \angle BCT

(ext. ∠ of

\triangle TBC

)

= 28^\circ + 35^\circ

\phantom{} = 63^\circ

\angle ACB = 180^\circ - 35^\circ - 63^\circ

(∠ sum of

\triangle ABC

)

= 82^\circ

Question 8

(a)[3]

\angle OAT = \angle OBT

\phantom{\angle OAT} = 90^\circ

(tangent ⊥ radius)

OT

is common (the hypotenuse of both triangles)

OA = OB

(radii)So

\triangle OAT \equiv \triangle OBT

(RHS).

(b)[2]

From part (a),

\triangle OAT \equiv \triangle OBT

.So

TA = TB

(corresponding sides).And

\angle AOT = \angle BOT

(corresponding angles), so

OT

bisects

\angle AOB

(c)Answer:

8

cm[1]

TA^{2} = 10^{2} - 6^{2}

(Pythagoras' theorem)

= 64

TA = 8

(d)Answer:

106.3^\circ

°[2]

\cos \angle AOT = \dfrac{OA}{OT}

\phantom{\cos \angle AOT} = \dfrac{6}{10}

\angle AOT = 53.13^\circ

\angle AOB = 2\angle AOT

\phantom{\angle AOB} = 106.3^\circ

(e)Answer:

48

cm²[1]

Area

= 2 \times

area of

\triangle OAT

= 2 \times \dfrac{1}{2} \times 6 \times 8

= 48

cm²