chapter 19 exercise 19 - biology leaving cert.shevlinbiology.webs.com/chapter 19 solutions.pdf ·...

Active Maths 2 (Strands 1–5): Ch 19 Solutions

Chapter 19 Exercise 19.1

1

Q. 1. (i) An axiom is a statement that is accepted but cannot be proven, e.g. x + 0 = x. (ii) A statement that can be proven logically: for example, Pythagoras’ Theorem. (iii) The logical steps used to prove a theorem, e.g. when proving two triangles are

similar, we show they have equal angles. (iv) A statement derived (found) from a theorem, e.g. it can be derived from Pythagoras’

Theorem that |A| + |B| > |C|, where A, B and C are the sides of a right-angled triangle and C is the hypotenuse.

(v) A theorem in reverse order, e.g. the angles in a quadrilateral sum to 360°. The converse is: a polygon whose angles sum to 360° is a quadrilateral.

Q. 2. (i) Statement: a = b Converse: b = a (true) (ii) Statement: when you stick two equal squares together you get a rectangle Converse: A rectangle is made up of two equal squares (false)

Square Square

Rectangle

Q. 3. Statement Converse Converse is True/False

If the largest angle in a triangle is obtuse, then the two smaller angles are acute.

If the two smaller angles in a triangle are acute, then the largest angle is obtuse.

False

If two triangles are congruent, then they are similar.

If two triangles are similar then they are congruent

False

If a triangle is equilateral, then it is also isosceles.

If a triangle is isosceles then it is also equilateral

False

If a number is divisible by 6, then it is also divisible by 3.

If a number is divisible by 1, then it is divisible by 3

False

Q. 4. Given: Triangles ABC and BDE with parallel lines l and m as shown below |AC| = |DE|.

B

C

m

I

ED

A1

3 4

2

To prove: ΔABC ≡ ΔDBE Construction: Label angles 1, 2, 3 and 4 Proof: |∠1| = |∠4| … alternate |AC| = |DE| … given |∠2| = |∠3| … alternate ⇒ Δ ABC ≡ ΔDBE …… ASA Q.E.D.

2 Active Maths 2 (Strands 1–5): Ch 19 Solutions

Q. 5. (i) Given: Circle with chords [PQ] and [ST] as shown below.

12

3

4

ST

R

Q

P

To prove: Δ PQR ≡ Δ SRT

Construction: Label angles 1, 2, 3, 4

Proof: |∠1| = |∠2| … vertically opposite

|RQ| = |SR| … radii

|∠4| = |∠3| … alternate

⇒ ΔPQR ≡ ΔSRT (ASA)

(ii) |PT| = |QS| … diameters

[|PR| = |RT| = |RQ| = |RS|]

Note: R is the centre of the circle.

Q. 6. (i) Given: Triangles DEF and LMO

|∠E| = 90°

|∠D| = 50°

|∠M| = 90°

|∠O| = 50°

|DF| = |LO|

To prove: ΔDEF = ΔLMO

Proof: |DF| = |∠LO| … given

|∠D| = |∠O| … given

|∠F| = 180° – (90° + 50°)

= 40° … angle is a Δ

|∠L| = 180° – (90° + 50°)


⇒ |∠F| = |∠L|

⇒ ΔDEF ≡ ΔLMO (ASA)

Q.E.D.

(ii) Given: Triangles ABC and BCD

|∠ABC| = |∠CBD| = 30°

|AB| = |BC| = |BD|

To prove: ΔABC ≡ ΔCBD

Proof: |AB| = |BD| … given

|∠ABC| = |∠CBD| = 30°

… given

BC is a common side

⇒ ΔABC ≡ ΔCBD (SAS)

(iii) Yes, by SSS

(iv) Yes, by ASA

Q. 7. (i) Yes |∠BAC| = |∠DAE|

… common angle

|∠ABC| = |∠ADE|

.… corresponding

|∠ACB| = |∠AED|

.… corresponding.

(ii) Yes |∠PSQ| = |∠RSQ|

.… both 90°

|∠PQS| = |∠QRS| … because triangle

PQR is similar to ΔPSQ and therefore

|∠PQS| = |∠PRQ|

Exercise 19.2

Q. 1. Given: Square ABCD as shown.

B 12

3

C

DA

To prove: |∠ACB| = 45°

Construction: Label angles 1, 2 and 3

3Active Maths 2 (Strands 1–5): Ch 19 Solutions

Proof: |∠2| = 90° … square

|AB| = |BC| … square

⇒ |∠1| = |∠3| … isosceles Δ

⇒ 90° + |∠1| + |∠3|


⇒ 2|∠1| = 90°

⇒ |∠1| = 45°

Q.E.D

Q. 2. Given: Diagram as shown

Y Z

X

W

To prove: |∠YZW| = 90°

Proof: |XY| = |XZ| = |XW|

So we can draw a circle around Δ WYZ with X as the centre with |XY|, |XZ| and |WX| as radii

⇒ |∠YZW| = 90° … angle in a semicircle.

Q. 3. Given: Diagram as shown.

B

C

DA1

2

3

To prove: |∠ABC| > |∠CDB|

Construction: Label angles 1, 2, 3

Proof: |∠1| = |∠2| + |∠3|

… exterior angle

⇒ |∠1| > |∠3|

Q.E.D.

Q. 4. Given: Triangle PQS as shown.

1

Q S

R

P

AA+B

B

To prove: Δ PQR is isosceles Construction: Label angle 1 Proof: |∠1| = A + B … exterior angle ⇒ Δ PQR is isosceles. Q.E.D.

Q. 5. Given: Triangle BCD with extension A and E as shown |∠ABC| = |∠CDE|.

B

C

EDA1 4 23

To prove: |∠3| = |∠4| Construction: Label angles 1, 2, 3 and 4 Proof: |∠3| = 180° – |∠1| … straight

angle |∠4| = 180° – |∠2| … straight

angle but |∠1| = |∠2| … given ⇒ |∠4| = 180° – |∠1| ⇒ |∠4| = |∠3| Q.E.D.


BC

E

DA

12

4

3

To prove: |∠BCE| = |∠CED| + |∠ABC| Construction: Label angles 1, 2, 3, 4 Proof: |∠1| = |∠2| + |∠4| …

exterior angle |∠4| = |∠3| … alternate ⇒ |∠1| = |∠2| + |∠3|

Q.E.D.


Q. 7. Given: Triangle PQR as shown.

P

X

Q R

1

2 3

To prove: |∠PXQ| >|PRQ|

Construction: Label angles 1, 2 and 3

Proof: |∠1| = |∠2| + |∠3|

… exterior angle

⇒ |∠1| > |∠3|

Q.E.D.

Q. 8. Given: Isosceles Δ ABC as shown.

DC

B

A

2

3 4 1

To prove: |∠DCB| = 90° + |∠ABC|

________ 2


Proof: |∠3| = |∠4| … isosceles Δ

⇒ |∠4| = 180° – |∠1|

… straight angle

|∠2| = 180° – (|∠3| + |∠4|)

… angles in a Δ

⇒ |∠2| = 180° – (180° – |∠1| + 180° – |∠1|)

⇒ |∠2| = 2 |∠1| – 180°

⇒ 2|∠1| = 180° + |∠2|

⇒ |∠1| = 90° + |∠2|

_____ 2

Q.E.D.


DCBE

A

7

51 4 6 2

8

3

To prove: Δ ABE ≡ Δ ACD

Construction: Label angles 1, 2, 3, 4, 5, 6, 7, 8

Proof: |AE| = |AD| … given S |AB| = |AC| … given S |∠1| = |∠2| … isosceles Δ

|∠3| = |∠4| … isosceles Δ

⇒ |∠5| = 180° – |∠3| = 180° – |∠4| = |∠6| ⇒ |∠7| = |∠8|

… 2 equal angles in a Δ implies the 3rd angles are equal A

⇒ Δ ABE ≡ Δ ACD (SAS)

Q.E.D.

Q. 10. (i) |AM| = |AM| (common)

|BM| = |MC| (given)

|AB| = |AC| (given)

∴ ΔABM ≡ ΔAMC (side, side, side)

(ii) ∠AMB in ΔAMB corresponds to ∠AMC in ΔAMC, Therefore |∠AMB| = |∠AMC|

(iii) |∠AMB| = |∠AMC| = 180° (Straight line)

but |∠AMB| = |∠AMC|

⇒|∠AMB| = |∠AMC| = 90°

Therefore AM ⊥ BC


Q. 11. Given: Δ XYZ as shown.

X

ZY

2

4 6 3

5

1

To prove: |∠1| + |∠2| + |∠3| = 360°

Construction: Label angels 4, 5 and 6

Proof: |∠1| = |∠5| + |∠6| … exterior angle |∠2| = |∠4| + |∠6| … exterior angle |∠3| = |∠4| + |∠5| … exterior angle ⇒ |∠1| + |∠2| + |∠3| = 2 ( |∠4| + |∠5| + |∠6| ) but |∠4| + |∠5| + |∠6| = 180° … angle in a Δ

⇒ |∠1| + |∠2| + |∠3| = 2(180°) = 360° Q.E.D.

Q. 12. (i) Given: Δ PQR as shown below |PQ| = |PR|

X

Y

RQ

P

1 23

4

5

6

To prove: Δ XYP is isosceles Construction: Label angles 1–6 Proof: |∠1| = |∠2| … isosceles Δ

|∠3| + |∠2| + 90° = 180° … angles in a Δ

⇒ |∠3| = 90° – |∠2| ⇒ |∠3| = 90° – |∠1| |∠4| = |∠3| … vertically

opposite

⇒ |∠4| = 90° – |∠1| |∠5| + |∠1| + 90° = 180° … angles in a Δ

⇒ |∠5| = 90° – |∠1| ⇒ |∠4| = |∠5| ⇒ Δ XYP is isosceles Q.E.D.

(ii) But Δ PXY may be equilateral

|∠6| = |∠1| + |∠2| … exterior angle ⇒ |∠6| = 2|∠1| If Δ PXY is equilateral |∠6| = |∠4| = |∠5| ⇒ 2|∠1| = 90° – |∠1| ⇒ 3|∠1| = 90°

⇒ |∠1| = 30° So when |∠PQR| = 30°

Δ XYP is equilateral


A C D

B

1 3

245

To prove: |∠ABC| = 2|∠CDB|

Construction: Label angles 1, 2, 3, 4, 5, 6



|∠1| + |∠2| + |∠3| + |∠5|

= 180° … angles in a Δ

⇒ |∠1| + 2|∠2| + |∠5| = 180°

|∠1| + |∠4| + |∠5| = 180°

… angles in a Δ

⇒ 2|∠1| + |∠5| = 180°

|∠1| + 2|∠2| + |∠5|

= 2|∠1| + |∠5|

⇒ |∠1| = 2|∠2|

Q.E.D.


Q. 14. Given: Triangles PQS and SQR as shown below

P Q R

S

21

3 4 5 6

To prove: |∠PSQ| + |∠QSR| = 90°

Construction: Label angles 1–6 as shown


|∠5| = |∠1| + |∠3|

… exterior angle

⇒ |∠5| = 2|∠1|


|∠4| = |∠2| + |∠6|

… exterior angle

⇒ |∠4| = 2|∠2|

|∠4| + |∠5| = 180°

… straight angle

⇒ 2|∠2| + 2|∠1| = 180°

⇒ 2(|∠1| + |∠2|) = 180°

⇒ |∠1| + |∠2| = 90°

Q.E.D.

Q. 15. (a) Given:

A

B C

D

Ezw

T

X Y

To prove: [AE bisects DAC.

Proof: |Y| = |W| Alternate

|Y| = |X| isosceles

|X| = |Z| Corresponding

⇒ |W| = |Z|

Therefore [AE bisects ∠DAC

Q.E.D.

(b) No, the result in (a) would not still apply. Angle Y would not be equal to angle X.

Exercise 19.3

Q. 1. Given: Parallelogram ABCD as shown.

X

CD

A B

Y 24

6

5

1

3

To prove: Δ ADY similar to Δ BXY


Proof: |∠1| = |∠5| … alternate |∠2| = |∠4| … vertically

opposite |∠3| = |∠6| … alternate ⇒ Δ ADY and Δ BXY are similar

Q.E.D.

Q. 2. Given: Rhombus PQRS as shown.

P S

RQ

21

34

To prove: [PR] bisects |∠QPS| Construction: Label angles 1– 4 Proof: |∠1| = |∠3| … alternate |∠3| = |∠2| … isosceles Δ

⇒ |∠1| = |∠2|

⇒ [PR] bisects |∠QPS|

Q.E.D.


Q. 3. Given: Parallelogram ABCD and ABEC as shown.

A

D C E

B

To prove: Δ ADC ≡ BCE

Construction: Δ ADC ≡ Δ BCE

Proof: |AD| = |BC| … parallelogram

|AC| = |BE| … parallelogram

|DC| = |AB| … parallelogram

|AB| = |CE| … parallelogram

⇒ |DC| = |CE|

⇒ Δ ADC ≡ Δ BCE (SSS)

Q.E.D.

Q. 4. Given: Rectangle DEFG as shown

D G

E F21

To prove: |DF| = |EG|

Construction: Label angles 1 and 2

Proof: |DE| = |GF| … opposite sides

|∠1| = |∠2| … right angles

EF is a common side in Δ DEF and Δ EFG

⇒ Δ DEF ≡ Δ EFG

⇒ |DF| = |EG|

… Corresponding sides

Q.E.D.

Q. 5. Given: Parallelogram LMOP as shown

L M

OP

1 2

3 4

To prove: |∠1| + |∠2| = 180°

Construction: Label angles 3 and 4

Proof: |∠3| = |∠2| … opposite angles

|∠4| = |∠1| … opposite angles

|∠1| + |∠2| + |∠3| + |∠4|

= 360° … quadrilateral

⇒ 2(|∠1| + |∠2|) = 360°

⇒ |∠1| + |∠2| = 180°

Q.E.D.

Q. 6. Given: Parallelogram ABCD as shown. |AD| = 2|AB|. M is the midpoint of [BC]

A D

B M C

21

3 47

56

To prove: |∠AMD| = 90°

Construction: Label angles 1, 2, 3, 4, 5, 6, 7

Proof: |AD| = |BC| … opposite sides

⇒ |AD| = 2|AB|

⇒ 1 __ 2

|AD| = |AB|

⇒ 1 __ 2 |BC| = |AB|

⇒ |BM| = |AB|

⇒ |∠1| = |∠3|

|∠2| = |∠3| … alternate

⇒ |∠1| = |∠2|

Similarly |CM| = |CD|

⇒ |∠4| = |∠6|

|∠5| = |∠4| … alternate

⇒ |∠5| = |∠6|

|∠1| + |∠2| + |∠5| + |∠6|

= 180°

⇒ 2(|∠2| + |∠5|) = 180°

⇒ |∠2| + |∠5| = 90°


|∠2| + |∠5| = |∠7| = 180° … angles in a Δ

⇒ 90° + |∠7| = 180° ⇒ |∠7| = 90°

Q.E.D.

Q. 7. (a) Given: Parallelogram PQRS as shown.

X

Y

Q R

SP

2

3

1 4

5

6

To prove: |SY| = |QX|


Proof: |∠1| = |∠6| … alternate |PQ| = |SR| … opposite sides |∠3| = |∠5| … given ⇒ Δ RSY ≡ Δ PXQ (ASA)

(b) ⇒ |SY| = |QX|

… corresponding sides

Q.E.D.

Q. 8. Given: Parallelogram ABCD. M is the midpoint of [AC].

D X A

M

C Y B

2

3

4

1

To prove: |MX| = |MY|

Construction: Label angles 1, 2, 3, 4 Proof: |∠1| = |∠4| … alternate |CM| = |AM| … given |∠2| = |∠3| … vertically

opposite ⇒ Δ CYM ≡ Δ AXM (ASA)

⇒ |MX| = |MY|


Q.E.D.

Q. 9. (i) Given: Parallelogram PQRS. T is the midpoint of [SR].

R

U

S T

P Q

213 4

To prove: Δ PST ≡ Δ RTU


Proof: |∠1| = |∠4|

… alternate

|ST| = |TR| … given

|∠2| = |∠3|

… vertically opposite

⇒ Δ PST ≡ Δ RTU (ASA)

Q.E.D.

(ii) |RU| = |PS| … corresponding sides

Q.E.D.

(iii) |QU| = |QR| + |RU|

|QR| = |PS| … opposite sides

|RU| = |PS| … part (ii)

⇒ |QU| = 2|PS|

Q.E.D.

Q. 10. (i) Given: Square DEFG. P, Q, R, S are the midpoints of the sides.

D P E

QSO

G R F14

572

3

68


To prove: Δ DOS ≡ Δ GRO


Proof: |∠1| = |∠2| … both right angles |DO| = |GO| … diagonal bisect |GR| = |SO| ⇒ Δ DOS ≡ Δ GRO (RHS) Q.E.D.

(ii) Given: Same as part (i)

To prove: Δ DOG ≡ Δ DOE

Construction: Label angles 7, 8

Proof: |∠7| = |∠8|

… diagonals meet at night angles

|DG| = |DE| … square

|GO| = |EO|

… diagonals bisected

⇒ Δ DOG ≡ Δ DOE (RHS)

Q.E.D.

(iii) It can be shown that

Δ ROF ≡ Δ QOF ≡ Δ QOE ≡ Δ POE ≡ etc.

Using a similar proof on part (i)

⇒ Total area = 8 × area Δ ROF

= 8 × 8

= 64 cm2

Q. 11. Given: Parallelogram ABCD as shown, [AX] ⊥ [BC] and [CY] ⊥ [AD]

B X C

P

A Y D

Q

13

5 42

6

To prove: Δ PAB ≡ Δ CDQ


Proof: |∠1| = |∠2|

… alternate

As AD || BC and AX ⊥ BC and CY ⊥ AD

⇒ AX || CY

⇒ |∠3| = |∠4|

… alternate

⇒ |∠5| = |∠6|

… 2 equal angles implies the 3rd angles are equal

|AB| = |CD|

… opposite side

⇒ Δ PAB ≡ Δ CDQ (ASA)

Q.E.D.

∴ |BP| = |DQ|


Q.E.D.

Exercise 19.4

Q. 1. Consider Δs BEF and DEA.

|∠FBE| = |∠ADE| … alternate

|∠BFE| = |∠DAE| .… alternate

|∠BEF| = |∠DEA|

.… vertically opposite

∴ ΔBEF is similar to ΔDEA.

∴ |BE|

_____ |DE|

= |EF|

_____ |EA|

|DE|.|EF| = |BE|.|AE|.

Q. 2. RQ || TS

∴ |∠PQR| = |∠PST|

.… corresponding

|∠PRQ| = |∠PTS|

.… corresponding

|∠TPS| = |∠RPQ|

… common angle.

∴ ΔRPQ is similar to ΔTPS

∴ |RQ|

_____ |PR|

= |TS|

_____ |PT|

====


Q. 3. (i)

C

1

2 3

4

DA

B

E

Label angles 1 → 4

|∠1| + |∠2| = 90° as ΔABC is right-angled.

|∠2| + |∠3| = 90° … straight angle

∴ |∠1| + |∠2| = |∠2| + |∠3|

∴ |∠1| = |∠3|

|∠3| + |∠4| = 90° as ΔCDE is right-angled

|∠2| + |∠3| = 90° .… straight angle

∴ |∠3| + |∠4| = |∠2| + |∠3|

∴ |∠4| = |∠2|

Also |∠BAC| = |∠EDC|

… both 90°

∴ ΔBAC is similar to ΔCDE.

(ii) Redrawing Δs

C

1

2

3

4D EA

B

C

We have |AC|

_____ |DE|

= |BC|

_____ |CE|

… similar triangles theorem.

Q. 4. We note that |AC| . |FG| = |AG| . |BC|

could also be written as |AC|

____ |BC| = |AG|

____ |FG| .

BD || CE

⇒ |AC|

____ |BC| = |AE|

____ |DE|

DF || EG

⇒ |AE|

____ |DE| = |AG|

_____ |FG|

⇒ |AC| ____

|BC| =

|AG| _____ |FG|

∴ |AC| . |FG| = |AG| . |BC| (cross-multiply)

Q. 5. (i) Consider ΔABE and ΔFCE

|∠ABE| = |∠FCE| .… alternate |∠BAE| = |∠CFE| .… alternate |∠BEA| = |∠CEF| … vertically

opposite ∴ ΔABC is similar to ΔFCE

∴ |AE|

_____ |FE|

= |BE|

_____ |CE|

(ii) From (i) ΔABE is similar to ΔFCE

∴ |AB|

_____ |FC|

= |BE|

_____ |CE|

|AB|.|CE| = |BE|.|FC|

But |AB| = |DC|

… opposite sides of a parallelogram

∴ |DC|.|CE| = |BE|.|FC|

Q. 6.

S

Q

T

R

Δ SRT is similar to Δ QRS

as |∠QRS| = |∠SRT| … common angle and |∠RQS| = |∠RST| … both 90° ⇒ |∠RSQ| = |∠RTS| Also |∠RQS| = |∠SQT| … both 90°

∴ Δ RQS is similar to Δ SQT.

∴ |RQ|

_____ |QS|

= |QS|

_____ |QT|

⇒ x __ y = y _____

|QT|

|QT|x = y2

|QT| = y2

__ x


Q. 7. Δs ADF and AEC

|AD|

_____ |DE|

= |AF|

_____ |FC|

… 1

Δs AEF and ABC

|AE|

_____ |EB|

= |AF|

_____ |FC|

… 2

1 = 2

|AD|

_____ |DE|

= |AE|

_____ |EB|

Q. 8. (i) |∠PTQ| = |∠TQS| ..… alternate

|∠TQP| = |∠QST| … both 90°

∴ Δ QTS is similar to Δ QTP.

(ii) Δ QTS is similar to Δ QTP

∴ |QT|

_____ |QS|

= |PT|

_____ |QT|

|QT|2 = |PT|.|QS|

Q. 9. In Δ ABC and Δ BDC

|∠ABC| = |∠BDC| … both 90° |∠ACB| = |∠BCD| … common angle. ∴ Δ ABC is similar to Δ BDC. ⇒ |∠CAB| = |∠CBD| also |∠ADB| = |∠BDC| … both 90° ∴ Δ ADB is similar to Δ BDC

and also Δ ABC is similar to Δ ADB

⇒ |AB|

_____ |AC|

= |AD|

_____ |AB|

⇒ |AB|2 = |AD|.|AC|

Q. 10. (a) In Δ ADB and Δ APC

|∠ADB| = |∠ACP|

... angles at the circle subtended by the same are

|∠DAB| = |∠CAP|

... as |AD| bisects ∠BAC

∴ Δ ABD is similar to Δ APC

(b) As Δ ABD is similar to Δ APC

|AC|

_____ |AD|

= |PC|

_____ |BD|

|AC|.|BD| = |AD|.|PC|

Q. 11. (i) In Δ PQT and Δ RWQ

|∠PQT| = |∠RWQ|

..… alternate angles

|∠TPQ| = |∠QRW|

… opposite angles in a parallelogram.

∴ ΔPQT is similar to ΔRWQ.

(ii)

RQ

P Sx2x

3x

T

W

Let |TS| = x, ⇒ |PT| = 2x

⇒ |QR| = 3x

From (i) |PQ|

______ |RW|

= |PT|

_____ |QR|

⇒ |PQ|

______ |RW|

= 2x ___ 3x = 2 __ 3

⇒ |PQ| = 2 __ 3 |RW|

But |PQ| = |RS|

∴ |RS| = 2 __ 3 |RW|

⇒ |RS| : |RW|

2 : 1

Exercise 19.5

Q. 1. Given: Square ABCD and DBFE.

A Bx

x xy

y y

y

x CD

E

F

To prove: Area DBEF = 2 × (area ABCD)

Construction: Mark sides of ABCD as x and DBEF as y.


Proof: Using Pythagoras’ Theorem x2 + x2 = y2

⇒ 2x2 = y2

⇒ 2(area of ABCD) = area of BDFE

Q.E.D.

Q. 2. In Δ PSQ , |PQ|2 = |PS|2 + |QS|2

In Δ RSQ , |QR|2 = |SR|2 + |QS|2

But |PQ| = |QR| ∴ |PQ|2 = |QR|2

∴ |PS|2 + |QS|2 = |SR|2 + |QS|2

|PS|2 = |SR|2

Q. 3. Given: Triangle ABC as shown.

A

B D C

To prove: |AB|2 + |DC|2 = |BD|2 + |AC|2

Proof: |AB|2 = |BD|2 + |AD|2 … Pythagoras’ Theorem ⇒ |AD|2 = |AB|2 – |BD|2

|AC|2 = |AD|2 + |CD|2 … Pythagoras’ Theorem ⇒ |AD|2 = |AC|2 – |CD|2

⇒ |AC|2 – |CD|2 = |AB|2 – |BD|2

⇒ |AB|2 + |DC|2 = |BD|2 + |AC|2

Q.E.D.

Q. 4. Given: Triangles ABC and APQ as shown.

C

A P B

Q

To prove: |BQ|2 + |CP|2

= |PQ|2 + |BC|2

Construction: Draw PC and BQ.

Proof: |AQ|2 + |AP|2 = |PQ|2

… Pythagoras

⇒ |AC|2 + |AB|2 = |BC|2

… Pythagoras

⇒ |AQ|2 + |AP|2 + |AC|2 + |AB|2

= |PQ|2 + |BC|2

⇒ (|AQ|2 + |AB|2) + (|AP|2 + |AC|2)

= |PQ|2 + |BC|2

⇒ |BQ|2 + |CP|2 = |PQ|2 + |BC|2

… Pythagoras

Q.E.D.

Q. 5. Given: Circle with centre O, tangents [AB] and [AC] as shown.

1

2

AO

B

C

To prove: |AB| = |AC| Construction: Draw the line AO.

Label angles 1 and 2 Proof: |∠1| = |∠2| … right angles AO is a common side |OB| = |OC| … radii ⇒ Δ AOB ≡ Δ AOC (RHS) ⇒ |AB| = |AC| … corresponding sides

Q.E.D.

Q. 6. In Δ ABC

|AC|2 = |BC|2 + |AB|2

In semicircle Q

Area = 1 __ 2 pr2

= 1 __ 2 p ( |AC| _____ 2 ) 2

= |AC|2 p

_______ 4


In semicircle R we similarly have

Area = |BC|2p

_______ 4

and in semicircle P, Area = |AB|2p

_______ 4

So is |AC|2p

_______ 4 = |BC|2p

_______ 4 + |AB|2p

_______ 4 ?

Is |AC|2 p = |BC|2 p + |AB|2 p ?

Is |AC|2 = |BC|2 + |AB|2 True

Q. 7. By Pythagoras’ Theorem

|AB|2 = |AC|2 + |BC|2

But |AC| = |BC|

so |AB|2 = 2|BC|2

|AB| = √__

2 |BC|

Q. 8.

RP S

Q

Construction: QS, the perpendicular bisector of PR

In Δ QSP, |QS|2 = |PQ|2 − |PS|2

But Δ QPR is equilateral so |PQ| = |PR| and also |PS| = 1 __ 2 |PR|

∴ |QS|2 = |PR|2 − ( 1 __ 2 |PR| ) 2 = |PR|2 − 1 __ 4 |PR|2

|QS|2 = 3 __ 4 |PR|2

∴ |QS| = √__

3 __ 4 |PR|

= √__

3 ___ 2 |PR|

Q. 9. Q R

P N

M

S

T

Construction: [MN] through T such that [MN] || [QP] and [MN] ⊥ [PS] Join T to Q, R, P and S

By Pythagoras’ Theorem: |TM|2 = |QT|2 – |QM|2 = |TR|2 – |MR|2

|QT|2 – |QM|2 = |TR|2 – |MR|2

|QT|2 – |TR|2 = |QM|2 – |MR|2

Similarly |TN|2 = |PT|2 – |PN|2 =|TS|2 – |NS|2

So |PT|2 – |TS|2 = |PN|2 – |NS|2

But |QM|2 – |MR|2 = |PN|2 – [NS]2

as |QM| = |PN| and |MR| = |NS| ∴|QT|2 – |TR|2 = |PT|2 – |TS|2

|QT|2 + |TS|2 = |PT|2 + |TR|2


Q. 10. By Pythagoras’ Theorem

|EC|2 = |EF|2 + |FC|2

Now |EF| = |AB|

∴ |EF|2 = |AB|2

And by Pythagoras’ Theorem |FC|2 = |AF|2 + |AC|2

∴ |EC|2 = |AB|2 + |AF|2 + |AC|2

Exercise 19.6

Q. 1. Given: Triangle EDF inside circle as shown.

D

O

FE1 23

4

56

To prove: Δ DEO ≡ Δ EFO


Proof: |∠5| = |∠6| = 90°

… given + straight angle

|OD| = |OF| … radii

EO is a common side

⇒ Δ DEO ≡ Δ EFO (SAS)

Q.E.D.


O3

4

2

1

To prove: |∠1| = |∠2|

Construction: Label angle 3. Draw extension from ∠3 as shown and label angle 4

Proof: |∠1| + |∠3| = 180° … opposite angles in cyclic

quadrilateral |∠2| = |∠4| … alternate angle |∠3| + |∠4| = 180° … straight

angle ⇒ |∠3| + |∠2| = 180° ⇒ |∠1| + |∠3| = |∠3| + |∠2| ⇒ |∠1| = |∠2|

Q.E.D.

Q. 3. Given: Diameter [BD]. Centre O.

B

A

O

D

13

2

To prove: |∠1| + |∠2| = 90°

Construction: Mark equal radii. Label angle 3

Proof: |∠1| + |∠3| = 90°

… angle in a semi circle


⇒ |∠1| + |∠2| = 90°

Q.E.D.

Q. 4. Given: Chords [PQ] and [RS] of a circle as shown.

P R

QS

X1

2

3 45

6

To prove: Δ PXS and Δ RXQ are similar


Construction: Mark angles 1, 2, 3, 4, 5, 6

Proof: |∠1| = |∠6| … angles standing on the same arc

|∠2| = |∠5|

… angles standing on the same arc

|∠3| = |∠4|

… vertically opposite

⇒ Δ PXS and Δ RXQ are similar

Q.E.D.

Q. 5. Given: Cyclic quadrilateral ABCD as shown.

A D E

B

C

1

2 3

To prove: |∠CDE| = |∠CBA|


Proof: |∠1| + |∠2| = 180°

… opposite

|∠2| + |∠3| = 180°

… straight angle

⇒ |∠1| + |∠2| = |∠2| + |∠3|

⇒ |∠1| = |∠3|

Q.E.D.

Q. 6. (i) Given: Diameter [AB] and [PQ] of circle C as shown.

A Q

P B

1 2

4 3

To prove: |∠APQ| ≡ |∠PQB|

Construction: Construct the quadrilateral AQBP and label angles 1, 2, 3, 4

Proof: |∠1| = 90°


|∠2| = 90°


also |∠3| = |∠4| = 90°


⇒ AQBP is a rectangle

⇒ |∠APQ| = |∠PQB|

… alternate angles

Q.E.D.

(ii) AP || BQ

This holds and AQBP is a rectangle

Q.E.D.

Q. 7. Given: Diagram as shown

O

CB

A

To Prove: |∠ABO| = |∠CBO|

Construction: Draw [OC] and [OA]

Proof: |OC| = |OA| … radii |CB| = |AB| … given [OB] is a common side ⇒ Δ OCB ≡ Δ OAB (SSS) ⇒ |∠ABO| = |∠CBO| … corresponding angle


Q. 8. (i) Given: Diagram as shown

A B1

23

4C

O

D

To Prove: |∠ABO| = 2q


Proof: |∠2| = q … isosceles Δ

|∠1| = |∠2| + q

… exterior angle

⇒ |∠1| = 2q

Q.E.D.

(ii) To prove: |∠AOD| = 4q


Proof: |∠3| = |∠1| + |∠4|

… exterior angle

but |∠1| = |∠4| = 2q

… isosceles Δ (radii)

⇒ |∠3| = 4q

Q.E.D.

Q. 9. Given: O is the centre of a circle. M is the midpoint of [AB], as shown.

A BM

O

P

1 2

3

To prove: |∠AOM| = |∠APB|


Proof: |AM| = |MB| … given

|AO| = |OB| … radii

OM is a common side

⇒ Δ AOM ≡ Δ BOM (SSS)

⇒ |∠1| = |∠2|

… corresponding angles

|∠1| + |∠2| = 2|∠3|

… angle at the centre

⇒ 2|∠1| = 2|∠3|

⇒ |∠1| = |∠3|

Q.E.D.

Q. 10. (i) Given: Diameter [AB] of circle with centre C. |AB| = |BQ| as shown.

AC B

P

Q

12

To prove: |∠BPQ| = 90°

Construction: Join B to P. Label angles 1, 2.

Proof: |∠1| = 90°

… angles in a semi circle

|∠2| = 180° – 90° = 90°

… straight angle

Q.E.D.

(ii) To prove: |AP| = |PQ| Proof: |∠1| = |∠2| … right angles |AB| = |BQ| … given PB is a common side ⇒ Δ APB ≡ Δ QPB

⇒ |AP| = |PQ| … corresponding sides ⇒ P is the midpoint of

[AQ]


Q. 11. Given: Quadrilateral ABCD as shown.

AB

CD

1 2

34

To prove: ABCD is a cyclic quadrilateral


Proof: |∠1| + |∠2| + |∠3| + |∠4|

= 360° … quadrilateral

|∠2| = |∠4| = 90° … given

⇒ |∠1| + 90°+ |∠3| + 90°

= 360°

⇒ |∠1| + |∠3| = 180°

So opposite sides sum to 180°

⇒ ABCD is a cyclic quadrilateral

Q.E.D.

Q. 12. (i) Given: Diagram as shown.

CA B

P

Q

1

2

435

To prove: PC || QB


Proof: |∠1| = |∠2| = 90°

… angles in a semicircle

⇒ PC || QB

Q.E.D.

(ii) To prove: |AP| = |PQ|

Proof: Label angles 3, 4, 5

|∠1| = |∠2| … above

|∠3| = |∠4|

… corresponding

|∠5| is common to Δ APC and Δ AQB

⇒ Δ APC and Δ AQB are similar

|AP|

_____ |PQ|

= |AC|

_____ |CB|

but |AC| = |CB| … radii

⇒ |AP|

_____ |PQ|

= 1

⇒ |AP| = |PQ|

Q.E.D.

chapter 19 exercise 19 - biology leaving cert.shevlinbiology.webs.com/chapter 19 solutions.pdf ·...

Documents